10 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 10
An astronaut drifts 3 m from a space station, far from Earth and all other bodies. Which statement about the gravitational interaction between the astronaut and the station is correct?
Answer and reasoning
ANeither pulls the other, since gravity is zero in empty space. A student who thinks there is no gravity in space picks this. Any two masses attract, wherever they are; empty space between them does not stop the interaction.
BThey attract each other with forces of equal magnitude.Correct The gravitational force is attractive and mutual: the station pulls the astronaut toward it and she pulls the station toward her. The magnitudes are both Gmams/r², although the force is far too small to notice.
CThe station pulls her, but she exerts no pull on the station. A student who thinks only the larger body attracts picks this. Gravity is an interaction between two masses, so if the station pulls the astronaut, she pulls the station with a force of equal magnitude.
DEach pulls the other, but the station’s pull on her is the larger. A student who thinks the more massive object exerts the larger force picks this. The two forces have the same magnitude, Gmams/r²; the station’s much larger mass gives it a much smaller acceleration, not a smaller force.
Point P is in the air 2 m above the ground near Earth’s surface, and no object is at P. Which statement about the gravitational field at P is correct?
Answer and reasoning
AThere is no field at P until an object is placed there to feel it. A student who thinks a field needs an object to exist picks this. The field is created by Earth at every point around it; a test object only reveals it.
BIt is 10 N/kg for a 1 kg object at P and stronger for a heavier one. A student who thinks the field depends on the test mass picks this. A heavier object has a proportionally larger force, so the force per unit mass, the field, is the same, about 10 N/kg.
CIt is about 10 N/kg toward Earth’s center, object or no object.Correct A field models the noncontact force at every point in space. Earth sets up its gravitational field at P whatever is there; an object placed at P would simply experience a force of about 10 N per kilogram of its mass.
DIt exists at P only because the air surrounding P transmits Earth’s pull. A student who thinks gravity needs air picks this. Gravity is a noncontact force that acts through empty space; the field at P would be the same with no air.
Working The field at P is set up by Earth whether or not anything is there: |g⃗| = GM/r² ≈ 10 N/kg near the surface, directed toward Earth’s center. An object of mass m placed at P would have a force of about (10 N/kg)m exerted on it; the ratio F/m does not depend on m, and no medium is needed.
A student models the flight of a ball thrown 30 m straight up from the ground by treating the gravitational force on the ball as constant for the whole flight. Which reasoning best justifies this model?
Answer and reasoning
AThe field g is a universal constant, with one value at every distance from Earth. A student who thinks g is the same everywhere picks this. The field is GM/r² and does fall off with distance; it is nearly constant here only because 30 m is tiny compared with Earth’s radius.
BIts distance from Earth’s center changes by a tiny fraction, so the force barely changes.Correct The force depends on the distance between the centers of mass, about 6.4 × 10⁶ m. Rising 30 m changes that distance by about 0.0005%, so the change in the force is negligible and it can be treated as constant.
CThe ball’s weight is a fixed property of the ball, just like its mass, so it stays constant. A student who thinks weight cannot change picks this. Weight depends on the local field, which changes with distance from Earth’s center; the model works because that distance hardly changes.
DGravity weakens only above the atmosphere, and the ball stays within the air. A student who links gravity to the atmosphere picks this. Gravity weakens steadily with distance from Earth’s center, inside or outside the atmosphere; air plays no part in it.
Near the surface of Earth, the gravitational field strength is g ≈ 10 N/kg. Which statement correctly interprets this value?
Answer and reasoning
AThe field there is about 10 N/kg for a 1 kg object, and is larger for heavier objects. A student who thinks the field depends on the object placed in it picks this. A heavier object has more force on it, but the force per kilogram, the field, is the same.
BEvery object near Earth’s surface has a gravitational force of about 10 N on it. A student who confuses the field with the force picks this. 10 N/kg is a force per kilogram; a 5 kg object has about 50 N on it, a 0.1 kg object about 1 N.
CThe field has this value of about 10 N/kg at any distance from Earth’s center, near or far. A student who thinks g is a universal constant picks this. The value applies near the surface; farther from Earth’s center the field is GM/r², which is smaller.
DEarth exerts about 10 N of gravitational force on each kilogram of an object there.Correct A field strength in N/kg is a force per unit mass: near the surface, an object of mass m has about (10 N/kg)m of gravitational force exerted on it, and if gravity is the only force it accelerates at about 10 m/s².
Working g = Fg/m, so 10 N/kg means 10 N of gravitational force per kilogram: a 5 kg object has about 50 N exerted on it. The ratio does not depend on the object’s mass, and the value holds only near the surface, since the field is GM/r².
A person of mass m stands on a scale in an elevator that is moving downward and slowing down. The magnitude of the elevator’s acceleration is g/4, where g is the gravitational field strength. What is the scale reading?
Answer and reasoning
A0.75mg A student who links the reading to the direction of motion picks this: moving down, so lighter, m(g − g/4). The reading depends on the acceleration, which is upward here because the downward motion is slowing.
B1.00mg A student who thinks the scale always reads the gravitational force picks this. The person is accelerating, so the normal force, and with it the scale reading, differs from mg.
C1.25mgCorrect Slowing while moving down means the acceleration is upward. Newton’s second law with up positive: FN − mg = m(g/4), so the scale, which reads FN, shows 1.25mg.
D0.25mg A student who thinks the scale reads the net force picks this: m(g/4). The net force is FN − mg; the scale reads FN itself, which must exceed mg to give an upward acceleration.
Working Moving down and slowing ⇒ acceleration upward, ay = +g/4. FN − mg = m(g/4) ⇒ FN = 1.25mg. Distractors: ‘moving down, so lighter’ → m(g − g/4) = 0.75mg; FN = mg; net force m(g/4) = 0.25mg.
An astronaut in a space station orbiting 400 km above Earth’s surface floats freely and appears weightless. Which reasoning correctly explains why she appears weightless?
Answer and reasoning
AAt that altitude she is beyond Earth’s gravity, so no force at all acts on her. A student who thinks there is no gravity in orbit picks this. At 400 km the distance from Earth’s center is only about 6% larger than at the surface, so the gravitational force on her is still large.
BEarth’s gravity on her is balanced by an outward force, so the net force on her is zero. A student who thinks floating means balanced forces picks this. No object exerts an outward force on her; gravity is the only force, so the net force on her is not zero.
COnly gravity acts on her: she falls along with the station, so no floor pushes on her.Correct A system appears weightless when gravity is the only force exerted on it. She and the station both move under gravity alone, so the floor does not push on her and a scale under her would read zero, even though Earth’s field there is still about 90% of its surface value.
DIn orbit she has no mass, so Earth’s gravity has nothing to pull on. A student who thinks weightless objects lose their mass picks this. Her mass is the same as on Earth; what is missing is the normal force from a floor.
In an orbiting space station, an astronaut gives a floating 10 kg crate and a floating 40 kg crate identical pushes: the same force for the same short time. Which statement describes the crates’ motion after the pushes?
Answer and reasoning
ABoth move off with the same speed, since neither crate has any mass in orbit. A student who thinks weightless objects have no mass picks this. The crates keep their inertial mass in orbit, so the 40 kg crate resists the change in motion four times as much.
BBoth move off, but each slows and stops soon after its push ends. A student who thinks motion needs a continuing force picks this. Once the push ends, no force acts along the crates’ motion, so each keeps a constant velocity.
CThe 10 kg crate ends up moving at four times the speed of the heavier crate.Correct Inertial mass does not depend on gravity, so each crate’s acceleration during the push is F/m, and its final speed is (F/m)Δt. With one-quarter of the mass, the 10 kg crate ends with four times the speed; each then keeps moving at constant speed.
DOnly the 10 kg crate moves; the push is too weak to move the 40 kg one. A student who thinks a push must exceed a threshold before an object moves picks this. Any net force on a floating crate accelerates it; the 40 kg crate simply ends up slower.
Working Same F for the same Δt: a = F/m, so v = (F/m)Δt; inertial mass is unchanged in orbit, so v₁₀/v₄₀ = 40/10 = 4. After the push no force acts along the motion, so each keeps its speed.
Ball 1 has mass m and ball 2 has mass 3m. They are released from rest at the same height near Earth’s surface, and air resistance is negligible. F₁ and F₂ are the magnitudes of the gravitational forces exerted on the balls by Earth, and a₁ and a₂ are the magnitudes of their accelerations. Which comparison is correct?
Answer and reasoning
AF₂ = 3F₁ and a₂ = a₁Correct The gravitational force is proportional to each ball’s gravitational mass, so F₂ = 3F₁. Each acceleration is that force divided by the ball’s inertial mass, which equals its gravitational mass, so both balls accelerate equally: a₂ = a₁.
BF₂ = 3F₁ and a₂ = 3a₁ A student who thinks heavier objects fall faster picks this. Ball 2 has three times the force but also three times the inertia, so its acceleration is the same as ball 1’s.
CF₂ = F₁ and so a₂ = a₁ A student who thinks Earth pulls every object with the same force picks this. The accelerations are equal, but the forces are not: the gravitational force is proportional to the ball’s mass.
DF₂ = 9F₁ while a₂ = 3a₁ A student who thinks the heavier ball makes the field at its location three times as strong picks this, so its force is 3m × 3g. The field near Earth does not depend on the object placed in it; F₂ = 3F₁ and a₂ = a₁.
Working Gravitational mass sets the force: F = GMEm/r² ∝ m, so F₂ = 3F₁. Inertial mass sets the response: a = F/m; with inertial and gravitational mass equal, a₂ = 3F₁/(3m) = a₁.
The figure shows a uniform sphere with center O, far from all other objects, and a small object P outside it. A and B are two small elements of the sphere with equal masses, placed symmetrically above and below line OP. Which statement describes the net gravitational force exerted on P by elements A and B together?
Answer and reasoning
AToward O, with magnitude equal to the sum of the two forces’ magnitudes A student who adds force magnitudes regardless of direction picks this. The two forces point at an angle to each other, so only their components along OP add; the net force is smaller than the sum.
BToward the bottom of the page, equal to the sum of their magnitudes A student who thinks gravity always points “down” picks this. Each element pulls P toward itself, so the forces point toward A and toward B, and their sum points toward O.
CZero, because elements A and B pull P in different directions A student who thinks equal pulls from symmetric masses cancel picks this. Only the components perpendicular to OP cancel; both forces have components toward O, and these add.
DToward O, with magnitude less than the sum of the two forces’ magnitudesCorrect Each element pulls P toward itself, along the dashed lines, with equal magnitudes. Their components perpendicular to OP cancel and their components along OP add, so the net force points from P toward O and is smaller than the sum of the two magnitudes. Adding all such pairs gives the sphere’s net force toward O.
Working Each element pulls P toward itself along the dashed lines, with equal magnitudes (equal masses, equal distances). The components perpendicular to OP cancel; the components along OP add. The net force points from P toward O and is less than the sum of the magnitudes. Summing all such pairs over the sphere gives a net force toward O.
The figure shows a cross section of a thin, uniform spherical shell with center O, far from all other objects, and a small object at point P inside the shell. What is the net gravitational force exerted on the object by the shell?
Answer and reasoning
AToward the right side of the shell, since that part is nearest to P A student who thinks the nearest part of the shell pulls hardest picks this. More of the shell’s mass is on the far side, and its larger mass exactly offsets its larger distance, so the net force is zero.
BToward the left side of the shell, since more of its mass is on that side A student who counts only how much mass is on each side picks this. That larger mass is also farther away; at every point inside a uniform shell the opposite pulls cancel exactly.
CZero, because the pulls of all the parts of the shell cancel outCorrect Newton’s shell theorem: the net gravitational force exerted by a thin uniform shell on an object anywhere inside it is zero. The nearer part of the shell is closer but has less mass in each direction, and the pulls cancel exactly.
DToward O, as if all of the shell’s mass were concentrated at point O A student who treats the shell as a point mass at its center everywhere picks this. That model holds only outside the shell; inside, the net force from the shell is zero.
In preparation: 0 of 15 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.6.A.1 Newton’s law of universal gravitation Fix
Newton’s law of universal gravitation
The gravitational force between two objects or systems has magnitude |F⃗g| = Gm₁m₂/r², where m₁ and m₂ are their masses and r is the distance between their centers of mass. The force is proportional to each mass and inversely proportional to r².
Universal gravitational constant, G
The constant in Newton’s law of gravitation, G = 6.67 × 10⁻¹¹ N·m²/kg². It is the same everywhere and should not be confused with g, the local gravitational field strength.
Attractive force
The gravitational force is always attractive: each of two interacting objects is pulled toward the other, and the two forces have equal magnitudes.
Line of action of the gravitational force
The gravitational force between two systems is exerted along the line connecting their centers of mass, directed toward the other system.
Center of mass as the point of action
The gravitational forces on all parts of a system can be considered a single force exerted on the system’s center of mass, which lies closer to the more massive parts.
Students often think r is the distance between the surfaces of the objects, or the height of an object above a planet’s surface. In fact r is the distance between the centers of mass of the two objects. For an object at height h above a planet of radius R, r = R + h.
Students often think The gravitational force is inversely proportional to the distance r, so doubling r halves the force. In fact It is inversely proportional to the square of that distance: doubling r makes the force one-quarter as large, and tripling r makes it one-ninth as large.
2.6.A.2 Field Fix
Field
A model that assigns a vector to every point in space to describe the noncontact force an object would experience at that point. A field exists at a point whether or not an object is placed there.
Gravitational field, g⃗
The gravitational force per unit mass on a test object, g⃗ = F⃗g/m, in N/kg. For a system of mass M, |g⃗| = GM/r² at distance r from its center of mass, directed toward that center. It does not depend on the test object’s mass.
Free-fall acceleration and field strength
If the gravitational force is the only force exerted on an object, its acceleration in m/s² is numerically equal to the gravitational field strength in N/kg at its location (1 N/kg = 1 m/s²).
Students often think The gravitational field at a point and the gravitational force on an object there are the same quantity, so either can be reported as the other. In fact No. The field g⃗ is the force per unit mass, in N/kg; the force on an object of mass m placed at that point is m g⃗, in N.
Students often think Gravity needs air or an atmosphere; where there is no air, such as on an airless moon or asteroid, there is no gravitational force. In fact No. Gravity is a noncontact force between masses and acts through empty space; the Moon and asteroids, which have no atmosphere, exert gravitational forces.
2.6.A.3 Weight Fix
Weight
The gravitational force exerted by an astronomical body on a relatively small nearby object, Fg = mg, in N. It depends on the local field, so it changes with location; mass does not.
Students often think An object’s weight is a fixed property of the object, the same on any planet and at any height. In fact No. Weight is the gravitational force exerted by an astronomical body, mg, so it changes with the local field: it is different on another planet and smaller far above Earth.
Students often think Mass and weight are the same quantity, so a value in kilograms can be used directly as a weight in newtons. In fact No. Mass, in kg, measures an object’s inertia and does not depend on location. Weight is the gravitational force on the object, in N, and equals mg, so it depends on the local field.
2.6.B.1 Constant-force approximation Fix
Constant-force approximation
Over a displacement that changes the center-to-center distance by a negligible fraction (for example, heights of meters to kilometers near Earth, compared with RE ≈ 6.4 × 10⁶ m), the gravitational force can be treated as constant.
Students often think g ≈ 10 N/kg is a universal constant of nature with the same value at every distance from Earth. In fact No. g ≈ 10 N/kg only near Earth’s surface. The field is GM/r², so it falls off with distance from Earth’s center and is one-quarter as large at r = 2RE.
2.6.B.2 Near-surface field strength, g ≈ 10 N/kg Fix
Near-surface field strength, g ≈ 10 N/kg
Near Earth’s surface the gravitational field strength is about 10 N/kg, so each kilogram of an object has about 10 N of gravitational force exerted on it (the Table of Information prints 9.8 N/kg).
2.6.C.1 Apparent weight Fix
Apparent weight
The magnitude of the normal force exerted on a system, for example the reading of a scale under it. For vertical motion, FN = m(g + ay) with ay positive upward, so it differs from mg whenever the system accelerates vertically.
Students often think A scale always reads the gravitational force mg on the object on it, so apparent weight and the gravitational force are always equal. In fact No. A scale reads the magnitude of the normal force it exerts. That equals mg only when the person’s acceleration is zero; if the person accelerates, FN = m(g + ay) with ay positive upward.
Students often think A scale under an accelerating person reads the net force ma on the person. In fact No. The scale reads the normal force it exerts. The net force is the normal force minus the gravitational force, may, which is much smaller than the reading.
2.6.C.2 Idea 7 Fix
Students often think In an elevator, a person feels heavier (the scale reads more) while moving upward and lighter while moving downward. In fact No. It depends on the acceleration, not the velocity. Moving down while slowing (acceleration upward) gives a reading above mg; moving up while slowing (acceleration downward) gives a reading below mg.
Students often think Only the elevator’s own mass is accelerated by the cable; the mass of the load inside can be left out of the system. In fact The whole system: the cable’s tension, minus the total gravitational force, accelerates the elevator and everything in it, so a = T/(M + m) − g.
2.6.C.3 Apparent weightlessness Fix
Apparent weightlessness
The condition in which a system’s apparent weight is zero: either no forces are exerted on it, or the gravitational force is the only force exerted on it (free fall).
Students often think Objects that appear weightless have no gravitational force exerted on them; in space, gravity is zero. In fact Yes. At typical station altitudes Earth’s gravitational field is still close to 90% of its surface value. The astronaut appears weightless because gravity is the only force exerted on her, so nothing, such as a floor, pushes on her.
Students often think A floating astronaut has zero net force on her: Earth’s gravity is balanced by an outward force. In fact No. The gravitational force is the only force exerted on her, so the net force on her is not zero and she accelerates with the station. She appears weightless because no floor or scale pushes on her.
2.6.C.4 Noninertial reference frame Fix
Noninertial reference frame
A reference frame that is accelerating, such as an accelerating rocket or elevator; Newton’s first law is not verified by an observer in it.
Equivalence principle
An observer in a noninertial reference frame cannot distinguish between an object’s apparent weight and the gravitational force exerted on it by a gravitational field; for example, a closed cabin accelerating at 10 m/s² in deep space is indistinguishable from one at rest on Earth.
2.6.D.1 Inertial mass Fix
Inertial mass
The property of an object, in kg, that determines how much its motion resists changes when it interacts with another object; it is the m in a⃗ = F⃗net/m and does not depend on gravity.
Students often think Objects that appear weightless have no mass, or no inertia, so they can be started or stopped with no effort. In fact No. Inertial mass does not depend on gravity. A floating 40 kg crate in an orbiting station is just as hard to accelerate as it would be on Earth.
Students often think An object moves only if the push on it exceeds some threshold, such as its weight; a small push on a heavy object produces no motion. In fact No. On a floating object with no other horizontal forces, any net force, however small, produces an acceleration a = Fnet/m.
2.6.D.2 Gravitational mass Fix
Gravitational mass
The property of an object, in kg, that determines the strength of the gravitational attraction between it and another system with mass; it is the m in Fg = GMm/r².
Students often think Earth pulls every object with the same gravitational force, whatever its mass; that is why objects fall together. In fact No. The gravitational force is proportional to the object’s mass, Fg = mg, so a 3 kg object has three times the gravitational force of a 1 kg object at the same place. Their free-fall accelerations are equal because the force per unit mass is the same.
2.6.D.3 Equivalence of inertial and gravitational mass Fix
Equivalence of inertial and gravitational mass
Experiments find inertial and gravitational mass equal to very high precision; as a result, all objects at the same location fall with the same acceleration when only gravity acts.
2.6.E.1 Superposition of mass elements Fix
Superposition of mass elements
The net gravitational force from an extended distribution is the vector sum of the forces from its small mass elements dm; for a uniform sphere, the components perpendicular to the line toward the center cancel in symmetric pairs.
Students often think The net gravitational force from several masses is the sum of the magnitudes of the individual forces, whatever their directions. In fact No. Forces are vectors. The net force is the vector sum, so components perpendicular to the line of symmetry cancel and the net magnitude is less than the sum of the magnitudes.
Students often think At any point equally distant from two equal masses, their gravitational fields cancel and the net field is zero. In fact No. It is zero only at the midpoint between them, where the two fields point in opposite directions. At other equidistant points the fields have the same magnitude but point at an angle, so their components along the symmetry line add.
2.6.E.2 Newton’s shell theorem Fix
Newton’s shell theorem
For a thin uniform spherical shell of mass M: at any point inside, the net gravitational force from the shell is zero; at any point outside, the shell acts as a single object of mass M at its center.
Partial mass, mpartial
For an object at distance r from the center of a uniform sphere of density ρ, the mass within radius r, mpartial = ρ(4/3)πr³ = M(r/R)³, in kg; only this mass contributes to the net gravitational force.
Density, ρ
Mass per unit volume, in kg/m³; for a uniform sphere, ρ = M/((4/3)πR³).
Students often think An object inside a hollow shell is pulled toward the nearer wall, because the nearer part of the shell pulls harder. In fact No. The nearer part of the shell is closer but has less mass in any given cone of directions, and the farther part has more mass but is farther away; the pulls cancel exactly at every point inside, so the net force is zero.
Students often think An off-center object inside a hollow shell is pulled toward the farther wall, because more of the shell’s mass lies on that side. In fact No. The far side does contain more mass, but it is farther away; at every point inside a thin uniform shell, the pulls of the near and far sides cancel exactly.
2.6.E.3 Linear force inside a uniform sphere Fix
Linear force inside a uniform sphere
Inside a uniform sphere the gravitational force on an object is Fg,partial = −kr, directed toward the center, with k = GMm/R³ = (4/3)πGρm (in N/m); its magnitude grows linearly from zero at the center to GMm/R² at the surface.
Students often think Gravity gets stronger the deeper an object goes into a planet and is strongest at the center. In fact No. Inside a uniform planet the field is proportional to the distance from the center: it is largest at the surface and zero at the center.
Students often think Inside a planet the gravitational force is proportional to the partial mass inside the object’s radius, so it grows as r³. In fact No. The force is Gmpartial m/r², so it depends on the partial mass and on r. Since mpartial ∝ r³, the force is proportional to r³/r² = r.
32 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 32
The figure shows two uniform metal spheres, with their masses, their radii and the gap between their surfaces. What is the magnitude of the gravitational force that each sphere exerts on the other?
Answer and reasoning
A1.7 × 10⁻⁶ N A student who uses the gap between the surfaces as r picks this: Gm₁m₂/(0.10 m)². The law uses the distance between the spheres’ centers, which adds both radii to the gap.
B2.7 × 10⁻⁷ NCorrect Spheres act as point masses at their centers, so r is the distance between centers: 0.10 m + 0.10 m + 0.050 m = 0.25 m. Then |Fg| = Gm₁m₂/r² = (6.67 × 10⁻¹¹)(50)(5.0)/(0.25)² N = 2.7 × 10⁻⁷ N.
C6.7 × 10⁻⁸ N A student who divides by r instead of r² picks this: Gm₁m₂/(0.25 m). The force falls off as the inverse square of the center-to-center distance.
D5.3 × 10⁻⁸ N A student who calculates the 50 kg sphere’s field at the small sphere, Gm₁/r², and reports it as the force picks this. The force on the 5.0 kg sphere is its mass times that field, Gm₁m₂/r².
Working The spheres can be treated as point masses at their centers, so r is the center-to-center distance: r = 0.10 m + 0.10 m + 0.050 m = 0.25 m. |Fg| = Gm₁m₂/r² = (6.67 × 10⁻¹¹ N·m²/kg²)(50 kg)(5.0 kg)/(0.25 m)² = 2.7 × 10⁻⁷ N. Distractors: gap used as r → 1.7 × 10⁻⁶ N; r not squared → 6.7 × 10⁻⁸ N; field of the 50 kg sphere at the small sphere, Gm₁/r², reported as the force → 5.3 × 10⁻⁸ N.
Two asteroids, far from all other objects, are released from rest a short distance apart. Asteroid 1 has mass m and asteroid 2 has mass 4m. F₁ is the magnitude of the gravitational force exerted on asteroid 1 by asteroid 2, and F₂ is the magnitude of the force exerted on asteroid 2 by asteroid 1; a₁ and a₂ are the magnitudes of the asteroids’ accelerations. Which comparison is correct?
Answer and reasoning
AF₁ = F₂ and a₁ = 4a₂Correct Both forces equal G(m)(4m)/r², because the law is symmetric in the two masses. Each acceleration is that force divided by the asteroid’s own mass, so asteroid 1, with one-quarter of the mass, has four times the acceleration.
BF₁ = 4F₂ and a₁ = 16a₂ A student who thinks the more massive asteroid exerts the larger force picks this, making the force on asteroid 1 four times as large. The two forces have the same magnitude, G(m)(4m)/r², so a₁ = 4a₂.
CF₁ = F₂ and a₁ = a₂ A student who thinks equal forces give equal accelerations picks this. The forces are equal, but each is divided by a different mass: a₁ = F/m is four times a₂ = F/(4m).
DF₁ > 0 while F₂ = a₂ = 0 A student who thinks only the larger body attracts picks this. Gravity is a mutual, attractive interaction: asteroid 1 pulls asteroid 2 with a force equal in magnitude to F₁, so asteroid 2 also accelerates.
Working F₁ = F₂ = G(m)(4m)/r²: the expression is symmetric in the two masses (the forces are a third-law pair). a₁ = F₁/m and a₂ = F₂/(4m), so a₁ = 4a₂.
Two uniform spheres, A and B, each of radius R, are placed so that the gap between their surfaces is 2R. The gravitational force that B exerts on A has magnitude F. Sphere A is replaced by a sphere of the same radius and three times the mass, and the spheres are moved apart until the gap between their surfaces is 4R. What is now the magnitude of the gravitational force exerted on sphere A by sphere B?
Answer and reasoning
A0.75F A student who uses the gaps between the surfaces as r picks this: 3 × (2R/4R)² = 0.75. The distance in the law is between the centers, so each gap must have both radii added to it.
B2.00F A student who treats the force as inversely proportional to r picks this: 3 × (4R/6R) = 2. The force depends on 1/r², so the distance factor is (4/6)², not 4/6.
C1.33FCorrect The law uses center-to-center distances: 4R before and 6R after. Tripling A’s mass triples the force, and the distance factor is (4R/6R)² = 4/9, so the new force is 3 × 4/9 = 4/3 of F, about 1.33F.
D0.44F A student who thinks the force on a sphere does not depend on its own mass picks this, keeping only the distance factor (4/6)². The force is proportional to each mass, so tripling A’s mass triples it.
Working Center-to-center distance: before, R + 2R + R = 4R; after, R + 4R + R = 6R. F ∝ mA mB/r²: Fnew/F = 3 × (4R/6R)² = 3 × 4/9 = 4/3 ≈ 1.33. Distractors: gaps used as r → 3 × (2/4)² = 0.75; r not squared → 3 × 4/6 = 2.00; A’s own mass ignored → (4/6)² = 0.44.
The figure shows two uniform spheres, far from all other objects, with their masses and the distance D between their centers. A small object is placed on the line between the centers. At what distance from the center of the sphere of mass M is the net gravitational force exerted on the small object by the two spheres zero?
Answer and reasoning
A0.20D A student who makes each force inversely proportional to distance picks this: M/x = 4M/(D − x) gives x = D/5. With the inverse-square law, the ratio of distances is the square root of the mass ratio, 2, not 4.
B0.80D A student who identifies the zero-force point with the center of mass picks this: xcm = 4M·D/(5M) = 0.80D from M. That is where the masses balance, not where their pulls on a third object cancel, which is much closer to M.
C0.50D A student who thinks the spheres pull equally on an object midway between them picks this. At the midpoint the distances are equal, so the 4M sphere pulls four times as hard; the pulls balance only nearer to M.
D0.33DCorrect The two forces on the small object point in opposite directions, so they cancel where GMm/x² = G(4M)m/(D − x)². Taking square roots, D − x = 2x, so x = D/3 ≈ 0.33D: the point is closer to the less massive sphere.
Working Let x be the distance from M. Forces on the small object m are opposite in direction, so zero net force needs GMm/x² = G(4M)m/(D − x)² ⇒ (D − x)² = 4x² ⇒ D − x = 2x (taking the root between the spheres) ⇒ x = D/3 ≈ 0.33D. Distractors: 1/r law → D − x = 4x → 0.20D; center of mass, 4M·D/(5M) = 0.80D from M; midpoint 0.50D.
A spacecraft coasts with its engines off past an asteroid, far from all other bodies. In which direction is the gravitational force exerted on the spacecraft by the asteroid?
Answer and reasoning
AToward the asteroid’s center, along the line joining themCorrect The gravitational force is attractive and is exerted along the line connecting the centers of mass of the two objects, so it points from the spacecraft toward the asteroid’s center.
BStraight down, in the one fixed direction of “down” everywhere A student who thinks gravity always points “down” picks this. Far from Earth there is no fixed down; the force points toward the center of the object exerting it, here the asteroid.
CAlong the spacecraft’s velocity, in its direction of motion A student who thinks a moving object has a force along its motion picks this. With the engines off, the only object interacting with the spacecraft is the asteroid, which pulls it toward the asteroid’s center.
DThere is none, since the asteroid has no atmosphere at all A student who thinks gravity needs air picks this. Every object with mass exerts a gravitational force; an airless asteroid attracts the spacecraft just as a planet would, only more weakly.
The figure shows a dumbbell made of two small uniform spheres joined by a light rigid rod; tick marks divide the line joining the spheres’ centers into four equal parts. The masses of the spheres are shown. Point W is the center of the larger sphere, Z is the lowest point of the smaller sphere, and X and Y are on the rod. At which point can the gravitational force exerted by Earth on the dumbbell be considered to be exerted?
Answer and reasoning
APoint Y A student who places the gravitational force at the geometric center picks this. The point midway between the centers would be the center of mass only if the two spheres had equal masses; here the center of mass is closer to the 3m sphere.
BPoint W A student who attaches the whole gravitational force to the heavier part picks this. Earth also pulls on the smaller sphere, so the single equivalent force acts at the center of mass, between the spheres.
CPoint Z A student who thinks gravity acts at an object’s lowest point picks this. Gravity acts on all of the dumbbell’s mass, and the total can be considered exerted at its center of mass, point X.
DPoint XCorrect The gravitational forces on all parts of a system can be considered a single force exerted at its center of mass. Measuring from the center of the 3m sphere in units of one part, xcm = (3m·0 + m·4)/(4m) = 1 part: point X.
Working Center of mass from the center of the 3m sphere along the rod: xcm = (3m·0 + m·4s)/(4m) = s, one of the four equal parts, which is point X. Y is the midpoint (geometric center); W is the heavier sphere; Z is the lowest point.
A planet of radius R has gravitational field strength gs at its surface. What is the magnitude of the planet’s gravitational field at a height h above its surface?
Answer and reasoning
AgsR²/(R+h)²Correct From the surface value, GM = gsR². The field depends on the distance from the planet’s center, r = R + h, so g = GM/r² = gsR²/(R+h)².
BgsR/(R+h) A student who makes the field inversely proportional to r picks this, scaling gs by R/(R+h). The field is GM/r², so the scale factor is (R/(R+h))².
CgsR²/h² A student who measures the distance from the surface picks this, using r = h. The distance in GM/r² is from the planet’s center, so r = R + h; with r = h the field would be infinite at the surface.
Dgs A student who treats the field as the same at every height picks this. A constant field is a good model only when h is negligible compared with R; in general the field falls off as 1/(R+h)².
Working At the surface, gs = GM/R², so GM = gsR². At height h the distance from the center is r = R + h: g = GM/r² = gsR²/(R+h)². Distractors: 1/r dependence → gsR/(R+h); distance measured from the surface, r = h → gsR²/h²; field taken as constant → gs.
A 70 kg astronaut appears weightless inside a space station 4.0 × 10⁵ m above Earth’s surface. Earth’s mass is 6.0 × 10²⁴ kg and its radius is 6.4 × 10⁶ m. What is the magnitude of the gravitational force exerted on the astronaut by Earth?
Answer and reasoning
A7.0 × 10² N A student who uses g = 10 N/kg at every distance from Earth picks this: (70 kg)(10 N/kg). At 4.0 × 10⁵ m the distance from Earth’s center is 6% larger, so the field is noticeably smaller, about 8.7 N/kg.
B1.8 × 10⁵ N A student who uses the altitude as r picks this: GMm/(4.0 × 10⁵ m)². The distance in the law is measured from Earth’s center, so the radius must be added to the altitude.
C4.1 × 10⁹ N A student who divides by r instead of r² picks this: GMm/(6.8 × 10⁶ m). The force depends on 1/r², and only the r² version gives a force in newtons.
D6.1 × 10² NCorrect r is measured from Earth’s center: r = 6.4 × 10⁶ m + 4.0 × 10⁵ m = 6.8 × 10⁶ m. |Fg| = GMm/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(70)/(6.8 × 10⁶)² N = 6.1 × 10² N. The force is far from zero; she appears weightless because it is the only force on her.
Working r = RE + h = 6.4 × 10⁶ m + 0.40 × 10⁶ m = 6.8 × 10⁶ m. |Fg| = GMm/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(70)/(6.8 × 10⁶)² N = 6.1 × 10² N, about 87% of her weight at the surface. She appears weightless because this is the only force exerted on her, not because it is zero. Distractors: g = 10 N/kg everywhere → 7.0 × 10² N; r = h → 1.8 × 10⁵ N; r not squared → 4.1 × 10⁹ N.
The figure shows two small spheres, with their masses, at two corners of an equilateral triangle, with its side length, far from all other objects. G is the universal gravitational constant. What is the magnitude of the net gravitational field at point P, the third corner?
Answer and reasoning
A2.00 GM/s² A student who adds the two field magnitudes picks this. The fields point in different directions, so only their components along the line of symmetry add; their sideways components cancel.
B1.73 GM/s²Correct Each field has magnitude GM/s² and points from P toward its sphere, at 30° to the vertical line of symmetry. The horizontal components cancel and the vertical ones add: 2(GM/s²)cos 30° = √3 GM/s² ≈ 1.73 GM/s², directed toward the midpoint of the base.
C1.00 GM/s² A student who takes the components with sin 30° picks this. The angle is measured from the line of symmetry, so the component along that line is GM/s² × cos 30°.
D0.00 GM/s² A student who thinks the fields of equal masses cancel at any point equally distant from them picks this. They cancel only at the midpoint between the masses; at P both fields have a component toward the base, and these add.
Working Each sphere’s field at P has magnitude GM/s², directed from P toward that sphere. Each field makes 30° with the line of symmetry (the altitude from P). Components perpendicular to it cancel; components along it add: g = 2(GM/s²)cos 30° = √3 GM/s² ≈ 1.73 GM/s². Distractors: magnitudes added → 2.00; sin 30° used → 1.00; ‘equidistant from equal masses’ taken to cancel → 0.
The figure shows three uniform spherical planets with their masses and radii. gX, gY and gZ are the magnitudes of the gravitational fields at the surfaces of planets X, Y and Z. Which comparison is correct?
Answer and reasoning
AgY > gX > gZ A student who thinks surface gravity depends only on the planet’s mass picks this, ranking by 4M, M, M/2. The radius matters too: Y’s doubled radius cuts its field by a factor of 4.
BgY > gX = gZ A student who uses M/R instead of M/R² picks this, giving 2 : 1 : 1 for Y, X and Z. With the inverse square, Y’s field equals X’s and Z’s is twice X’s.
CgZ > gX = gYCorrect The surface field is GM/R². For Y, four times the mass at twice the radius gives 4/2² = 1 times gX. For Z, half the mass at half the radius gives (1/2)/(1/2)² = 2 times gX. So gZ is largest, and gX = gY.
DgX = gY = gZ A student who thinks every planet has the same surface field picks this. The field GM/R² depends on each planet’s mass and radius, and Z’s is twice the others’.
An astronaut on the Moon throws a 0.20 kg ball straight up. The Moon’s mass is 7.3 × 10²² kg and its radius is 1.7 × 10⁶ m. Only the gravitational force is exerted on the ball after it leaves her hand. What is the ball’s acceleration at the highest point of its flight?
Answer and reasoning
AZero, since the ball is momentarily at rest there A student who thinks zero velocity means zero acceleration picks this. The gravitational force is still exerted at the top, so the velocity is still changing: an instant later the ball is moving down.
B10 m/s², directed toward the Moon’s center A student who uses Earth’s value of g everywhere picks this. The Moon’s field, GM/R² with the Moon’s own mass and radius, is about 1.7 N/kg.
C0.34 m/s², directed toward the Moon’s center A student who confuses force and field picks this: (0.20 kg)(1.7 N/kg) = 0.34 N is the gravitational force on the ball, not its acceleration. Dividing the force by the mass gives back 1.7 m/s².
D1.7 m/s², directed toward the Moon’s centerCorrect The Moon’s field at its surface is GM/R² = (6.67 × 10⁻¹¹)(7.3 × 10²²)/(1.7 × 10⁶)² N/kg = 1.7 N/kg. With gravity the only force, the ball’s acceleration is numerically equal to the field: 1.7 m/s² toward the Moon’s center, at every point of the flight. At the top only the velocity is zero.
Working Field at the surface: g = GM/R² = (6.67 × 10⁻¹¹)(7.3 × 10²²)/(1.7 × 10⁶)² N/kg = 1.7 N/kg. With gravity the only force, a = Fg/m = g = 1.7 m/s² toward the Moon’s center, at every point of the flight, including the top where only the velocity is zero. Distractors: a = 0 at the top; Earth’s 10 m/s²; the force mg = (0.20 kg)(1.7 N/kg) = 0.34 N reported as the acceleration.
An astronaut’s weight at Earth’s surface is W. Planet Q has three times Earth’s mass and twice Earth’s radius. What is the astronaut’s weight at the surface of planet Q?
Answer and reasoning
A1.50W A student who makes the surface field proportional to M/R picks this: 3/2 = 1.5. The field is GM/R², so the doubled radius divides it by 4, not 2.
B1.00W A student who thinks weight is a fixed property of the astronaut picks this. Her mass is the same on Q, but her weight is the gravitational force there, which depends on Q’s field.
C0.75WCorrect Weight is mg, and the surface field is GM/R². Three times the mass at twice the radius gives a field 3/2² = 0.75 times Earth’s; the astronaut’s mass is unchanged, so her weight is 0.75W.
D3.00W A student who thinks surface gravity depends only on the planet’s mass picks this. Q’s larger radius puts its surface twice as far from its center, which reduces the field by a factor of 4.
Working Weight = mg with g = GM/R². gQ/gE = 3/2² = 0.75, and the mass m is unchanged, so WQ = 0.75W. Distractors: 1/R → 3/2 = 1.50; weight unchanged → 1.00; mass only → 3.00.
A satellite is at an altitude RE above Earth’s surface, where RE is Earth’s radius, and the gravitational force exerted on it by Earth has magnitude F. The satellite is moved to an altitude of 3RE. What is the magnitude of the gravitational force exerted on it by Earth at the new altitude?
Answer and reasoning
A0.25FCorrect The distance from Earth’s center goes from RE + RE = 2RE to RE + 3RE = 4RE, so it doubles and the force falls to (1/2)² = 0.25 of its value. Over such a large change in r the force is far from constant.
B0.11F A student who uses the altitudes as r picks this: (RE/3RE)² = 1/9. The law uses the distance from Earth’s center, which is RE larger than each altitude.
C0.50F A student who treats the force as inversely proportional to r picks this: 2RE/4RE = 1/2. Doubling r reduces the force by a factor of 2² = 4.
D1.00F A student who thinks the gravitational field is the same at every distance picks this. A constant-force model works only when r changes by a negligible fraction; here r doubles.
Working Distances from Earth’s center: 2RE before, 4RE after. F ∝ 1/r²: Fnew/F = (2/4)² = 0.25. Distractors: altitudes as r → (1/3)² = 0.11; 1/r → 2/4 = 0.50; constant force → 1.00.
A 60 kg student stands on a scale in an elevator. The graph shows the elevator’s vertical velocity v as a function of time t, with upward positive. What does the scale read at t = 7 s? Use g = 10 m/s².
Answer and reasoning
A480 NCorrect The scale reads the normal force. At t = 7 s the slope of the graph is (0 − 4 m/s)/(2 s) = −2 m/s², so the acceleration is 2 m/s² downward. FN − mg = may gives FN = (60 kg)(10 m/s² − 2 m/s²) = 480 N, less than mg.
B600 N A student who thinks a scale always reads mg picks this. The elevator is accelerating at t = 7 s, so the normal force differs from the gravitational force.
C720 N A student who links the reading to the direction of motion picks this: moving up, so heavier, (60 kg)(10 + 2) m/s². The elevator is moving up but slowing, so its acceleration is downward and the reading is below mg.
D120 N A student who thinks the scale reads the net force picks this: (60 kg)(2 m/s²). The net force is the normal force minus the gravitational force; the scale reads the normal force itself.
Working From 6 s to 8 s the slope is (0 − 4 m/s)/(2 s) = −2 m/s², so a = 2 m/s² downward at t = 7 s. Scale reading = normal force: FN − mg = may ⇒ FN = m(g + ay) = (60 kg)(10 − 2) m/s² = 480 N. Distractors: FN = mg = 600 N; ‘moving up, so heavier’ → (60)(10 + 2) = 720 N; net force |ma| = 120 N.
A box of mass m rests on the floor of an elevator car of mass M. The cable pulls up on the car with a constant tension of magnitude T, which is greater than (M + m)g, where g is the gravitational field strength. What is the apparent weight of the box?
Answer and reasoning
Amg A student who thinks apparent weight always equals the gravitational force picks this. The box accelerates upward, so the floor must push on it with more than mg.
BmT/(M+m)Correct For the car and box together, T − (M + m)g = (M + m)a, so a = T/(M + m) − g, upward. For the box, FN − mg = ma, so FN = m(g + a) = mT/(M + m). The apparent weight is the magnitude of this normal force, which exceeds mg because the box accelerates upward.
CmT/M A student who leaves the box out of the accelerating system picks this, using a = T/M − g. The tension accelerates the car and the box together, so the total mass M + m belongs in the denominator.
DmT/(M+m)−mg A student who identifies the apparent weight with the net force on the box picks this: ma = mT/(M + m) − mg. The apparent weight is the normal force alone, which is the net force plus mg.
Working System car + box: T − (M + m)g = (M + m)a ⇒ a = T/(M + m) − g (upward). Box alone: FN − mg = ma ⇒ FN = m(g + a) = mT/(M+m). Distractors: FN = mg; system without the box, a = T/M − g ⇒ FN = mT/M; net force on the box, ma = mT/(M+m) − mg.
An elevator moving downward slows to a stop. A student claims that while the elevator is slowing, a scale under a passenger reads more than the magnitude of the gravitational force exerted on her. Which reasoning correctly supports the claim?
Answer and reasoning
AThe gravitational force on her increases while the elevator is slowing down. A student who thinks the scale reads the gravitational force picks this, explaining the larger reading as larger gravity. The gravitational force is unchanged; the normal force changes because she is accelerating.
BHer downward motion carries a force that the scale must stop, as well as her weight. A student who thinks a moving object carries a force in its direction of motion picks this. Only Earth and the scale exert forces on her; the larger reading comes from the upward acceleration.
CThe net force on her points upward, and the scale reading equals that net force. A student who thinks a scale reads the net force picks this. The net force is FN − mg, the difference between two forces; the scale reads FN itself, which exceeds mg because the net force is upward.
DHer acceleration is upward, so the upward normal force must exceed the gravitational force.Correct Slowing while moving down means the velocity is changing upward, so the acceleration is upward. Newton’s second law then needs an upward net force: FN − mg = ma with a > 0, so the normal force, which the scale reads, is greater than mg.
In which situation does the person appear weightless? Air resistance is negligible unless it is mentioned.
Answer and reasoning
AA skydiver falling at a constant terminal speed, with air resistance acting on her A student who thinks every falling object is weightless picks this. At terminal speed the air resistance balances gravity, so the skydiver has an apparent weight equal to mg.
BA diver in the air after leaving a springboard, before reaching the waterCorrect Once the diver leaves the board, the gravitational force is the only force exerted on her (air resistance is negligible), which is the condition for appearing weightless—whether she is moving up, down or momentarily at rest.
CAn astronaut standing in the air-filled cabin of a rocket hovering 400 km above Earth A student who thinks there is no gravity in space picks this. At 400 km Earth’s field is still close to 90% of its surface value; the rocket hovers because its thrust balances gravity, and the cabin floor pushes up on the astronaut, so she has an apparent weight.
DAn astronaut standing at rest on the airless surface of the Moon, far from Earth A student who thinks gravity needs air picks this. The Moon pulls the astronaut and its surface pushes up on her, so she has an apparent weight, about one-sixth of what it is on Earth.
Cabin 1 is a closed, windowless room at rest on Earth’s surface. Cabin 2 is an identical room in a rocket far from all planets and stars, accelerating at 10 m/s² in the direction from its floor to its ceiling. In each cabin an astronaut stands on a scale and releases a heavy ball and a light ball together from the same height. Air resistance is negligible. Which prediction is correct?
Answer and reasoning
AIn cabins 1 and 2, the balls land together and the scales give the same reading.Correct In cabin 2 the released balls move at constant velocity while the floor accelerates up to them at 10 m/s², so both land together, just as both fall at about 10 m/s² in cabin 1; the floor also pushes up on the astronaut with the same force in both. By the equivalence principle she cannot tell the cabins apart.
BIn cabin 2, both balls float beside her hand; there is no gravity there. A student who reasons that without gravity nothing can fall picks this. The balls do stay where they were released, but the floor accelerates up to meet them, so to the astronaut they appear to fall at 10 m/s².
CIn cabin 1 the heavy ball lands first; in cabin 2 both balls land together. A student who thinks heavier objects fall faster picks this. With air resistance negligible, both balls fall with the same acceleration on Earth too, so the balls land together in both cabins.
DHer scale reads her weight in cabin 1, but it reads zero in cabin 2. A student who thinks a scale reads the gravitational force picks this. A scale reads the normal force; in cabin 2 the floor must push her with m(10 m/s²) to accelerate her, the same as in cabin 1.
Working Cabin 1: both balls fall with acceleration g ≈ 10 m/s² (equal for all masses), and the scale reads FN = mg. Cabin 2: the gravitational force is negligible; the released balls move at constant velocity while the floor accelerates toward them at 10 m/s², so both reach the floor together, and the floor pushes the astronaut with FN = m(10 m/s²). Same observations: the equivalence principle.
Which observation provides evidence that an object’s inertial mass and its gravitational mass are equivalent?
Answer and reasoning
AIn a vacuum, a lead ball falls with greater acceleration than a wooden ball of equal size. A student who thinks heavier objects fall faster picks this. In a vacuum, lead and wooden balls fall with the same acceleration; a difference would show that the two kinds of mass are not equivalent.
BIn a vacuum, objects of different mass and material fall with equal accelerations.Correct The gravitational force is proportional to gravitational mass and the acceleration is that force divided by inertial mass. Equal accelerations for every object, whatever its mass or composition, show that gravitational mass is proportional to inertial mass for every object; with both measured in kilograms they are equal, and experiments confirm this to very high precision.
CA 1 kg object has the same weight on the Moon as it has on Earth’s surface. A student who thinks weight is a fixed property picks this. The object’s weight on the Moon is about one-sixth of its weight on Earth, and weight alone says nothing about inertial mass.
DObjects inside an orbiting space station lose their mass, and so they float freely. A student who thinks weightless objects have no mass picks this. Objects in orbit keep both their inertial and their gravitational mass; they float because gravity is the only force exerted on them.
The figure shows a uniform solid sphere and a thin, uniform spherical shell concentric with it, far from all other objects, with their masses and radii, and a point P with its distance from the common center. G is the universal gravitational constant. What is the magnitude of the gravitational field at P?
Answer and reasoning
A0.44 GM/R²Correct P is inside the shell, so by the shell theorem the shell exerts no net force there. P is outside the solid sphere, which acts as a point mass at the center: g = GM/(1.5R)² = GM/(2.25R²) ≈ 0.44 GM/R².
B1.33 GM/R² A student who treats the shell as a point mass at the center even inside it picks this: G(3M)/(1.5R)². The shell’s own field is zero everywhere inside it, so only the sphere’s mass M counts.
C4.00 GM/R² A student who measures the distance from the sphere’s surface picks this: GM/(0.5R)². The sphere acts as a point mass at its center, so the distance is 1.5R.
D0.00 GM/R² A student who thinks the field anywhere inside a shell is zero picks this. Only the shell’s own contribution is zero inside it; the solid sphere inside P’s radius still pulls P toward the center.
Working P is inside the shell, so the shell contributes zero (shell theorem). P is outside the solid sphere, which acts as a point mass M at the center: g = GM/(1.5R)² = GM/(2.25R²) ≈ 0.44 GM/R². Distractors: shell treated as a point mass inside too → 3GM/(2.25R²) ≈ 1.33; distance from the sphere’s surface, 0.5R → GM/(0.25R²) = 4.00; ‘inside a shell the field is zero’ applied to everything → 0.
The figure shows a cross section of a thin, uniform spherical shell, with its mass and radius, and a small object outside it, with its mass and its distance from the shell’s surface. What is the magnitude of the gravitational force exerted on the small object by the shell?
Answer and reasoning
A6.7 × 10⁻⁸ N A student who uses the distance from the shell’s surface picks this: GMm/(1.0 m)². The shell acts as if its mass were at its center, 3.0 m from the object.
B7.4 × 10⁻⁹ NCorrect Outside a thin uniform shell, the shell acts as a single object at its center, so r = 2.0 m + 1.0 m = 3.0 m. |Fg| = GMm/r² = (6.67 × 10⁻¹¹)(500)(2.0)/(3.0)² N = 7.4 × 10⁻⁹ N.
C2.2 × 10⁻⁸ N A student who divides by r instead of r² picks this: GMm/(3.0 m). The force falls off as 1/r², which also gives the correct unit, N.
D3.7 × 10⁻⁹ N A student who calculates the shell’s field at the object, GM/r², and reports it as the force picks this. The force is the object’s mass times that field: (2.0 kg)(3.7 × 10⁻⁹ N/kg).
Working Outside a thin uniform shell, treat the shell as a point mass at its center: r = 2.0 m + 1.0 m = 3.0 m. |Fg| = GMm/r² = (6.67 × 10⁻¹¹)(500)(2.0)/(3.0)² N = 7.4 × 10⁻⁹ N. Distractors: distance from the surface, 1.0 m → 6.7 × 10⁻⁸ N; r not squared → 2.2 × 10⁻⁸ N; field GM/r² reported as the force → 3.7 × 10⁻⁹ N.
A spherical planet of radius R has a density that depends only on the distance r from its center: ρ(r) = ρ₀(1 − r/R), where ρ₀ is a constant. G is the universal gravitational constant. What is the magnitude of the gravitational field at the planet’s surface?
Answer and reasoning
A4.19 Gρ₀R A student who treats the planet as uniform with its central density picks this: M = ρ₀(4/3)πR³. The density falls to zero at the surface, so the true mass is only one-quarter of that.
B2.09 Gρ₀R A student who averages ρ(r) over the radius, ρ₀/2, and multiplies by the volume picks this. Outer shells hold more volume, where the density is low, so the mass-weighted average density is ρ₀/4, not ρ₀/2.
C6.28 Gρ₀R A student who uses 4πR² dr for every shell picks this: M = 4πR² ∫₀ᴿ ρ₀(1 − r/R) dr = 2πρ₀R³. A shell at radius r has area 4πr², which must stay inside the integral.
D1.05 Gρ₀RCorrect Add thin shells: M = ∫₀ᴿ ρ₀(1 − r/R)4πr² dr = 4πρ₀R³(1/3 − 1/4) = πρ₀R³/3. Each uniform shell acts on an outside point as a point mass at the center, so g = GM/R² = (π/3)Gρ₀R ≈ 1.05 Gρ₀R.
Working M = ∫₀ᴿ ρ₀(1 − r/R)4πr² dr = 4πρ₀[r³/3 − r⁴/(4R)]₀ᴿ = 4πρ₀R³(1/3 − 1/4) = πρ₀R³/3. Each thin shell is uniform and acts as a point mass at the center for a point outside it, so at the surface g = GM/R² = (π/3)Gρ₀R ≈ 1.05 Gρ₀R. Distractors: uniform with ρ₀ → (4π/3) ≈ 4.19; average of ρ over r (ρ₀/2) times volume → (2π/3) ≈ 2.09; dV = 4πR² dr → M = 2πρ₀R³ → 2π ≈ 6.28.
A small probe is at the bottom of a deep shaft inside a uniform spherical planet, a distance r from the planet’s center. Which part of the planet’s mass determines the net gravitational force exerted on the probe?
Answer and reasoning
AThe mass within a sphere of radius r centered on the planet’s centerCorrect The planet can be divided into thin shells. Shells farther from the center than the probe exert zero net force on it; the mass at distances less than r acts as a point mass at the center. So only the partial mass within radius r determines the force.
BThe planet’s whole mass, treated as if it were all at the planet’s center A student who applies GMm/r² with the whole mass inside the planet picks this. The layers farther out than the probe surround it like shells and exert zero net force.
CNone of it, as the net force is zero everywhere inside the planet A student who extends the hollow-shell result to a solid planet picks this. The mass closer to the center than the probe still pulls it toward the center.
DThe outer layer of mass between the probe and the surface above it A student who thinks the nearest mass pulls hardest picks this. The layers above the probe surround it on all sides like shells, and their net pull on it is zero.
The figure shows a cross section of a uniform spherical planet, with its radius, and a probe at the bottom of a shaft, with its depth below the surface. The gravitational field at the planet’s surface is 4.0 N/kg. What is the magnitude of the gravitational field at the probe?
Answer and reasoning
A7.1 N/kg A student who uses the planet’s whole mass at the smaller distance picks this: (4.0 N/kg)(4/3)². Only the mass closer to the center than the probe contributes; the outer layers add zero net force.
B5.3 N/kg A student who takes the partial mass as proportional to r picks this: g = (4.0 N/kg)(R/r). The partial mass is proportional to r³, which makes the field inside proportional to r.
C3.0 N/kgCorrect The probe is r = 4.0 × 10⁶ m − 1.0 × 10⁶ m = 3.0 × 10⁶ m from the center. Only the mass within r counts, M(r/R)³, so g = GM(r/R)³/r² = (GM/R²)(r/R) = (4.0 N/kg)(3/4) = 3.0 N/kg.
D1.0 N/kg A student who uses the depth as the distance r picks this: (4.0 N/kg)(1.0/4.0). The partial mass and the field depend on the distance from the planet’s center, 3.0 × 10⁶ m.
Working Distance from the center: r = 4.0 × 10⁶ m − 1.0 × 10⁶ m = 3.0 × 10⁶ m. Only the partial mass M(r/R)³ contributes: g = G M(r/R)³/r² = (GM/R²)(r/R) = (4.0 N/kg)(3.0/4.0) = 3.0 N/kg. Distractors: whole mass, gs(R/r)² = 7.1 N/kg; partial mass ∝ r, gs(R/r) = 5.3 N/kg; depth used as r, gs(1.0/4.0) = 1.0 N/kg.
A uniform spherical planet has density 5.0 × 10³ kg/m³ and radius 4.0 × 10⁶ m. A probe is 3.0 × 10⁶ m below the planet’s surface. What is the partial mass of the planet that contributes to the net gravitational force exerted on the probe?
Answer and reasoning
A5.7 × 10²³ kg A student who uses the depth below the surface as rpartial picks this. The partial mass is the mass closer to the center than the probe, within a radius equal to the probe’s distance from the center.
B3.4 × 10²³ kg A student who scales the planet’s mass by r/R picks this: M(1.0/4.0). Mass scales with volume, so the partial mass is M(r/R)³, one sixty-fourth of M here.
C1.3 × 10²⁴ kg A student who thinks the whole planet’s mass acts on an object inside it picks this. The layers farther from the center than the probe exert zero net force on it.
D2.1 × 10²² kgCorrect The probe is rpartial = 4.0 × 10⁶ m − 3.0 × 10⁶ m = 1.0 × 10⁶ m from the center. Only the mass within that radius contributes: mpartial = ρ(4/3)πrpartial³ = (5.0 × 10³)(4/3)π(1.0 × 10⁶)³ kg = 2.1 × 10²² kg.
Working rpartial = 4.0 × 10⁶ m − 3.0 × 10⁶ m = 1.0 × 10⁶ m. mpartial = ρ(4/3)π(rpartial)³ = (5.0 × 10³)(4/3)π(1.0 × 10⁶)³ kg = 2.1 × 10²² kg. Distractors: depth as rpartial → (5.0 × 10³)(4/3)π(3.0 × 10⁶)³ = 5.7 × 10²³ kg; partial mass ∝ r, M(r/R) = 3.4 × 10²³ kg; whole mass M = 1.3 × 10²⁴ kg.
A uniform spherical planet has density ρ and radius R. A small object of mass m is inside the planet, in a shaft, a distance r from the planet’s center. G is the universal gravitational constant. What is the magnitude of the gravitational field at the object’s location?
Answer and reasoning
A(4/3)πGρR²/r A student who scales the partial mass in proportion to r, M(r/R) = ρ(4/3)πR²r, picks this. Mass grows with volume, as r³, so the field inside grows in proportion to r.
B(4/3)πGρrCorrect Only the mass within radius r contributes: mpartial = ρ(4/3)πr³. It acts as a point mass at the center, so g = Gmpartial/r² = Gρ(4/3)πr³/r² = (4/3)πGρr, proportional to r.
C(4/3)πGρmr A student who confuses the field with the force picks this: it is the gravitational force on the object. The field is the force per unit mass, so dividing by m gives (4/3)πGρr.
D0 A student who extends the hollow-shell result to a solid planet picks this. The shells outside radius r add nothing, but the mass within r still pulls the object toward the center.
Working mpartial = ρ(4/3)πr³; g = Gmpartial/r² = Gρ(4/3)πr³/r² = (4/3)πGρr. Distractors: partial mass ∝ r, ρ(4/3)πR³(r/R) = ρ(4/3)πR²r → (4/3)πGρR²/r; the force on the object, m times the field → (4/3)πGρmr; ‘zero anywhere inside a planet’ → 0.
A uniform spherical planet has radius R. An object can be moved from the planet’s center, along a narrow shaft, to the surface and then far out into space. Which of the graphs shown best represents the magnitude F of the gravitational force on the object as a function of its distance r from the planet’s center?
Answer and reasoning
AGraph 1 A student who uses the whole planet’s mass inside it picks the graph in which F keeps rising as r decreases toward zero. Only the partial mass within r counts, and it shrinks as r³, so F falls to zero at the center.
BGraph 2 A student who thinks the force is zero everywhere inside a planet picks the graph that is zero up to R. The mass closer to the center than the object still pulls it, so F grows steadily from zero to its surface value.
CGraph 3Correct Inside the planet only the mass within radius r pulls on the object, so F = GMmr/R³, which grows in proportion to r from zero at the center. The force is largest at the surface, GMm/R², and outside it falls off as 1/r².
DGraph 4 A student who thinks gravity is strongest at the center picks the graph that is largest at r = 0. At the center the surrounding mass pulls equally in all directions, so the force there is zero.
Working Inside (r < R): only the partial mass M(r/R)³ counts, so F = GMmr/R³, a straight line through the origin (Fg,partial = −kr). At r = R, F = GMm/R², its maximum. Outside: F = GMm/r², decreasing as 1/r². Whole mass inside → F diverges as r → 0; zero inside → F = 0 for r < R; strongest at the center → F largest at r = 0.
A small object inside a uniform spherical planet of radius R is at a distance R/4 from the planet’s center, where the gravitational force exerted on it by the planet has magnitude F₁. The object is moved along a shaft to a distance R/2 from the center. What is now the magnitude of the gravitational force on it?
Answer and reasoning
A0.25F₁ A student who uses the whole planet’s mass with the inverse-square law inside the planet picks this. Only the mass within r counts, and it grows faster than r², so the force increases outward.
B0.50F₁ A student who makes the partial mass proportional to r picks this, which gives F ∝ r/r² = 1/r. The partial mass grows as r³, so F ∝ r³/r² = r.
C8.00F₁ A student who makes the force proportional to the partial mass alone picks this: (2)³ = 8. The distance in the law also doubles, dividing the force by 2², so F doubles.
D2.00F₁Correct Inside a uniform sphere, F = Gmpartial m/r² with mpartial ∝ r³, so F ∝ r (Fg,partial = −kr). Doubling the distance from R/4 to R/2 doubles the force: 2.00F₁.
Working Inside a uniform sphere F = kr (Fg,partial = −kr), so doubling r doubles F: 2.00F₁. Distractors: whole mass, F ∝ 1/r² → 0.25F₁; partial mass ∝ r, F ∝ 1/r → 0.50F₁; force ∝ partial mass alone, ∝ r³ → 8.00F₁.
Planets A and B are uniform spheres with the same density. Planet A has radius R and planet B has radius 2R. Identical probes are placed in shafts inside the planets, each at a distance R/2 from its planet’s center. FA and FB are the magnitudes of the gravitational forces exerted on the probes by their planets. Which is correct?
Answer and reasoning
AFB/FA = 1.0Correct Inside a uniform sphere only the mass within radius r counts: F = Gρ(4/3)πr³m/r² = (4/3)πGρmr. That depends on the density and on r, not on the planet’s radius; with the same ρ and r the forces are equal.
BFB/FA = 8.0 A student who uses each planet’s whole mass picks this: B has 2³ = 8 times A’s mass. The mass farther from the center than R/2 adds zero net force in either planet.
CFB/FA = 4.0 A student who takes the partial mass as M(r/R) picks this: 8M(1/4) for B against M(1/2) for A. The partial mass is ρ(4/3)πr³, the same in both planets.
DFB/FA = 3.0 A student who uses each probe’s depth below the surface as r picks this: in F ∝ ρr the depths are R/2 for A and 3R/2 for B, a ratio of 3. The distance in the law is measured from the planet’s center, R/2 for both probes.
Working Inside a uniform sphere, F = Gρ(4/3)πr³m/r² = (4/3)πGρmr: it depends on ρ and r but not on the planet’s radius. Same ρ, same r ⇒ FB/FA = 1. Distractors: whole masses (8 : 1) at the same r → 8; partial mass ∝ r: A gives MA/2, B gives 8MA/4 = 2MA → 4; depth below the surface used as r → 1.5R/0.5R = 3.0.
An astronaut wakes in a closed, windowless cabin. Her scale gives the same reading as at home on Earth, and a ball she releases accelerates toward the floor at 10 m/s². She concludes that she cannot tell whether the cabin is at rest on Earth’s surface or in a rocket accelerating at 10 m/s² far from all planets and stars. Which reasoning supports her conclusion?
Answer and reasoning
AEarth’s field is 10 N/kg at any distance from Earth, so the rocket cabin has the same gravity. A student who thinks g has the same value at every distance from Earth picks this. Far from all planets and stars the gravitational field is negligible; the matching readings come from the rocket’s acceleration.
BA floor accelerating at 10 m/s² pushes on her and rises to meet the ball, mimicking gravity exactly.Correct In the rocket the floor must push her with m(10 m/s²) to accelerate her, the same push as on Earth, and the released ball moves at constant velocity while the floor catches up with it at 10 m/s². Every observation inside matches Earth’s gravity, as the equivalence principle states.
CHer scale reads the gravitational force on her, which has the same value in both cabins. A student who thinks a scale reads the gravitational force picks this. A scale reads the normal force; in the rocket there is almost no gravitational force, yet the floor pushes on her as hard as on Earth.
DHer mass is the same in both cabins, so her scale has to give the same reading. A student who treats mass and weight as the same quantity picks this. The scale reads a force, which depends on how hard the floor pushes; the same mass gives a reading of zero in a freely falling cabin.
Working In the rocket, far from all masses, the gravitational force is negligible. To accelerate her at 10 m/s² the floor must push her with m(10 m/s²), the same normal force as on Earth, so the scale reads the same. The released ball has no force on it and moves at constant velocity while the floor accelerates toward it at 10 m/s², so relative to her it ‘falls’ at 10 m/s². By the equivalence principle no observation inside distinguishes the two cases.
Two small uniform spheres, of masses m and 3m, are far from all other objects and are released from rest with their centers a distance d apart. G is the universal gravitational constant, and r is the distance between the spheres' centers. At the instant of release, what is the magnitude of d²r/dt²?
Answer and reasoning
A3Gm/d² A student who thinks only the more massive sphere pulls, so that sphere 3m stays at rest, picks this: only sphere m's acceleration, 3Gm/d², changes r. Sphere m pulls sphere 3m with a force of the same magnitude, so sphere 3m also accelerates toward sphere m, at Gm/d².
B6Gm/d² A student who thinks equal forces give equal accelerations gives sphere 3m the same acceleration as sphere m, 3Gm/d², and picks this: 3Gm/d² + 3Gm/d². With the same force, the sphere with three times the mass has one-third the acceleration, Gm/d².
C2Gm/d² A student who subtracts the magnitudes of the two accelerations, 3Gm/d² − Gm/d², picks this. The spheres accelerate in opposite directions, toward each other, so both accelerations reduce r and their magnitudes add.
D4Gm/d²Correct The spheres pull each other with forces of equal magnitude, 3Gm²/d². Sphere m accelerates at 3Gm/d² and sphere 3m at Gm/d², toward each other along the line of centers, so both accelerations shrink r: |d²r/dt²| = 3Gm/d² + Gm/d² = 4Gm/d².
Working Each sphere is pulled toward the other by a force of magnitude F = G(m)(3m)/d² = 3Gm²/d². Sphere m: a₁ = F/m = 3Gm/d², toward sphere 3m. Sphere 3m: a₂ = F/(3m) = Gm/d², toward sphere m. With x measured along the line from sphere m to sphere 3m, r = x₂ − x₁, so d²r/dt² = a₂ₓ − a₁ₓ = (−Gm/d²) − (3Gm/d²) = −4Gm/d²; magnitude 4Gm/d². (Sphere 3m taken as unaccelerated: 3Gm/d². Both spheres given sphere m's acceleration: 6Gm/d². Magnitudes subtracted: 2Gm/d².) Checked with sympy.
A planet far from all other objects is modeled as a hollow, uniform thick spherical shell of mass M, with inner radius R and outer radius 2R; the region within radius R of its center is empty. A small probe is at the bottom of a shaft, a distance 1.5R from the planet's center. G is the universal gravitational constant. What is the magnitude of the gravitational field at the probe's location?
Answer and reasoning
A0.44 GM/R² A student who treats the planet's whole mass as a point at its center, as for a point outside the planet, picks this: GM/(1.5R)². The probe is inside the planet, and the material farther from the center than the probe exerts zero net force on it.
B0.33 GM/R² A student who takes the partial mass to be proportional to the distance from the center, M(1.5R/2R) = 0.75M, picks this. Mass grows with volume, and the partial mass is only the material between R and 1.5R: the fraction (1.5³ − 1)/(2³ − 1) of M.
C0.15 GM/R²Correct Only the shell material closer to the center than the probe contributes; the material farther out forms thin shells around the probe that exert zero net force on it. At uniform density that material is the fraction (1.5³ − 1³)/(2³ − 1³) ≈ 0.339 of M, so g = G(0.339M)/(1.5R)² ≈ 0.15 GM/R².
D0.00 GM/R² A student who applies the hollow-shell result, zero net force, everywhere inside the planet picks this. The probe is within the shell's material, not in the empty cavity: the material between R and 1.5R is closer to the center than the probe and pulls it toward the center.
Working Density: ρ = M/[(4/3)π((2R)³ − R³)] = M/[(4/3)π(7R³)]. Only the material at distances less than 1.5R from the center contributes; the material farther out is a set of thin shells around the probe, each exerting zero net force on it. mpartial = ρ(4/3)π[(1.5R)³ − R³] = M(3.375 − 1)/7 = 19M/56 ≈ 0.339M. g = Gmpartial/(1.5R)² = (19/126)GM/R² ≈ 0.15 GM/R². (Whole mass treated as a point at the center: GM/(1.5R)² ≈ 0.44 GM/R². Partial mass taken in proportion to r, M(1.5R/2R) = 0.75M: 0.75GM/(2.25R²) ≈ 0.33 GM/R². Zero everywhere inside the planet: 0.) Checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
