AP Physics C Mechanics Quiz: Gravitational Force

20 questions · exam conditions

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Gravitational ForceQuestion 1 of 20

In a laboratory on Earth, a student measures the mass of a block to be $m_g$ using a spring scale and $m_i$ using an inertial balance. The experiment is then moved to a spaceship accelerating at $a=g$ in deep space. How will the new measurements, $m'_g$ and $m'_i$ , compare to the original measurements?

m'_g = m_g

and

m'_i = m_i

m'_g = 0

and

m'_i = m_i

m'_g = m_g

and

m'_i = 0

m'_g = 0

and

m'_i = 0

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Gravitational Force

Practice Gravitational Force in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Gravitational Force, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$m'_g = m_g$ and $m'_i = m_i$ (correct answer)
$m'_g = 0$ and $m'_i = m_i$
$m'_g = m_g$ and $m'_i = 0$
$m'_g = 0$ and $m'_i = 0$

Explanation: Mass is an intrinsic property of an object. The spring scale measures apparent weight, and in the accelerating spaceship, the apparent gravity is the same as on Earth, so it will measure the same gravitational mass. The inertial balance measures resistance to acceleration (inertial mass), which is also an intrinsic property. Therefore, both measurements will yield the same results.

Question 2

An object of mass $m$ is placed inside a uniform solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center, where $r < R$ . The magnitude of the gravitational force on the object is proportional to which of the following quantities?

$r$ (correct answer)
$r^2$
$1/r$
$1/r^2$

Explanation: The gravitational force inside a uniform solid sphere is given by the expression

F_g = G \frac{Mmr}{R^3}

. Since

G, M, m,

and

R

are constants for this situation, the force

F_g

is directly proportional to the distance

r

from the center.

Question 3

Planet X has mass $M$ and radius $R$ . Planet Y has mass $2M$ and radius $4R$ . An object of mass $m$ is placed on the surface of each planet. What is the ratio of the gravitational force on the object on Planet Y to the force on the object on Planet X, $F_Y / F_X$ ?

$1/8$ (correct answer)
$1/4$
$1/2$
$2$

Explanation: The gravitational force on the surface of a planet is given by

F = G \frac{M_{planet} m}{R_{planet}^2}

. For Planet X,

F_X = G \frac{M m}{R^2}

. For Planet Y,

F_Y = G \frac{(2M) m}{(4R)^2} = G \frac{2Mm}{16R^2} = \frac{1}{8} G \frac{Mm}{R^2}

. The ratio is

\frac{F_Y}{F_X} = \frac{\frac{1}{8} G \frac{Mm}{R^2}}{G \frac{Mm}{R^2}} = \frac{1}{8}

Question 4

A thin uniform rod has mass $M$ and length $L$ . A small object of mass $m$ is placed on the axis of the rod at a distance $d$ from the nearest end. Which of the following integrals correctly represents the magnitude of the gravitational force exerted by the rod on the object?

$G \frac{Mm}{L} \int_d^{d+L} \frac{1}{x^2} dx$ (correct answer)
$G \frac{Mm}{L} \int_d^{d+L} \frac{1}{x} dx$
$G m \int_d^{d+L} \frac{1}{x^2} dx$
$G \frac{Mm}{(d+L/2)^2}$

Explanation: To find the total force, we must integrate the forces from all infinitesimal mass elements of the rod. Let the object

m

be at the origin. The rod lies along an axis from

x=d

x=d+L

. The linear mass density is

\lambda = M/L

. A mass element is

dm = \lambda dx = (M/L)dx

. The force from this element is

dF = G \frac{m \, dm}{x^2} = G \frac{m(M/L)dx}{x^2}

. Integrating this from

x=d

x=d+L

gives the total force.

Question 5

A thin rod of length $L$ has a non-uniform linear mass density given by $\lambda(x) = \beta x$ , where $x$ is the distance from one end at $x=0$ and $\beta$ is a positive constant. Which expression gives the magnitude of the gravitational force on a point mass $m$ located at $x=-d$ ?

$G m \beta \int_0^L \frac{x}{(d+x)^2} dx$ (correct answer)
$G m \beta \int_0^L \frac{1}{(d+x)^2} dx$
$G m \beta \int_0^L \frac{x^2}{(d+x)^2} dx$
$G m \frac{\beta L^2/2}{(d+L/2)^2}$

Explanation: Consider an infinitesimal segment of the rod of length

dx

at position

x

on the rod (

0 \le x \le L

). Its mass is

dm = \lambda(x) dx = \beta x dx

. The distance between this segment and the point mass

m

x=-d

r = x - (-d) = x+d

. The gravitational force exerted by this segment on

m

dF = G \frac{m \, dm}{r^2} = G \frac{m (\beta x dx)}{(x+d)^2}

. To find the total force, this expression must be integrated over the entire length of the rod, from

x=0

x=L

Question 6

A planet of uniform density has mass $M$ and radius $R$ . A tunnel is drilled to a point a distance $r = R/3$ from the center. What is the magnitude of the gravitational force on a small object of mass $m$ at this location?

$G \frac{Mm}{3R^2}$ (correct answer)
$G \frac{Mm}{9R^2}$
$G \frac{Mm}{R^2}$
$G \frac{(M/27)m}{(R/3)^2}$

Explanation: Inside a uniform solid sphere, the gravitational force at a distance

r

from the center is due only to the mass enclosed within that radius,

M_{enclosed} = M(r/R)^3

. The force is

F_g = G \frac{M_{enclosed} m}{r^2} = G \frac{M(r^3/R^3) m}{r^2} = G \frac{Mmr}{R^3}

. Substituting

r = R/3

, the force is

F_g = G \frac{Mm(R/3)}{R^3} = G \frac{Mm}{3R^2}

Question 7

Three objects, each of mass $M$ , are located at the vertices of an equilateral triangle with side length $L$ . What is the magnitude of the net gravitational force on one of the masses due to the other two?

$\sqrt{3} G \frac{M^2}{L^2}$ (correct answer)
$2 G \frac{M^2}{L^2}$
$G \frac{M^2}{L^2}$
$\frac{\sqrt{3}}{2} G \frac{M^2}{L^2}$

Explanation: The force on one mass is the vector sum of the forces from the other two. Each individual force has a magnitude of

F = G \frac{M^2}{L^2}

. The angle between these two force vectors is 60 degrees. Using the law of cosines for vector addition, the resultant force magnitude is

F_{net} = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)} = \sqrt{2F^2 + 2F^2(1/2)} = \sqrt{3F^2} = \sqrt{3} F

. Thus, the net force is

\sqrt{3} G \frac{M^2}{L^2}

Question 8

A binary star system consists of two stars of masses $M$ and $2M$ separated by distance $d$ . At what distance from the star of mass $M$ is the gravitational field zero?

$\frac{d}{1+\sqrt{2}}$ (correct answer)
$\frac{d}{3}$
$\frac{d}{2}$
$\frac{2d}{3}$

Explanation: When dealing with gravitational fields from multiple sources, you're looking for the point where the net gravitational field equals zero. This occurs where the gravitational fields from both masses have equal magnitudes but opposite directions. The gravitational field from a point mass is

g = \frac{GM}{r^2}

, and it points toward the mass. For the net field to be zero, you need a point between the two stars where the fields cancel out. Let's call the distance from mass

M

to this point

x

, so the distance from mass

2M

(d-x)

. Setting the field magnitudes equal:

\frac{GM}{x^2} = \frac{G(2M)}{(d-x)^2}

Simplifying:

\frac{1}{x^2} = \frac{2}{(d-x)^2}

Cross-multiplying:

(d-x)^2 = 2x^2

Taking the square root:

d-x = x\sqrt{2}

Solving for

x

d = x + x\sqrt{2} = x(1+\sqrt{2})

Therefore:

x = \frac{d}{1+\sqrt{2}}

This confirms answer (A) is correct. Answer (B)

\frac{d}{3}

would result from incorrectly assuming the masses are equal. Answer (C)

\frac{d}{2}

is the midpoint, which ignores that the larger mass creates a stronger field. Answer (D)

\frac{2d}{3}

places the zero point closer to the larger mass, which is backwards—the zero point should be closer to the smaller mass. Strategy tip: Remember that the gravitational field zero point is always closer to the smaller mass in a two-body system. Set up your equation by equating field magnitudes, not forces.

Question 9

Two spherical objects, one with mass $M$ and the other with mass $4M$ , are separated by a center-to-center distance $d$ . What is the magnitude of the gravitational force exerted by the larger object on the smaller object?

$G \frac{4M^2}{d^2}$ (correct answer)
$G \frac{M^2}{d^2}$
$G \frac{5M^2}{d^2}$
$G \frac{4M^2}{d}$

Explanation: Newton's Law of Universal Gravitation is

F_g = G \frac{m_1 m_2}{r^2}

. Here,

m_1=M

m_2=4M

, and

r=d

. The force is

F_g = G \frac{(M)(4M)}{d^2} = G \frac{4M^2}{d^2}

. By Newton's third law, the force exerted by the larger object on the smaller one is equal in magnitude to the force exerted by the smaller object on the larger one.

Question 10

A planet has a mass of $6.0 \times 10^{24}$ kg and a radius of $6.4 \times 10^6$ m. What is the magnitude of the gravitational field strength at an altitude of $3.2 \times 10^6$ m above the planet's surface? The universal gravitational constant is $G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$ .

$4.3 \, \text{N/kg}$ (correct answer)
$9.8 \, \text{N/kg}$
$39.1 \, \text{N/kg}$
$2.4 \, \text{N/kg}$

Explanation: The gravitational field strength is

g = G \frac{M}{r^2}

. The distance

r

from the center of the planet is the sum of the planet's radius and the altitude:

r = 6.4 \times 10^6 \, \text{m} + 3.2 \times 10^6 \, \text{m} = 9.6 \times 10^6 \, \text{m}

. Substituting the values,

g = (6.67 \times 10^{-11}) \frac{6.0 \times 10^{24}}{(9.6 \times 10^6)^2} \approx 4.3

N/kg.

Question 11

An astronaut is in a spacecraft that is accelerating away from a planet in deep space, far from any significant gravitational fields. The acceleration of the spacecraft is constant and has a magnitude of $9.8 \, \text{m/s}^2$ . Which of the following best describes the astronaut's experience inside the spacecraft?

The astronaut feels a sensation of weight similar to standing on Earth's surface. (correct answer)
The astronaut feels weightless because there is no significant gravity.
The astronaut feels a force pushing them toward the front of the spacecraft.
The astronaut's apparent weight is zero, but their mass is unchanged.

Explanation: According to the principle of equivalence, the effects of a uniform gravitational field are indistinguishable from the effects of a constant, uniform acceleration. An upward acceleration of

9.8 \, \text{m/s}^2

creates an apparent gravitational field, causing the floor of the spacecraft to push on the astronaut with a force equal to their weight on Earth. This is felt as normal weight.

Question 12

A uniform spherical shell has mass $M$ and radius $R$ . What is the magnitude of the gravitational force exerted by the shell on a point mass $m$ located a distance $3R$ from the center of the shell?

$G \frac{Mm}{9R^2}$ (correct answer)
$G \frac{Mm}{3R^2}$
$G \frac{Mm}{R^2}$
$0$

Explanation: According to Newton's shell theorem, for any point outside a uniform spherical shell, the gravitational force exerted by the shell is the same as if all the shell's mass were concentrated at its center. The distance from the center is

r=3R

. Using the law of universal gravitation,

F = G \frac{Mm}{(3R)^2} = G \frac{Mm}{9R^2}

Question 13

A point mass $m$ is located inside a uniform spherical shell of mass $M$ and radius $R$ . The distance of the point mass from the center of the shell is $R/2$ . What is the magnitude of the net gravitational force exerted by the shell on the point mass?

$0$ (correct answer)
$G \frac{Mm}{(R/2)^2}$
$G \frac{Mm}{R^2}$
$G \frac{(M/2)m}{(R/2)^2}$

Explanation: A fundamental result of Newton's shell theorem is that the net gravitational force exerted by a uniform spherical shell on any point mass located inside the shell is zero. The gravitational forces from all parts of the shell cancel each other out at any interior point.

Question 14

The magnitude of the gravitational force exerted by Earth on a satellite is $F_0$ when the satellite is at an altitude equal to Earth's radius, $R_E$ . When is it appropriate to approximate the gravitational force on the satellite using $F_g = mg$ , where $g$ is the acceleration due to gravity at Earth's surface?

When the satellite is in a low Earth orbit with an altitude much less than $R_E$ . (correct answer)
It is always appropriate because the mass of the satellite does not change.
When the satellite is in a geostationary orbit where its period matches Earth's rotation.
It is never appropriate because the satellite is not on the surface of Earth.

Explanation: The approximation

F_g = mg

, where

g

is the surface gravity value, is valid only when the distance from Earth's center is approximately equal to Earth's radius,

R_E

. This condition is met for objects on or very near the surface. For a low Earth orbit, the altitude

h

is much smaller than

R_E

, so

r = R_E + h \approx R_E

, and the approximation is reasonable.

Question 15

Planet A has radius $R$ and density $\rho$ . Planet B has radius $2R$ and density $\rho/2$ . Let $g_A$ and $g_B$ be the gravitational accelerations at the surfaces of Planet A and Planet B, respectively. What is the ratio $g_B / g_A$ ?

$1$ (correct answer)
$1/2$
$2$
$4$

Explanation: The gravitational acceleration at the surface of a planet is proportional to the product of its density and radius,

g \propto \rho R

. For Planet A,

g_A \propto \rho R

. For Planet B,

g_B \propto (\rho/2)(2R) = \rho R

. Since both expressions are proportional to the same value, the surface gravities are equal, and the ratio

g_B / g_A

is 1.

Question 16

A space probe travels from Earth's surface (radius $R$ ) to a point where the gravitational field strength is $\frac{1}{16}$ of its value at Earth's surface. At what distance from Earth's center is the probe located?

$4R$ (correct answer)
$2R$
$8R$
$16R$

Explanation: This question tests your understanding of how gravitational field strength varies with distance from a massive object like Earth. When you encounter gravitational field problems, always remember that the field strength follows an inverse square law with distance. The gravitational field strength at any distance

r

from Earth's center is given by

g = \frac{GM}{r^2}

, where

G

is the gravitational constant and

M

is Earth's mass. At Earth's surface (distance

R

), the field strength is

g_0 = \frac{GM}{R^2}

. Since the probe experiences a field strength that is

\frac{1}{16}

of the surface value, we can write:

\frac{GM}{r^2} = \frac{1}{16} \cdot \frac{GM}{R^2}

Simplifying by canceling

GM

\frac{1}{r^2} = \frac{1}{16R^2}

Cross-multiplying gives us

r^2 = 16R^2

, so

r = 4R

. Looking at the wrong answers: Choice B (

2R

) would give a field strength of

\frac{1}{4}

the surface value, not

\frac{1}{16}

. Choice C (

8R

) would result in

\frac{1}{64}

the surface strength. Choice D (

16R

) would give

\frac{1}{256}

the surface strength. These all stem from incorrectly applying the inverse square relationship. Remember that gravitational field strength decreases as the square of the distance increases. If you want the field to be

\frac{1}{n^2}

times weaker, you need to be

n

times farther away. Since

\frac{1}{16} = \frac{1}{4^2}

, the distance must be

4R

Question 17

Two identical spherical masses are initially separated by distance $r$ . If one sphere is moved to a new position where the gravitational force between them is reduced to $\frac{1}{9}$ of its original value, and the masses remain unchanged, what is the ratio of the final separation distance to the initial separation distance?

$\frac{1}{3}$
$3$ (correct answer)
$\frac{1}{9}$
$9$

Explanation: Using Newton's law of universal gravitation,

F = \frac{Gm_1m_2}{r^2}

. When the force is reduced to

\frac{1}{9}

of its original value:

\frac{F_{final}}{F_{initial}} = \frac{1}{9} = \frac{r_{initial}^2}{r_{final}^2}

. Solving for the ratio:

\frac{r_{final}^2}{r_{initial}^2} = 9

, so

\frac{r_{final}}{r_{initial}} = 3

. Choice A confuses the inverse relationship. Choice C uses the force ratio directly. Choice D squares the correct answer.

Question 18

A satellite orbits Earth at a distance of $2R$ from Earth's center, where $R$ is Earth's radius. If the satellite's orbital radius is increased to $4R$ , by what factor does the gravitational force on the satellite change?

Decreases by a factor of $2$
Decreases by a factor of $4$ (correct answer)
Decreases by a factor of $8$
Increases by a factor of $2$

Explanation: The gravitational force follows

F = \frac{GMm}{r^2}

. Initially,

F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}

. Finally,

F_2 = \frac{GMm}{(4R)^2} = \frac{GMm}{16R^2}

. The ratio is

\frac{F_2}{F_1} = \frac{4R^2}{16R^2} = \frac{1}{4}

, so the force decreases by a factor of 4. Choice A uses only the linear change in radius. Choice C incorrectly cubes the radius change. Choice D incorrectly suggests an increase.

Question 19

A uniform spherical planet has mass $M$ and radius $R$ . A small mass $m$ is placed at a distance $\frac{R}{2}$ below the planet's surface (at distance $\frac{R}{2}$ from the planet's center). Assuming the planet has uniform density, what is the gravitational force on mass $m$ ?

$\frac{GMm}{4R^2}$
$\frac{GMm}{R^2}$
$\frac{GMm}{2R^2}$ (correct answer)
$\frac{2GMm}{R^2}$

Explanation: For a uniform sphere, only the mass within radius

r

from the center contributes to the gravitational force at that point. The mass within radius

\frac{R}{2}

M_{inside} = M \cdot \frac{(\frac{R}{2})^3}{R^3} = \frac{M}{8}

. The gravitational force is

F = \frac{G(\frac{M}{8})m}{(\frac{R}{2})^2} = \frac{GMm}{8} \cdot \frac{4}{R^2} = \frac{GMm}{2R^2}

. Choice A uses only the volume ratio. Choice B ignores the internal structure. Choice D incorrectly doubles the surface value.

Question 20

A hollow spherical shell of mass $M$ and inner radius $a$ and outer radius $b$ has uniform density. A point mass $m$ is located at the geometric center of the shell. What is the gravitational force on the point mass?

$\frac{GMm}{a^2}$
$\frac{GMm}{b^2}$
$\frac{GMm}{(\frac{a+b}{2})^2}$
$0$ (correct answer)

Explanation: According to the shell theorem, a spherically symmetric mass distribution exerts zero net gravitational force on a point mass located at its center. This is because every element of mass in the shell has a corresponding element on the opposite side, and these paired elements exert equal and opposite forces on the central point mass, resulting in zero net force. Choices A, B, and C incorrectly apply gravitational force formulas using different characteristic distances of the shell.