3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A satellite of mass 500 kg moves in a circular orbit around a planet of mass 6.0 × 10²⁴ kg. The only forces are the gravitational forces between the two objects. Which statement correctly compares the gravitational forces that the planet and the satellite exert on each other, and their effects?
Answer and reasoning
AThe planet exerts the larger force, so the satellite has the larger acceleration. A student who thinks the more massive object exerts the larger force picks this. The two forces have equal magnitudes; the satellite accelerates far more because the same force acts on a far smaller mass.
BThe satellite is too small to exert any gravitational force on the planet. A student who thinks only massive bodies exert gravitational forces picks this. The satellite attracts the planet with a force equal in magnitude to the planet's force on it.
CThe forces are equal in magnitude, so the objects have equal accelerations. A student who thinks equal forces give equal motions picks this. The forces are equal, but a = F/m: the planet's mass is about 10²² times larger, so its acceleration is negligible.
DThe forces are equal in magnitude, and the planet's acceleration is negligible.Correct The forces are a Newton's third-law pair, so they have equal magnitudes. By a = F/m, the planet's acceleration is smaller than the satellite's by the ratio 500 kg/(6.0 × 10²⁴ kg), about 10⁻²², so the planet's motion is negligible.
A satellite moves in a circular orbit around a planet. Only the planet's gravitational force is exerted on the satellite, and the planet's motion is negligible. Which quantity changes as the satellite moves around its orbit?
Answer and reasoning
AThe satellite's momentum, since the direction of its velocity changes.Correct In a circular orbit the satellite's speed is constant but the direction of its velocity turns continually, so its momentum mv⃗, a vector, changes. Its kinetic energy and angular momentum and the system's Ug and mechanical energy are all constant.
BThe satellite's angular momentum, since its direction of motion changes. A student who treats angular momentum as momentum by another name picks this. L = rmv sin θ about the planet's center has r, v and θ = 90° all constant in a circular orbit, so L is constant even though the velocity's direction changes.
CThe system's gravitational potential energy, since the satellite keeps falling. A student who thinks an orbiting satellite keeps losing Ug because it is falling picks this. Ug = −GMm/r depends only on the separation, which is constant in a circular orbit.
DThe satellite's kinetic energy, since the planet's gravity pulls on it. A student who thinks any force on a moving object changes its speed picks this. The gravitational force is perpendicular to the velocity at every instant, so it does no work and the kinetic energy is constant.
A probe is launched from the surface of a planet with exactly the escape speed. The planet has no atmosphere, its motion is negligible, and the only force exerted on the probe after launch is the planet's gravitational force. Which statement about the mechanical energy of the probe–planet system is correct?
Answer and reasoning
AIt is zero, and it stays zero at every point of the probe's motion after launch.Correct Escape speed is defined as the speed that makes the system's mechanical energy zero. Only the internal gravitational force acts, so the mechanical energy stays at that value, zero, throughout the motion.
BIt is positive, since the probe needs extra energy to break free of gravity. A student who thinks escape needs positive mechanical energy picks this. Escape speed is the speed that makes K + Ug exactly zero; a positive value would mean a speed greater than the escape speed.
CIt is negative at launch and rises to zero as the probe gets very far away. A student who thinks the system gains energy as the probe climbs picks this. With only gravity acting, K + Ug is constant: Ug rises toward zero and K falls by the same amount.
DIt equals the probe's kinetic energy at launch, since Ug is zero at the planet's surface. A student who puts the zero of Ug at the planet's surface picks this. For a satellite–planet system Ug is defined to be zero at infinite separation, so at the surface Ug = −GMm/R, which exactly cancels the launch kinetic energy.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.6.A.1 Central object and satellite Fix
Central object and satellite
In a system of two objects interacting only through gravitational forces, the central object is the far more massive one (a star or planet) and the satellite is the object orbiting it (a moon, spacecraft or probe), whose mass is negligible in comparison.
Negligible motion of the central object
The satellite and the central object exert gravitational forces of equal magnitude on each other, but the central object's acceleration is smaller than the satellite's by the ratio of their masses, m/M. When m is negligible compared with M, the central object can be treated as at rest and only the satellite's motion is analyzed.
Students often think In a gravitational interaction, the object with more mass exerts the larger force on the other object. In fact No. The two forces form one interaction: they have equal magnitudes, GMm/r², and opposite directions, whatever the masses.
Students often think Only the massive body in a system exerts a gravitational force; a small object such as a satellite does not attract the planet. In fact Yes. Every object with mass attracts every other; the satellite pulls on the planet with a force equal in magnitude to the planet's pull on it. The planet's motion is negligible because of its large mass, not because it feels no force.
6.6.A.2 Conservation laws for an orbit Fix
Conservation laws for an orbit
When the only force is the gravitational force between the satellite and the central object, the system's mechanical energy is constant (no external work, no dissipation) and the satellite's angular momentum about the central object's center is constant (the force points along the line to that center, so it exerts no torque about it).
Circular orbit
An orbit at a constant distance r from the central object's center. The gravitational force is always perpendicular to the satellite's velocity, so the satellite's speed and kinetic energy are constant; the system's gravitational potential energy and mechanical energy and the satellite's angular momentum are also constant. The direction of the satellite's velocity, and so its momentum, changes continually.
Elliptical orbit
An orbit in which the satellite's distance from the central object varies between a closest point and a farthest point. The system's mechanical energy and the satellite's angular momentum are constant, but Ug and K change: K is greatest where the satellite is closest (Ug most negative) and least where it is farthest.
Angular momentum of a satellite
L = rmv sin θ about the central object's center, where θ is the angle between the position vector from that center and the velocity. It is constant in any orbit. At the closest and farthest points of an elliptical orbit θ = 90°, so r₁v₁ = r₂v₂ there. Unit: kg·m²/s.
Gravitational potential energy of a satellite–central-object system
Ug = −G(m₁m₂)/r, with r the distance between the centers and Ug defined to be zero when the objects are infinitely far apart. Ug is negative at every finite separation and increases toward zero as the separation increases. Unit: J.
Mechanical energy of the system
E = K + Ug, the sum of the satellite's kinetic energy and the system's gravitational potential energy. With Ug zero at infinite separation, E is negative for a satellite in orbit, zero for one that just escapes, and positive for one that escapes with speed to spare. Unit: J.
Students often think A satellite's angular momentum is its momentum mv by another name: it changes whenever the direction of the velocity changes, and keeping it constant means keeping the speed constant. In fact No. The angular momentum about the central object's center is L = rmv sin θ: it depends on the distance r and the angle as well as on mv. In a circular orbit L is constant while the momentum's direction changes; in an elliptical orbit L is constant while the speed changes.
Students often think An orbiting satellite is continually falling toward the planet, so the system's gravitational potential energy keeps decreasing even in a circular orbit. In fact No. In a circular orbit the separation r is constant, so Ug = −GMm/r is constant. The satellite 'falls' only in the sense that its velocity keeps turning toward the planet; it never gets closer.
6.6.A.3 Escape velocity Fix
Escape velocity
The velocity a satellite must have at a given distance from the central object so that the mechanical energy of the satellite–central-object system is zero: (1/2)mv² − GMm/r = 0. Its magnitude is the escape speed.
Motion at escape speed
A satellite launched with exactly the escape speed, with gravity from the central object the only force on it, slows down continually as it moves away; its speed approaches zero only as its distance approaches infinity, so it never stops and never returns.
Escape speed equation
From (1/2)mv² = GMm/r, vesc = √(2GM/r), with M the central object's mass and r the distance from its center. The satellite's mass cancels; vesc is larger for a more massive central object and smaller at a larger distance. Unit: m/s.
Students often think To escape, the system's mechanical energy must be positive, so a probe launched with exactly the escape speed does not quite escape and eventually falls back. In fact No. Escape requires the mechanical energy to be at least zero. With exactly zero, the probe still escapes: its speed approaches zero only as its distance approaches infinity. Positive energy means it escapes with speed to spare.
Students often think The mechanical energy of a probe–planet system increases as the probe moves away, because the probe gains potential energy as it escapes gravity. In fact No. With only gravity acting between them, the system's mechanical energy is constant: Ug increases as the probe moves away and K decreases by the same amount.
10 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 10
A satellite moves in a circular orbit around a planet, and the planet's motion is negligible. A student claims: "The planet's gravitational force pulls on the satellite at every instant, so the satellite's kinetic energy must be increasing." Which reasoning correctly shows that the satellite's kinetic energy is constant?
Answer and reasoning
AAn outward centrifugal force balances gravity, so the net force on the satellite is zero. A student who believes in an outward centrifugal force picks this. The only force on the satellite is gravity; it is not balanced, and the unbalanced force is what gives the satellite its centripetal acceleration.
BThe satellite is weightless in orbit, so the planet exerts no gravitational force on it. A student who thinks there is no gravity in orbit picks this. The planet's gravitational force is exerted on the satellite at every instant; the satellite is weightless because gravity is the only force on it.
CThe force is perpendicular to the satellite's velocity at every instant, so it does no work on the satellite.Correct Only a force component along the velocity does work. In a circular orbit the gravitational force points toward the center and the velocity is tangent to the circle, so the two are always perpendicular; the work done is zero, and the kinetic energy is constant.
DThe system's mechanical energy is constant, and a constant total keeps each part constant. A student who thinks a constant total means constant parts picks this. A constant K + Ug allows K to change if Ug changes, as in an elliptical orbit; it is the zero work done by the perpendicular force that keeps K constant here.
The diagram shows the elliptical orbit of a satellite around a planet whose motion is negligible, with the satellite's distances from the planet's center at its closest point P and its farthest point A, and its speed at P. At P and at A the satellite's velocity is perpendicular to the line from the planet's center. What is the satellite's speed at A?
Answer and reasoning
A8.0 × 10³ m/s A student who thinks constant angular momentum means constant speed picks this. L = rmv sin θ includes the distance: at A the satellite is twice as far away, so to keep L constant its speed must be half as large.
B4.0 × 10³ m/sCorrect The gravitational force points along the line to the planet's center, so it exerts no torque about that point and the satellite's angular momentum is constant. With the velocity perpendicular to the radius at P and A, rP m vP = rA m vA, so vA = (8.0 × 10³ m/s)(7.0 × 10⁶ m)/(1.4 × 10⁷ m) = 4.0 × 10³ m/s.
C2.0 × 10³ m/s A student who scales the speed with the inverse square of the distance, like the gravitational force, gets (8.0 × 10³ m/s)(1/2)² = 2.0 × 10³ m/s. Angular momentum rmv is constant, so v is inversely proportional to r, not to r², at these two points.
D1.6 × 10⁴ m/s A student who treats the satellite as a point on a rotating object, with v = rω and the same ω, doubles the speed at twice the distance. Nothing keeps the satellite's angular speed constant; its angular momentum is what is constant, so it moves more slowly farther out.
Working The gravitational force is directed along the line to the planet's center, so it exerts no torque about that point, and the satellite's angular momentum about it is constant. At P and A, sin θ = 1: rP m vP = rA m vA. vA = vP rP/rA = (8.0 × 10³ m/s)(7.0 × 10⁶ m)/(1.4 × 10⁷ m) = 4.0 × 10³ m/s.
A satellite moves in an elliptical orbit around a planet whose motion is negligible. The graph shows the gravitational potential energy Ug of the satellite–planet system as a function of time t, with three instants marked t₁, t₂ and t₃. Which statement about the satellite's kinetic energy K is correct?
Answer and reasoning
AK is greatest at t₂, where Ug is closest to zero. A student who thinks a satellite moves fastest where it is farthest away picks this. At t₂, Ug is at its highest, so the satellite is farthest from the planet; K + Ug is constant, so K is at its lowest there.
BK has the same value at t₁, at t₂ and at t₃. A student who thinks a constant mechanical energy means constant kinetic energy picks this. The graph shows Ug changing, so K must change by the same amount in the opposite direction to keep K + Ug constant.
CK is greatest at t₁, where Ug is most negative.Correct Only gravity acts between the satellite and the planet, so K + Ug is constant. K is therefore greatest where Ug is lowest (most negative), at t₁, when the satellite is closest to the planet, and least at t₂, when it is farthest.
DK is zero at t₂, where the satellite is farthest away. A student who pictures the satellite stopping at its farthest point, like a ball at the top of its flight, picks this. The satellite's angular momentum is constant and not zero, so it is still moving at t₂; K is at a minimum there, not zero.
Satellites X and Y have equal masses and move in circular orbits around the same planet. The radius of X's orbit is 2R and the radius of Y's orbit is 4R. How does the gravitational potential energy UX of the X–planet system compare with UY of the Y–planet system?
Answer and reasoning
AUX is greater, since the planet's gravitational pull on X is the stronger. A student who links potential energy to the strength of the force picks this. Ug = −GMm/r: UX = −GMm/(2R) and UY = −GMm/(4R). Moving a satellite outward takes positive work against the attraction, so Ug is greater farther out even though the force is weaker.
BUY is greater, since Ug increases toward zero as the distance grows.Correct Ug = −GMm/r, zero at infinite separation. UX = −GMm/(2R) and UY = −GMm/(4R), and −GMm/(4R) is closer to zero, so it is the greater value: UY > UX.
CUX is greater, since its value has the larger magnitude of the pair. A student who compares negative energies by their sizes picks this. UX = −GMm/(2R) has twice the magnitude of UY = −GMm/(4R), but because both are negative, UX is the lower value.
DThey cannot be compared, since the zero of Ug may be chosen at any point. A student who thinks an arbitrary zero makes values incomparable picks this. Moving the zero adds the same constant to both values, so which is greater does not change; with the usual zero at infinity, UY > UX.
The gravitational potential energy of a satellite–planet system is −4.0 × 10⁹ J when the satellite is in a circular orbit of radius R. The satellite is then moved to a circular orbit of radius 2R. What is the change in the gravitational potential energy of the system?
Answer and reasoning
A+3.0 × 10⁹ J A student who takes Ug to fall off as 1/r², like the force, divides −4.0 × 10⁹ J by 4 and gets a change of −1.0 × 10⁹ J − (−4.0 × 10⁹ J) = +3.0 × 10⁹ J. Ug = −GMm/r varies as 1/r, so doubling r halves it, to −2.0 × 10⁹ J, a change of +2.0 × 10⁹ J.
B−4.0 × 10⁹ J A student who thinks Ug is lower where the satellite is farther away, because the gravitational force is weaker there, doubles the value to −8.0 × 10⁹ J and gets a change of −4.0 × 10⁹ J. Ug = −GMm/r increases toward zero as r increases: doubling r halves it, to −2.0 × 10⁹ J, a change of +2.0 × 10⁹ J.
C−2.0 × 10⁹ J A student who drops the negative sign of Ug sees its size fall from 4.0 × 10⁹ J to 2.0 × 10⁹ J and gets a change of −2.0 × 10⁹ J. With the zero at infinite separation, Ug rises from −4.0 × 10⁹ J to −2.0 × 10⁹ J, toward zero: a change of +2.0 × 10⁹ J.
D+2.0 × 10⁹ JCorrect Ug = −GMm/r with the zero at infinite separation, so Ug is inversely proportional to r. Doubling r halves Ug, from −4.0 × 10⁹ J to −2.0 × 10⁹ J, so ΔUg = −2.0 × 10⁹ J − (−4.0 × 10⁹ J) = +2.0 × 10⁹ J: the gravitational force does negative work as the satellite moves away, so Ug rises.
Working Ug = −GMm/r, with Ug = 0 at infinite separation, so Ug ∝ 1/r. Doubling r halves Ug: (−4.0 × 10⁹ J)/2 = −2.0 × 10⁹ J. ΔUg = −2.0 × 10⁹ J − (−4.0 × 10⁹ J) = +2.0 × 10⁹ J.
A probe is launched straight up from the surface of a planet of radius R. The planet has no atmosphere and its motion is negligible. The graph shows the probe's kinetic energy K and the gravitational potential energy Ug of the probe–planet system as functions of the distance r from the planet's center, after launch. Which claim, including its reasoning, is supported by the graph?
Answer and reasoning
AThe probe escapes, since its kinetic energy at launch is positive. A student who thinks any probe moving away with kinetic energy escapes picks this. Escape needs K + Ug ≥ 0; here K + Ug = 2 − 6 = −4 (× 10⁸ J), and the graph shows K falling to zero at 1.5R.
BThe probe escapes, since the system's mechanical energy increases with r. A student who thinks the system gains energy as the probe climbs picks this. Read the graph at R and at 1.5R: K + Ug is −4 × 10⁸ J at both, so the mechanical energy is constant, and negative.
CThe probe does not escape, since Ug is negative at its launch point. A student who thinks a negative Ug always traps the probe picks this. Ug is negative at every finite distance, even for a probe launched faster than the escape speed; what decides escape is the sign of K + Ug.
DThe probe does not escape, since K + Ug is less than zero.Correct At launch, r = R, K = 2 × 10⁸ J and Ug = −6 × 10⁸ J, so K + Ug = −4 × 10⁸ J, and it stays at that value. A probe escapes only if K + Ug ≥ 0; the graph confirms that K reaches zero at r = 1.5R, where the probe stops and turns back.
A probe is launched straight up from the surface of a planet with exactly the escape speed. The planet has no atmosphere, its motion is negligible, and the only force exerted on the probe after launch is the planet's gravitational force. Which statement describes the probe's motion after launch?
Answer and reasoning
AIt slows down until it is beyond the reach of gravity, then moves at a constant speed. A student who thinks gravity stops acting at some distance picks this. The gravitational force weakens as 1/r² but is never zero, so the probe keeps slowing at every distance.
BIt slows to a stop at a large but finite distance, then falls back to the planet. A student who thinks escape needs positive mechanical energy picks this. With K + Ug = 0, K = GMm/r is positive at every finite distance, so the probe never stops and never returns.
CIt slows down all the time, its speed approaching zero as its distance approaches infinity.Correct With exactly the escape speed, K + Ug = 0, so K = GMm/r, which is positive at every finite distance: the probe never stops. The planet's gravitational force keeps slowing it, so its speed approaches zero only as r approaches infinity.
DIt speeds up as it moves away, since Ug decreases as its distance increases. A student who thinks Ug is lower farther from the planet picks this. Ug = −GMm/r increases toward zero as the probe moves away, so its kinetic energy decreases and it slows down.
A satellite moves in a circular orbit of radius r around a planet with speed v₀. The planet's motion is negligible. Which expression gives the escape speed at the distance r from the planet's center, in terms of v₀?
Answer and reasoning
A1.41v₀Correct For the circular orbit, GMm/r² = mv₀²/r gives GM/r = v₀². Escape needs zero mechanical energy: (1/2)mv² = GMm/r, so v² = 2GM/r = 2v₀² and v = √2 v₀ = 1.41v₀.
B1.00v₀ A student who thinks the orbital speed is already enough to escape picks this. A satellite in a circular orbit has K = (1/2)mv₀² = GMm/(2r), so K + Ug = −GMm/(2r) < 0: it is bound. Escape needs K = GMm/r, twice as much.
C2.00v₀ A student who leaves out the square roots writes v₀ = GM/r for the orbit and v = 2GM/r for escape, and so gets v = 2v₀. Both equations give v², not v: v₀² = GM/r and v² = 2GM/r, so v² = 2v₀² and v = √2 v₀ = 1.41v₀.
D2.41v₀ A student who adds the escape speed, √2 v₀, to the speed the satellite already has gets (1 + √2)v₀. The escape speed is the total speed the satellite must have at that distance: its speed must be raised to √2 v₀, not increased by √2 v₀.
Working Circular orbit: gravity provides the centripetal force, GMm/r² = mv₀²/r, so GM/r = v₀². Escape: the mechanical energy must be zero, (1/2)mv² − GMm/r = 0, so v² = 2GM/r = 2v₀² and v = √2 v₀ = 1.41v₀.
A 500 kg probe is launched from the surface of a moon of mass 7.3 × 10²² kg and radius 1.7 × 10⁶ m. The moon has no atmosphere and its motion is negligible. Use G = 6.67 × 10⁻¹¹ N·m²/kg². What is the escape speed from the moon's surface?
Answer and reasoning
A1.7 × 10³ m/s A student who uses the speed of a circular orbit at the surface, √(GM/R), picks this. A probe at that speed would orbit, with negative mechanical energy; escape needs √2 times as much speed.
B2.4 × 10³ m/sCorrect Escape requires zero mechanical energy: (1/2)mv² = GMm/R. The probe's mass cancels, so v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 7.3 × 10²² / 1.7 × 10⁶) m/s = 2.4 × 10³ m/s.
C5.7 × 10⁶ m/s A student who stops at 2GM/R = 5.7 × 10⁶ picks this. That is v², in m²/s²; the speed is its square root, 2.4 × 10³ m/s.
D5.4 × 10⁴ m/s A student who thinks the probe's mass affects its escape speed puts the 500 kg into the formula. The probe's mass appears in both its kinetic energy and Ug, so it cancels out of the escape speed.
Working (1/2)mv² − GMm/R = 0, so v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ N·m²/kg² × 7.3 × 10²² kg / 1.7 × 10⁶ m) = √(5.73 × 10⁶ m²/s²) = 2.4 × 10³ m/s. The probe's mass cancels.
A satellite of mass m moves in a circular orbit at a height R above the surface of a planet of radius R. The gravitational field strength at the planet's surface is g, and the planet's motion is negligible. Which expression gives the total mechanical energy of the satellite–planet system?
Answer and reasoning
A−0.50mgR A student who uses the height, R, as r in the force law and in Ug = −GMm/r gets K = (1/2)mgR and Ug = −mgR, and picks this. r is measured from the planet's center: r = R + R = 2R.
B+0.75mgR A student who drops the negative sign of Ug adds (1/2)mgR to K = (1/4)mgR and picks this. Ug = −GMm/r is negative, and it must be added with its sign.
C+2.00mgR A student who applies the near-surface model at the orbit takes the gravitational force as mg, so mg = mv²/(2R) gives K = mgR, and takes Ug = mgΔy = mgR with zero at the surface; the sum is +2.00mgR. At a height equal to the planet's radius g is not constant: the force is GMm/r² and Ug = −GMm/r, with zero at infinite separation.
D−0.25mgRCorrect The orbit radius is 2R and GM = gR². From GMm/(2R)² = mv²/(2R), v² = gR/2, so K = (1/4)mgR. Ug = −GMm/(2R) = −(1/2)mgR. E = K + Ug = −0.25mgR; it is negative because the satellite is bound to the planet.
Working At the surface g = GM/R², so GM = gR². The orbit radius is r = R + R = 2R. Newton's second law for the circular orbit: GMm/r² = mv²/r, so v² = GM/r = gR²/(2R) = gR/2 and K = (1/2)mv² = (1/4)mgR. Ug = −GMm/r = −mgR²/(2R) = −(1/2)mgR. E = K + Ug = (1/4 − 1/2)mgR = −0.25mgR. Errors: r taken as the height R → K = (1/2)mgR, Ug = −mgR, E = −0.50mgR; sign of Ug dropped → E = (1/4 + 1/2)mgR = +0.75mgR; near-surface model applied at the orbit (constant g, so mg = mv²/(2R) and K = mgR; Ug = mgΔy = mgR with zero at the surface) → E = +2.00mgR.
Compiled from the AP Physics 1 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
