AP Physics C: Mechanics Gravitation — Worked Answer Explanations

Unit 7 · 12 questions explained

Below is a complete answer key for our AP Physics C: Mechanics Gravitation practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Gravitation practice test and come back here to review, or head back to the Gravitation unit overview.

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Question 1 · Easy
Two point masses $M$ and $m$ are separated by distance $r$ . Which expression gives the magnitude of the gravitational force between them?
- A
  $\frac{GM m}{r}$
  Why not A: Missing the square on $r$ .
- B
  $\frac{GM m}{r ^{2}}$ Correct
- C
  $\frac{GM}{r ^{2}}$
  Why not C: Gravitational field, not force (missing $m$ ).
- D
  $GM m r^{2}$
  Why not D: Multiplied instead of dividing by $r^{2}$ .
Explanation
Newton's law of universal gravitation: $F = \frac{GM m}{r ^{2}}$ . The force is attractive and acts along the line joining the two masses. Note $G \approx 6.67 \times 1 0^{- 11}$ N·m²/kg².

Key takeaway
$F = GMm/r^2$. Force falls off as the inverse square of distance — doubling $r$ quarters the force.
Question 2 · Easy
A satellite of mass $m$ orbits Earth (mass $M_{E}$ , radius $R_{E}$ ) at altitude $h$ above the surface in a circular orbit. What is the orbital speed?
- A
  $\frac{G M _{E}}{R _{E}}$
  Why not A: Used Earth's radius instead of orbital radius $r = R_{E} + h$ .
- B
  $\frac{G M _{E}}{R _{E} + h}$ Correct
- C
  $\frac{G M _{E}}{R _{E} + h}$
  Why not C: Forgot the square root.
- D
  $\frac{G M _{E} m}{R _{E} + h}$
  Why not D: Included satellite mass $m$ — orbital speed is independent of $m$ .
Explanation
Set gravitational force equal to centripetal force: $\frac{G M _{E} m}{r ^{2}} = \frac{m v ^{2}}{r}$ where $r = R_{E} + h$ . Cancel $m$ and $r$ : $v = \frac{G M _{E}}{r} = \frac{G M _{E}}{R _{E} + h}$ .

Key takeaway
Orbital speed $v = \sqrt{GM_E/r}$ is independent of satellite mass. The higher the orbit, the slower the satellite.
Question 3 · Easy
The gravitational potential energy of mass $m$ at distance $r$ from mass $M$ is $U (r) = - \frac{GM m}{r}$ . Why is $U$ negative?
- A
  Because $G$ , $M$ , and $m$ are all negative quantities.
  Why not A: $G$ , $M$ , and $m$ are all positive.
- B
  Because we define $U = 0$ at infinity, and work must be done on the system to separate the masses.Correct
- C
  Because the force is repulsive at close range.
  Why not C: Gravity is always attractive.
- D
  Because kinetic energy is always greater than potential energy.
  Why not D: This is not a general statement and does not explain the sign convention.
Explanation
We set $U (\infty) = 0$ as the reference. Since gravity is attractive, the system is in a bound (lower-energy) state when the masses are close together, requiring $U < 0$ . Separating the masses to infinity requires adding energy: $W = - U = GM m / r > 0$ .

Key takeaway
$U = -GMm/r < 0$ because the zero reference is at $r = \infty$. Bound systems have $E = KE + U < 0$.
Question 4 · Easy
Using Kepler's third law $T^{2} \propto r^{3}$ , if Earth's orbital radius is $1$ AU and period is $1$ yr, what is the orbital period of a planet at $4$ AU?
- A
  $4$ yr
  Why not A: Used $T \propto r$ (linear) rather than $T \propto r^{3/2}$ .
- B
  $8$ yrCorrect
- C
  $16$ yr
  Why not C: Used $T^{2} = r^{3}$ without taking the square root (forgot $T = r^{3/2}$ ).
- D
  $64$ yr
  Why not D: Used $T = r^{3}$ directly.
Explanation
$T \propto r^{3/2}$ . $\frac{T _{P}}{T _{E}} = (\frac{r _{P}}{r _{E}})^{3/2} = 4^{3/2} = (2^{2})^{3/2} = 2^{3} = 8$ . So $T_{P} = 8$ yr.

Key takeaway
Kepler's third law: $T^2 = \frac{4\pi^2}{GM}r^3$, so $T \propto r^{3/2}$. Quadrupling $r$ gives $4^{3/2} = 8$ times longer period.
Question 5 · Medium
Derive the gravitational potential energy by integrating the gravitational force. Starting from $F (r) = - \frac{GM m}{r ^{2}}$ (toward $M$ ), which integral correctly gives $U (r)$ with $U (\infty) = 0$ ?
- A
  $U (r) = \int_{r}^{\infty} \frac{GM m}{r ^{'2}} d r^{'}$ Correct
- B
  $U (r) = - \int_{r}^{\infty} \frac{GM m}{r ^{'2}} d r^{'}$
  Why not B: Extra negative sign gives $U = + GM m / r > 0$ — wrong sign for attractive gravity.
- C
  $U (r) = \int_{0}^{r} \frac{GM m}{r ^{'2}} d r^{'}$
  Why not C: Wrong limits: integrating from 0 diverges and the zero reference is at $r = 0$ , not $\infty$ .
- D
  $U (r) = \int_{\infty}^{r} \frac{GM m}{r ^{'2}} d r^{'}$
  Why not D: Swapped limits give an opposite sign; this equals $- GM m / r$ (positive), not $- GM m / r$ .
Explanation
$U (r) = - \int_{\infty}^{r} F_{e x t} d r^{'} = - \int_{\infty}^{r} (- \frac{GM m}{r ^{'2}}) d r^{'} = \int_{r}^{\infty} \frac{GM m}{r ^{'2}} d r^{'}$ .

Evaluating: $U (r) = GM m [- \frac{1}{r ^{'}}]_{r}^{\infty} = GM m (0 - (- \frac{1}{r})) = - \frac{GM m}{r}$ .

The work definition $U (r) = - W_{\infty}^{r} = - \int_{\infty}^{r} F_{g} \cdot d r^{'}$ confirms the result.

Key takeaway
Gravitational PE via integration: $U(r) = \int_r^\infty \frac{GMm}{r'^2}\,dr' = -\frac{GMm}{r}$. The key step is choosing limits so $U(\infty) = 0$.
Question 6 · Medium
A rocket launches from Earth's surface (mass $M_{E}$ , radius $R_{E}$ ). What is the minimum launch speed needed to escape Earth's gravity entirely?
- A
  $\frac{G M _{E}}{R _{E}}$
  Why not A: Circular orbit speed — not enough to escape.
- B
  $\frac{2 G M _{E}}{R _{E}}$ Correct
- C
  $\frac{2 G M _{E}}{R _{E}}$
  Why not C: Forgot the square root.
- D
  $\frac{G M _{E}}{2 R _{E}}$
  Why not D: Off by a factor of $4$ (missing the factor of 2 in the numerator).
Explanation
Set total mechanical energy to zero (barely escaping): $\frac{1}{2} m v_{esc}^{2} - \frac{G M _{E} m}{R _{E}} = 0 \Rightarrow v_{esc} = \frac{2 G M _{E}}{R _{E}}$ . For Earth, $v_{esc} \approx 11.2$ km/s. Note $v_{esc} = 2 v_{or bi t}$ at the surface.

Key takeaway
Escape velocity: set $KE + U = 0$ (zero total energy) and solve: $v_{esc} = \sqrt{2GM/R}$. It is $\sqrt{2}$ times the circular orbit speed at the same radius.
Question 7 · Medium
A satellite in circular orbit of radius $r$ around Earth (mass $M_{E}$ ) has total mechanical energy $E$ . Which expression is correct?
- A
  $E = - \frac{G M _{E} m}{r}$
  Why not A: This equals the potential energy $U$ , not the total energy.
- B
  $E = \frac{G M _{E} m}{2 r}$
  Why not B: Positive sign error — bound orbits have negative total energy.
- C
  $E = - \frac{G M _{E} m}{2 r}$ Correct
- D
  $E = 0$
  Why not D: Only true at escape velocity; a bound orbit has $E < 0$ .
Explanation
Circular orbit: $v^{2} = G M_{E} / r$ , so $K E = \frac{1}{2} m v^{2} = \frac{G M _{E} m}{2 r}$ . Potential energy: $U = - \frac{G M _{E} m}{r}$ . Total: $E = K E + U = \frac{G M _{E} m}{2 r} - \frac{G M _{E} m}{r} = - \frac{G M _{E} m}{2 r}$ . Equivalently, $E = U /2 = - K E$ .

Key takeaway
For a circular orbit: $E = -GM_E m/(2r) = U/2 = -KE$. The virial theorem: total energy equals half the potential energy.
Question 8 · Medium
Using the gravitational field $g = - \frac{GM}{r ^{2}} \overset{r}{^}$ , find the work done by gravity on mass $m$ moving from $r_{1} = 2 R_{E}$ to $r_{2} = 4 R_{E}$ .
- A
  $W = \frac{G M _{E} m}{4 R _{E}}$
  Why not A: Computed only $U (r_{2})$ instead of $- Δ U = U (r_{1}) - U (r_{2})$ .
- B
  $W = - \frac{G M _{E} m}{4 R _{E}}$ Correct
- C
  $W = \frac{G M _{E} m}{2 R _{E}}$
  Why not C: Computed $U (r_{1})$ only: $- G M_{E} m / (2 R_{E})$ with wrong sign.
- D
  $W = 0$
  Why not D: Gravity is conservative but does non-zero work when $r$ changes.
Explanation
$W_{g r a v i t y} = - Δ U = U (r_{1}) - U (r_{2}) = (- \frac{G M _{E} m}{2 R _{E}}) - (- \frac{G M _{E} m}{4 R _{E}}) = - \frac{G M _{E} m}{2 R _{E}} + \frac{G M _{E} m}{4 R _{E}} = - \frac{G M _{E} m}{4 R _{E}}$ .

Negative work because moving away from Earth increases potential energy.

Key takeaway
$W_{gravity} = -\Delta U = U_i - U_f$. Moving outward (increasing $r$) means gravity does negative work.
Question 9 · Hard
Apply energy conservation to derive the orbital speed of a satellite in a circular orbit of radius $r$ around a planet of mass $M$ . What is $v$ ?
- A
  $v = \frac{2 π r}{T}$ only
  Why not A: Correct definition of orbital speed but not derived from energy conservation.
- B
  $v = \frac{GM}{2 r}$
  Why not B: Off by factor of $2$ ; this comes from equating $K E = E_{t o t a l}$ rather than force balance.
- C
  $v = \frac{GM}{r}$ Correct
- D
  $v = \frac{GM}{r ^{2}}$
  Why not D: Dimensional error; $GM / r^{2}$ has units m/s², not m/s.
Explanation
Centripetal acceleration equals gravitational field: $\frac{v ^{2}}{r} = \frac{GM}{r ^{2}} \Rightarrow v^{2} = \frac{GM}{r} \Rightarrow v = \frac{GM}{r}$ .

Equivalently: the total mechanical energy $E = - GM m / (2 r)$ and $K E = \frac{1}{2} m v^{2} = GM m / (2 r)$ , yielding the same result. The two methods agree because the virial theorem holds.

Key takeaway
Circular orbital speed $v = \sqrt{GM/r}$ follows from equating gravity to centripetal acceleration. Higher orbits are slower.
Question 10 · Hard
The gravitational potential inside a uniform solid sphere of mass $M$ and radius $R$ at distance $r < R$ from the center is $Φ (r) = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})$ . What is the gravitational field $g (r)$ at $r < R$ ?
- A
  $g = - \frac{GM}{R ^{3}} r$ (toward center)Correct
- B
  $g = - \frac{GM}{R ^{2}}$ (constant)
  Why not B: The surface value; inside, less mass contributes (shell theorem).
- C
  $g = - \frac{GM}{r ^{2}}$ (toward center)
  Why not C: Inverse-square law applies outside the sphere only.
- D
  $g = 0$
  Why not D: $g = 0$ only at the center; elsewhere inside, only the enclosed mass contributes.
Explanation
Gravitational field from potential: $g (r) = - \frac{d Φ}{d r} = - \frac{d}{d r} [- \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})] = - \frac{GM}{2 R ^{3}} (2 r) = - \frac{GM}{R ^{3}} r$ .

Alternative via Gauss's law for gravity: enclosed mass at radius $r < R$ is $M_{e n c} = M (r / R)^{3}$ . $g = - G M_{e n c} / r^{2} = - GM r / R^{3}$ . Same result.

Key takeaway
Inside a uniform sphere, $g(r) = -GMr/R^3$ — linear in $r$ (like a spring). At the surface, $g = -GM/R^2$; at the center, $g = 0$.
Question 11 · Hard
Derive Kepler's third law ( $T^{2} \propto r^{3}$ ) for a circular orbit from Newton's law of gravitation and the definition of circular motion.
- A
  $T^{2} = \frac{4 π ^{2} r ^{3}}{GM}$ Correct
- B
  $T^{2} = \frac{2 π r ^{3}}{GM}$
  Why not B: Missing the extra factor of $2 π$ ; comes from writing $v = 2 π r / T$ incorrectly.
- C
  $T^{2} = \frac{4 π ^{2} r ^{2}}{GM}$
  Why not C: Power of $r$ is 2 instead of 3 — forgot one factor of $r$ when substituting $v$ .
- D
  $T = \frac{2 π r}{GM / r}$
  Why not D: Correct intermediate step ( $T = 2 π r / v$ ) but not simplified to $T^{2} \propto r^{3}$ form.
Explanation
Circular motion: $v = 2 π r / T$ . Gravitational force provides centripetal acceleration:
$\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} = \frac{m ( 2 π r / T ) ^{2}}{r} = \frac{4 π ^{2} m r}{T ^{2}} .$
Solve for $T^{2}$ :
$T^{2} = \frac{4 π ^{2} r ^{3}}{GM} .$
This is Kepler's third law: $T^{2} \propto r^{3}$ (independent of mass $m$ ), with proportionality constant $4 π^{2} / (GM)$ .

Key takeaway
Kepler III follows from $F_g = F_c$: substitute $v = 2\pi r/T$, cancel $m$, and solve for $T^2$ to get $4\pi^2 r^3/(GM)$.
Question 12 · Hard
A satellite is transferred from a circular orbit of radius $r_{1}$ to a circular orbit of radius $r_{2} > r_{1}$ via a Hohmann transfer (half-ellipse). During the transfer, which statement about speed is correct at the outer transfer point?
- A
  The satellite must decelerate to enter the ellipse and decelerate again to circularize at $r_{2}$ .
  Why not A: Entering the ellipse from a circular orbit requires a prograde burn (speed increase), not deceleration.
- B
  The satellite accelerates at $r_{1}$ , coasts on the ellipse, and decelerates at $r_{2}$ to reach the circular orbit speed.
  Why not B: The satellite's ellipse speed at $r_{2}$ is less than the circular orbit speed there, so it must accelerate (prograde burn) to circularize.
- C
  The satellite accelerates at $r_{1}$ , coasts on the ellipse, then accelerates again at $r_{2}$ to match the circular orbit speed.Correct
- D
  The satellite's speed at $r_{2}$ on the ellipse equals the circular orbit speed at $r_{2}$ , so no second burn is needed.
  Why not D: At apoapsis of the transfer ellipse, speed is less than circular orbit speed ( $v_{a p o} < v_{c i r c} (r_{2})$ ); a prograde burn is required.
Explanation
Hohmann transfer: two prograde burns. First burn at $r_{1}$ increases speed from $v_{c i r c} (r_{1}) = GM / r_{1}$ to the perigee speed of the transfer ellipse $v_{p} = 2 GM r_{2} / [r_{1} (r_{1} + r_{2})]$ . The satellite coasts to apogee at $r_{2}$ where its speed is $v_{a} = 2 GM r_{1} / [r_{2} (r_{1} + r_{2})] < GM / r_{2} = v_{c i r c} (r_{2})$ . The second burn accelerates the satellite from $v_{a}$ up to $v_{c i r c} (r_{2})$ .

Key takeaway
Hohmann transfer uses two prograde burns: speed up at $r_1$ to enter the ellipse, then speed up again at $r_2$ to circularize. Moving to a higher orbit always requires adding energy.