College Physics Quiz: Gravitational Force

20 questions · exam conditions

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Gravitational ForceQuestion 1 of 20

A satellite orbits Earth at an altitude equal to Earth's radius. Compared to the gravitational acceleration at Earth's surface, the gravitational acceleration at the satellite's orbital position is:

\frac{1}{4}

of the surface value

\frac{1}{2}

of the surface value

\frac{1}{\sqrt{2}}

of the surface value

\frac{1}{8}

of the surface value

\frac{1}{3}

of the surface value

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College Physics Quiz

College Physics Quiz: Gravitational Force

Practice Gravitational Force in College Physics 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 College Physics.

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

A satellite orbits Earth at an altitude equal to Earth's radius. Compared to the gravitational acceleration at Earth's surface, the gravitational acceleration at the satellite's orbital position is:

$\frac{1}{4}$ of the surface value (correct answer)
$\frac{1}{2}$ of the surface value
$\frac{1}{\sqrt{2}}$ of the surface value
$\frac{1}{8}$ of the surface value
$\frac{1}{3}$ of the surface value

Explanation: This question tests your understanding of how gravitational acceleration varies with distance from Earth's center. When dealing with satellites at different altitudes, you need to apply Newton's law of universal gravitation and recognize how the inverse square relationship affects gravitational field strength. Gravitational acceleration follows the formula

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

, where r is the distance from Earth's center. At Earth's surface,

r = R

(Earth's radius), so

g_{surface} = \frac{GM}{R^2}

. The satellite orbits at altitude

R

, meaning it's at distance

r = R + R = 2R

from Earth's center. At the satellite's position:

g_{satellite} = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{1}{4} \cdot \frac{GM}{R^2} = \frac{1}{4}g_{surface}

Therefore, answer A is correct. Answer B (

\frac{1}{2}

) incorrectly assumes gravitational acceleration decreases linearly with distance rather than following the inverse square law. Answer C (

\frac{1}{\sqrt{2}}

) might result from mistakenly taking the square root of the distance factor instead of squaring it. Answer D (

\frac{1}{8}

) could come from incorrectly cubing the distance ratio or making multiple calculation errors. Study tip: Always identify what distance you're measuring from Earth's center, not just the altitude. The key pattern is: if distance from center increases by factor n, gravitational acceleration decreases by factor

n^2

. Draw a diagram showing Earth's center, surface, and satellite position to visualize the geometry clearly.

Question 2

On planet X, an object weighs $60\text{ N}$ . On planet Y, the same object weighs $90\text{ N}$ . If planet X has twice the radius of planet Y, what is the ratio of planet X's mass to planet Y's mass?

$\frac{8}{3}$ (correct answer)
$\frac{3}{2}$
$\frac{2}{3}$
$\frac{4}{3}$
$\frac{3}{8}$

Explanation: When you encounter planetary weight problems, you're dealing with gravitational force relationships. The key insight is that weight equals mass times gravitational acceleration, and gravity depends on both planetary mass and radius. The gravitational acceleration on any planet is

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

, where G is the gravitational constant, M is the planet's mass, and R is its radius. Since weight

W = mg

, we can write: For planet X:

60 = m \cdot \frac{GM_X}{R_X^2}

For planet Y:

90 = m \cdot \frac{GM_Y}{R_Y^2}

Taking the ratio of these equations eliminates the object's mass:

\frac{60}{90} = \frac{GM_X/R_X^2}{GM_Y/R_Y^2} = \frac{M_X}{M_Y} \cdot \frac{R_Y^2}{R_X^2}

Since

R_X = 2R_Y

, we have:

\frac{2}{3} = \frac{M_X}{M_Y} \cdot \frac{R_Y^2}{(2R_Y)^2} = \frac{M_X}{M_Y} \cdot \frac{1}{4}

Solving for the mass ratio:

\frac{M_X}{M_Y} = \frac{2}{3} \times 4 = \frac{8}{3}

Answer A is correct. Answer B (

\frac{3}{2}

) ignores the radius relationship entirely. Answer C (

\frac{2}{3}

) represents the weight ratio but forgets that gravity also depends on radius squared. Answer D (

\frac{4}{3}

) incorrectly applies the radius factor as linear rather than squared. Remember: gravitational problems always involve the inverse square relationship with distance. When comparing planetary gravities, both mass and radius matter, and radius effects are squared.

Question 3

The gravitational field strength at the surface of a planet is $12\text{ m/s}^2$ . At what altitude above the surface will the gravitational field strength be reduced to $3\text{ m/s}^2$ ? Express your answer in terms of the planet's radius $R$ .

$R$ (correct answer)
$2R$
$3R$
$0.5R$
$4R$

Explanation: When you encounter gravitational field problems involving altitude changes, remember that gravitational field strength follows an inverse square relationship with distance from the planet's center. The gravitational field strength is given by

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

, where

r

is the distance from the planet's center. At the surface,

r = R

and

g = 12 \text{ m/s}^2

. At altitude

h

, the distance becomes

r = R + h

and

g = 3 \text{ m/s}^2

. Setting up the ratio:

\frac{g_{\text{altitude}}}{g_{\text{surface}}} = \frac{R^2}{(R+h)^2}

Substituting values:

\frac{3}{12} = \frac{R^2}{(R+h)^2}

This simplifies to:

\frac{1}{4} = \frac{R^2}{(R+h)^2}

Taking the square root:

\frac{1}{2} = \frac{R}{R+h}

Cross-multiplying:

R + h = 2R

, so

h = R

Choice A (

R

) is correct. Choice B (

2R

) represents a common error where students confuse the altitude with the total distance from center. At altitude

2R

, you'd be at distance

3R

from center, giving field strength

\frac{12}{9} = 1.33 \text{ m/s}^2

. Choice C (

3R

) would put you at distance

4R

from center, yielding field strength

\frac{12}{16} = 0.75 \text{ m/s}^2

. Choice D (

0.5R

) gives distance

1.5R

from center, producing field strength

\frac{12}{2.25} = 5.33 \text{ m/s}^2

. Strategy tip: Always distinguish between altitude above surface and total distance from planet center. Set up ratios using the inverse square law to avoid calculation errors.

Question 4

A space probe is equidistant from Earth and the Moon. The probe experiences gravitational forces from both bodies. Given that Earth's mass is approximately $81$ times the Moon's mass, and the Earth-Moon distance is $d$ , at what distance from Earth's center should the probe be located so that the net gravitational force on it is zero?

$\frac{9d}{10}$ (correct answer)
$\frac{d}{2}$
$\frac{3d}{4}$
$\frac{4d}{5}$
$\frac{2d}{3}$

Explanation: This problem tests your understanding of gravitational equilibrium points, where competing gravitational forces cancel out. When you see a question about finding where gravitational forces balance, you need to set up an equation where the forces from different masses are equal and opposite. Let's say the probe is at distance

x

from Earth's center. Since the Earth-Moon distance is

d

, the probe is at distance

(d-x)

from the Moon's center. For zero net force, the gravitational pulls must be equal:

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

Since

M_E = 81M_M

, this becomes:

\frac{81M_M}{x^2} = \frac{M_M}{(d-x)^2}

Simplifying:

\frac{81}{x^2} = \frac{1}{(d-x)^2}

Taking the square root:

\frac{9}{x} = \frac{1}{d-x}

Cross-multiplying:

9(d-x) = x

, so

9d - 9x = x

, giving us

9d = 10x

Therefore:

x = \frac{9d}{10}

Looking at the wrong answers: B)

\frac{d}{2}

assumes equal masses, ignoring that Earth is much more massive. C)

\frac{3d}{4}

might result from taking the square root incorrectly or confusing the mass ratio. D)

\frac{4d}{5}

could come from using the mass ratio directly without proper force calculations. Study tip: In gravitational equilibrium problems, always remember that the forces depend on

\frac{1}{r^2}

, not just

\frac{1}{r}

. The equilibrium point is closer to the less massive object, but you must account for the inverse-square relationship when setting up your force balance equation.

Question 5

A pendulum clock that keeps perfect time on Earth is taken to a planet where the gravitational acceleration is $\frac{g}{4}$ . How does the period of the pendulum change?

The period doubles (correct answer)
The period is halved
The period increases by a factor of $4$
The period increases by a factor of $\sqrt{2}$
The period decreases by a factor of $\sqrt{2}$

Explanation: When you encounter pendulum problems involving different gravitational environments, you need to recall that the period of a simple pendulum depends on both length and gravitational acceleration through the formula

T = 2\pi\sqrt{\frac{L}{g}}

. Since the pendulum clock is simply moved to a different planet, its physical length

L

remains unchanged. The only variable that changes is the gravitational acceleration, which decreases from

g

\frac{g}{4}

. Let's compare the periods. On Earth:

T_{Earth} = 2\pi\sqrt{\frac{L}{g}}

. On the new planet:

T_{planet} = 2\pi\sqrt{\frac{L}{g/4}} = 2\pi\sqrt{\frac{4L}{g}} = 2\pi \cdot 2\sqrt{\frac{L}{g}} = 2T_{Earth}

The period doubles, confirming answer A is correct. Now for the wrong answers: B suggests the period halves, which would occur if gravity increased by a factor of 4, not decreased. C claims the period increases by a factor of 4 – this misconception likely comes from directly using the gravity ratio without considering the square root relationship. D suggests an increase by

\sqrt{2}

, which doesn't match any reasonable manipulation of our gravity change. Remember this key pattern: in the pendulum period formula, gravitational acceleration appears in the denominator under a square root. When gravity decreases by a factor of 4, you take the square root of the inverse ratio (

\sqrt{4} = 2

) to find how the period increases. This inverse square root relationship is crucial for pendulum problems across different gravitational environments.

Question 6

A binary star system consists of two stars of equal mass $M$ separated by distance $2d$ . A small test mass $m$ is placed at the midpoint between the stars. If the test mass is displaced slightly toward one star along the line connecting the stars, what happens?

The test mass experiences a net force toward the star it was displaced toward, making the equilibrium unstable (correct answer)
The test mass experiences a net force back toward the midpoint, making the equilibrium stable
The test mass experiences zero net force at any point along the line, making all positions equivalent
The test mass oscillates with simple harmonic motion about the midpoint with a predictable frequency
The gravitational forces remain balanced regardless of the displacement direction along the line

Explanation: When analyzing equilibrium in gravitational systems, you need to examine what happens to forces when the system is slightly perturbed. This reveals whether the equilibrium is stable (restoring forces) or unstable (amplifying forces). At the midpoint between two equal masses, the gravitational forces from each star exactly cancel, creating equilibrium. However, when you displace the test mass slightly toward one star by distance

x

, the force balance changes dramatically. The displaced test mass is now closer to one star (distance

d-x

) and farther from the other (distance

d+x

). Since gravitational force follows

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

, the force from the nearer star becomes

\frac{GMm}{(d-x)^2}

while the force from the farther star becomes

\frac{GMm}{(d+x)^2}

. The inverse-square relationship means the stronger force from the nearer star dominates, creating a net force that pulls the test mass further toward that star, away from the original equilibrium position. Option B incorrectly suggests stable equilibrium with restoring forces back toward the midpoint. Option C misunderstands that gravitational forces depend on distance—positions are definitely not equivalent. Option D assumes the restoring force needed for harmonic motion exists, but we've shown the opposite occurs. The correct answer is A: the test mass experiences amplifying forces that make the equilibrium unstable. Study tip: For gravitational equilibrium problems, always check the stability by examining how forces change with small displacements. Unstable equilibria are common in gravitational systems, unlike spring systems where equilibria are typically stable.

Question 7

The gravitational potential energy between two masses $m_1$ and $m_2$ separated by distance $r$ is $U = -\frac{Gm_1m_2}{r}$ . If the distance between the masses is tripled, by what factor does the gravitational force between them change?

Decreases by a factor of $9$ (correct answer)
Decreases by a factor of $3$
Decreases by a factor of $6$
Increases by a factor of $3$
Decreases by a factor of $27$

Explanation: When you encounter gravitational problems, remember that force and potential energy are related but follow different distance dependencies. The key is knowing how to derive one from the other. To find how gravitational force changes with distance, you need to use the relationship between force and potential energy:

F = -\frac{dU}{dr}

. Taking the derivative of the given potential energy

U = -\frac{Gm_1m_2}{r}

F = -\frac{d}{dr}\left(-\frac{Gm_1m_2}{r}\right) = -Gm_1m_2 \frac{d}{dr}(r^{-1}) = -Gm_1m_2(-r^{-2}) = \frac{Gm_1m_2}{r^2}

This gives you Newton's law of universal gravitation, showing that gravitational force is inversely proportional to

r^2

. When distance triples (

r \rightarrow 3r

), the new force becomes

\frac{Gm_1m_2}{(3r)^2} = \frac{Gm_1m_2}{9r^2}

, which is the original force divided by 9. Choice A is correct—the force decreases by a factor of 9. Choice B incorrectly applies the inverse relationship from potential energy (

r^{-1}

) directly to force, missing the derivative step. Choice C might come from confusing this with some other physics relationship or calculation error. Choice D suggests the force increases, which violates the basic principle that gravitational force weakens with distance. Study tip: Always remember the "inverse square law" pattern in physics—gravitational force, electric force, and light intensity all follow

1/r^2

relationships. When distance increases by factor

n

, these quantities decrease by factor

n^2

Question 8

A planet has the same density as Earth but twice Earth's radius. An object that weighs $100\text{ N}$ on Earth will weigh how much on this planet?

$200\text{ N}$ (correct answer)
$400\text{ N}$
$50\text{ N}$
$800\text{ N}$
$100\text{ N}$

Explanation: When you encounter problems comparing gravitational effects between planets, focus on how mass and radius changes affect surface gravity through Newton's law of universal gravitation. Surface gravity depends on a planet's mass divided by the square of its radius:

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

. Since weight equals mass times gravity (

W = mg

), you need to find how this planet's surface gravity compares to Earth's. First, determine the planet's mass. Since density equals mass divided by volume, and the planet has Earth's density but twice the radius:

\rho_{planet} = \rho_{Earth}

. The volume scales as

R^3

, so with radius doubled, volume increases by

2^3 = 8

. Therefore, the planet's mass is

8M_{Earth}

. Now calculate the gravity ratio:

\frac{g_{planet}}{g_{Earth}} = \frac{GM_{planet}/R_{planet}^2}{GM_{Earth}/R_{Earth}^2} = \frac{8M_{Earth}}{(2R_{Earth})^2} \times \frac{R_{Earth}^2}{M_{Earth}} = \frac{8}{4} = 2

The object weighs

100 \times 2 = 200\text{ N}

, making (A) correct. (B) 400 N incorrectly assumes gravity scales directly with mass increase (factor of 8) without accounting for radius effects. (C) 50 N mistakenly thinks doubling radius halves the weight, ignoring the mass change entirely. (D) 800 N wrongly uses only the mass factor of 8 while completely ignoring the radius increase. Study tip: For planetary comparison problems, always work through both mass and radius effects systematically. Mass changes affect gravity proportionally, while radius changes affect it by the inverse square.

Question 9

Two spherical masses $m_1 = 8.0\text{ kg}$ and $m_2 = 2.0\text{ kg}$ are initially at rest and separated by $1.0\text{ m}$ . If they are released and allowed to move under their mutual gravitational attraction only, what is the ratio of distances they each travel when they meet?

$m_1$ travels $\frac{1}{4}$ the distance that $m_2$ travels (correct answer)
$m_1$ travels $\frac{1}{2}$ the distance that $m_2$ travels
$m_1$ travels the same distance as $m_2$
$m_1$ travels $2$ times the distance that $m_2$ travels
$m_1$ travels $4$ times the distance that $m_2$ travels

Explanation: When you encounter problems involving two objects attracting each other gravitationally, the key principle is conservation of momentum. Since the system starts at rest with zero total momentum, the momentum must remain zero throughout the motion. Let's work through this systematically. If

m_1

travels distance

d_1

and

m_2

travels distance

d_2

, then at any moment their momenta must be equal in magnitude but opposite in direction:

m_1 v_1 = m_2 v_2

. Since both objects start from rest and experience the same time interval, their velocities are proportional to their distances traveled:

v_1 = d_1/t

and

v_2 = d_2/t

. Substituting:

m_1 \frac{d_1}{t} = m_2 \frac{d_2}{t}

, which simplifies to

m_1 d_1 = m_2 d_2

. Therefore:

\frac{d_1}{d_2} = \frac{m_2}{m_1} = \frac{2.0}{8.0} = \frac{1}{4}

. This means

m_1

travels

\frac{1}{4}

the distance that

m_2

travels, confirming answer A. Answer B (

\frac{1}{2}

) incorrectly uses the square root of the mass ratio. Answer C (same distance) ignores the mass difference entirely—a common misconception that heavier objects somehow "fall faster" in this context. Answer D (2 times) inverts the correct ratio, perhaps confusing which mass should be in the numerator. Remember: in gravitational problems with two bodies, the lighter object always moves farther because momentum conservation requires

m_1 d_1 = m_2 d_2

. The inverse relationship between mass and distance traveled is crucial.

Question 10

A geosynchronous satellite orbits Earth at a radius where the orbital period equals Earth's rotation period (24 hours). If a satellite orbits at half this radius, what will be its orbital period?

$6.0\text{ hours}$
$8.5\text{ hours}$ (correct answer)
$12\text{ hours}$
$17\text{ hours}$
$3.0\text{ hours}$

Explanation: This question tests your understanding of Kepler's Third Law, which relates orbital period to orbital radius. When you see problems about satellite orbits at different distances, immediately think about how period scales with radius. Kepler's Third Law states that the square of the orbital period is proportional to the cube of the orbital radius:

T^2 \propto r^3

. For two satellites, this gives us

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}

. Let's call the geosynchronous period

T_1 = 24

hours at radius

r_1

. The new satellite orbits at

r_2 = \frac{r_1}{2}

. Substituting into Kepler's Third Law:

\frac{T_2^2}{24^2} = \frac{(\frac{r_1}{2})^3}{r_1^3} = \frac{1}{8}

Therefore:

T_2^2 = \frac{24^2}{8} = 72

, so

T_2 = \sqrt{72} = 8.5

hours. Looking at the wrong answers: Choice A (6.0 hours) likely comes from incorrectly assuming period is inversely proportional to radius (

T \propto \frac{1}{r}

), giving

24 \times \frac{1}{2} = 12

, then making an arithmetic error. Choice C (12 hours) results from the common misconception that period is directly proportional to radius. Choice D (17 hours) might come from incorrectly applying

T \propto \sqrt{r}

, giving

24 \times \frac{1}{\sqrt{2}} \approx 17

. The answer is B (8.5 hours). Remember: For orbital mechanics problems, always start with Kepler's Third Law (

T^2 \propto r^3

). The period decreases dramatically as you move closer to the central body—cutting the radius in half reduces the period by a factor of

2^{3/2} = 2.83

Question 11

An object is placed at different distances from a massive sphere. At distance $r$ from the center, the gravitational potential energy is $U_1$ . At distance $2r$ , the potential energy is $U_2$ . What is the relationship between $U_1$ and $U_2$ ?

$U_1 = 2U_2$ (correct answer)
$U_1 = 4U_2$
$U_1 = \frac{U_2}{2}$
$U_1 = \frac{U_2}{4}$
$U_1 = 8U_2$

Explanation: When you encounter gravitational potential energy problems involving different distances from a massive object, focus on the inverse relationship between potential energy and distance. Gravitational potential energy is given by

U = -\frac{GMm}{r}

, where

G

is the gravitational constant,

M

is the mass of the sphere,

m

is the mass of the object, and

r

is the distance from the center. The key insight is that potential energy is inversely proportional to distance. At distance

r

U_1 = -\frac{GMm}{r}

At distance

2r

U_2 = -\frac{GMm}{2r}

To find the relationship, divide

U_1

U_2

\frac{U_1}{U_2} = \frac{-\frac{GMm}{r}}{-\frac{GMm}{2r}} = \frac{-\frac{GMm}{r}} \times \frac{2r}{-GMm} = 2

Therefore,

U_1 = 2U_2

, confirming answer A. Looking at the wrong answers: B)

U_1 = 4U_2

incorrectly applies an inverse-square relationship, which applies to gravitational force, not potential energy. C)

U_1 = \frac{U_2}{2}

gets the relationship backwards—this would mean the object has less potential energy when closer to the massive sphere. D)

U_1 = \frac{U_2}{4}

combines both errors: wrong direction and wrong power relationship. Remember: gravitational potential energy has an inverse relationship with distance (

\propto \frac{1}{r}

), while gravitational force has an inverse-square relationship (

\propto \frac{1}{r^2}

). Don't confuse these two fundamental relationships.

Question 12

A uniform solid sphere of mass $M$ and radius $R$ has a small cavity of radius $\frac{R}{4}$ carved out from its center. What is the gravitational field at a point on the surface of the original sphere, directly above the center of the cavity?

$\frac{63GM}{64R^2}$ (correct answer)
$\frac{GM}{R^2}$
$\frac{15GM}{16R^2}$
$\frac{7GM}{8R^2}$
$\frac{31GM}{32R^2}$

Explanation: When you encounter a gravitational field problem involving a cavity, use the superposition principle: treat the cavity as the original solid sphere minus a smaller sphere of material that was removed. The original uniform sphere would create a gravitational field of

\frac{GM}{R^2}

at its surface. However, we need to account for the missing mass in the cavity. The cavity has radius

\frac{R}{4}

, so its volume is

\frac{1}{64}

of the original sphere's volume, meaning it contained

\frac{M}{64}

of mass. The distance from the cavity's center to the point on the surface is

R

(the full radius). The "missing mass"

\frac{M}{64}

would have contributed a field of

\frac{G(M/64)}{R^2} = \frac{GM}{64R^2}

at this point. Using superposition, the actual field equals the original field minus the contribution from the removed mass:

\frac{GM}{R^2} - \frac{GM}{64R^2} = \frac{64GM - GM}{64R^2} = \frac{63GM}{64R^2}

Answer A is correct. Answer B (

\frac{GM}{R^2}

) ignores the cavity entirely. Answer C (

\frac{15GM}{16R^2}

) likely results from incorrectly calculating the cavity's mass fraction as

\frac{1}{16}

instead of

\frac{1}{64}

. Answer D (

\frac{7GM}{8R^2}

) suggests using

\frac{1}{8}

as the mass fraction, confusing linear and volume ratios. Study tip: For cavity problems, always remember that volume scales as the cube of linear dimensions, so a cavity with radius

\frac{R}{4}

has volume fraction

(\frac{1}{4})^3 = \frac{1}{64}

Question 13

Consider a uniform ring of mass $M$ and radius $R$ . A point mass $m$ is located on the axis of the ring at distance $h$ from the center. What is the magnitude of the gravitational force on the point mass?

$\frac{GMmh}{(h^2 + R^2)^{3/2}}$ (correct answer)
$\frac{GMm}{h^2 + R^2}$
$\frac{GMm}{(h^2 + R^2)^2}$
$\frac{GMmR}{(h^2 + R^2)^{3/2}}$
$\frac{GMm}{h^2}$

Explanation: When you encounter gravitational force problems involving extended objects like rings or disks, the key is using symmetry to simplify the vector addition of forces from all mass elements. To find the gravitational force, consider how each small mass element

dm

on the ring exerts a force on the point mass. Every element is at the same distance

\sqrt{h^2 + R^2}

from the point mass (using the Pythagorean theorem). The force from each element has magnitude

\frac{G \cdot dm \cdot m}{h^2 + R^2}

. Here's the crucial insight: due to the ring's symmetry, all horizontal force components cancel out. Only the vertical components (along the axis) survive. Each force makes an angle with the axis where

\cos\theta = \frac{h}{\sqrt{h^2 + R^2}}

. The net force is therefore:

F = \int \frac{G \cdot dm \cdot m}{h^2 + R^2} \cdot \frac{h}{\sqrt{h^2 + R^2}} = \frac{GMmh}{(h^2 + R^2)^{3/2}}

This confirms answer A is correct. Answer B,

\frac{GMm}{h^2 + R^2}

, ignores the angular factor and assumes all force acts along the axis. Answer C,

\frac{GMm}{(h^2 + R^2)^2}

, incorrectly squares the distance term. Answer D,

\frac{GMmR}{(h^2 + R^2)^{3/2}}

, mistakenly uses

R

instead of

h

in the numerator, confusing the geometry. Remember: in symmetric mass distribution problems, always identify which force components cancel due to symmetry, then apply the appropriate geometric factor to the surviving components.

Question 14

A uniform solid sphere and a uniform thin spherical shell, both with the same mass $M$ and radius $R$ , are placed at the same distance from a point mass $m$ . The center-to-center distance is $3R$ for both configurations. Which statement is correct about the gravitational forces?

Both objects exert identical gravitational forces on the point mass (correct answer)
The solid sphere exerts a stronger force because its mass is more concentrated
The spherical shell exerts a stronger force because its mass is distributed at the surface
The forces differ by a factor related to the moment of inertia difference
The solid sphere exerts a weaker force due to internal mass distribution effects

Explanation: When you encounter gravitational force problems involving extended objects, the key principle to remember is Newton's shell theorem. This theorem states that a spherically symmetric mass distribution exerts the same gravitational force on an external point as if all the mass were concentrated at the center. Both the solid sphere and spherical shell are spherically symmetric, meaning their mass is distributed symmetrically around their centers. Since the point mass

m

is located outside both objects (at distance

3R > R

), Newton's shell theorem applies to both configurations. The gravitational force in each case is simply

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

, where the center-to-center distance determines the force magnitude, not the internal mass distribution. Choice A is correct because both objects exert identical forces due to this fundamental principle. Choice B incorrectly assumes that mass concentration affects external gravitational fields—while the solid sphere does have more concentrated mass, this doesn't change the external field. Choice C makes the opposite error, suggesting surface distribution matters for external points, which it doesn't. Choice D incorrectly connects moment of inertia to gravitational force; while these objects do have different moments of inertia, this rotational property has no bearing on their gravitational effects on external masses. Remember: for any spherically symmetric object, external gravitational effects depend only on total mass and distance to the center, regardless of internal mass distribution. This makes many gravitational calculations much simpler than they initially appear.

Question 15

The gravitational acceleration on the surface of a neutron star is approximately $10^{12}\text{ m/s}^2$ . If a neutron star has a radius of $10\text{ km}$ , what is the approximate escape velocity from its surface?

$1.4 \times 10^8\text{ m/s}$ (correct answer)
$1.0 \times 10^8\text{ m/s}$
$2.0 \times 10^8\text{ m/s}$
$7.1 \times 10^7\text{ m/s}$
$3.2 \times 10^8\text{ m/s}$

Explanation: When you encounter escape velocity problems, you're dealing with the minimum speed needed for an object to break free from a gravitational field. The escape velocity formula is

v_e = \sqrt{2gR}

, where

g

is surface gravity and

R

is the radius. With the given values:

g = 10^{12} \text{ m/s}^2

and

R = 10 \text{ km} = 10^4 \text{ m}

, you can substitute directly:

v_e = \sqrt{2 \times 10^{12} \times 10^4} = \sqrt{2 \times 10^{16}} = \sqrt{2} \times 10^8

Since

\sqrt{2} \approx 1.41

, this gives

v_e \approx 1.4 \times 10^8 \text{ m/s}

, confirming answer (A). Looking at the wrong answers: (B)

1.0 \times 10^8 \text{ m/s}

would result from forgetting the factor of 2 in the escape velocity formula, using

v_e = \sqrt{gR}

instead. (C)

2.0 \times 10^8 \text{ m/s}

comes from incorrectly using 2 instead of

\sqrt{2}

as the coefficient. (D)

7.1 \times 10^7 \text{ m/s}

appears to involve calculation errors, possibly confusing the relationship between the given parameters. Remember that escape velocity problems on physics exams often test whether you know the correct formula and can handle the square root properly. The factor of 2 under the square root is crucial—it comes from energy conservation between kinetic and gravitational potential energy. Always double-check your formula before plugging in numbers.

Question 16

Two objects with masses $m_1 = 3.0\text{ kg}$ and $m_2 = 7.0\text{ kg}$ are separated by a distance of $2.0\text{ m}$ . A third object with mass $m_3 = 2.0\text{ kg}$ is placed at a point such that the net gravitational force on $m_3$ is zero. How far from $m_1$ should $m_3$ be placed?

$0.76\text{ m}$ (correct answer)
$1.24\text{ m}$
$1.50\text{ m}$
$0.50\text{ m}$
$1.00\text{ m}$

Explanation: When you encounter gravitational force problems involving a point where forces balance, you're dealing with the principle that gravitational forces are vector quantities that can cancel each other out. For the net gravitational force on

m_3

to be zero, the gravitational forces from

m_1

and

m_2

must be equal in magnitude but opposite in direction. This means

m_3

must lie on the line connecting

m_1

and

m_2

, and since

m_2 > m_1

, the equilibrium point will be closer to the smaller mass

m_1

. Let

x

be the distance from

m_1

m_3

. Then the distance from

m_2

m_3

(2.0 - x)

. Setting the gravitational forces equal:

\frac{Gm_1m_3}{x^2} = \frac{Gm_2m_3}{(2.0-x)^2}

The

G

and

m_3

terms cancel, leaving:

\frac{m_1}{x^2} = \frac{m_2}{(2.0-x)^2}

Substituting values:

\frac{3.0}{x^2} = \frac{7.0}{(2.0-x)^2}

Cross-multiplying and solving:

3.0(2.0-x)^2 = 7.0x^2

This yields

x = 0.76

m, confirming answer A. Answer B (1.24 m) would place

m_3

closer to

m_2

, making the force from

m_2

too strong. Answer C (1.50 m) places

m_3

too close to

m_2

for equilibrium. Answer D (0.50 m) makes

m_3

too close to

m_1

, creating imbalance. Remember: in gravitational equilibrium problems, the test mass always sits closer to the smaller mass, and you solve by setting force magnitudes equal.

Question 17

Two identical satellites orbit a planet. Satellite A orbits at radius $r$ and satellite B orbits at radius $4r$ . What is the ratio of the gravitational force on satellite A to the gravitational force on satellite B?

$16:1$ (correct answer)
$4:1$
$8:1$
$2:1$
$12:1$

Explanation: When you encounter orbital mechanics problems involving gravitational force, the key principle is Newton's law of universal gravitation:

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

, where the force is inversely proportional to the square of the distance. Since both satellites are identical (same mass

m

) orbiting the same planet (same

M

), you can set up a ratio. For satellite A at radius

r

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

. For satellite B at radius

4r

F_B = \frac{GMm}{(4r)^2} = \frac{GMm}{16r^2}

. Taking the ratio:

\frac{F_A}{F_B} = \frac{\frac{GMm}{r^2}}{\frac{GMm}{16r^2}} = \frac{GMm}{r^2} \times \frac{16r^2}{GMm} = 16

Therefore, the ratio is

16:1

, making (A) correct. (B) 4:1 represents the common error of forgetting to square the radius ratio. Students often think "4r is 4 times larger, so force is 4 times smaller," but this ignores the inverse square relationship. (C) 8:1 has no clear physical basis and likely results from confused arithmetic, perhaps mixing up the radius factor with the inverse square law incorrectly. (D) 2:1 suggests taking the square root of the correct answer, which doesn't correspond to any relevant physical relationship in gravitational problems. Study tip: Always remember that gravitational force follows an inverse square law. When the distance increases by a factor of

n

, the force decreases by a factor of

n^2

. This pattern appears frequently in physics problems involving fields that spread out in three dimensions.

Question 18

A spacecraft traveling from Earth to Mars must pass through the point where Earth's and Mars' gravitational influences are equal. If Mars has $0.11$ times Earth's mass and the Earth-Mars distance is $2.3 \times 10^8\text{ km}$ , approximately how far from Earth's center is this equilibrium point?

$1.8 \times 10^8\text{ km}$ (correct answer)
$1.2 \times 10^8\text{ km}$
$1.5 \times 10^8\text{ km}$
$2.0 \times 10^8\text{ km}$
$1.0 \times 10^8\text{ km}$

Explanation: When spacecraft travel between planets, there's a special point called the gravitational equilibrium point where the gravitational forces from both celestial bodies are equal. This is where you need to set up an equation using Newton's law of universal gravitation. At the equilibrium point, the gravitational force from Earth equals the gravitational force from Mars. Let's call the distance from Earth's center to this point

x

, so the distance from Mars is

(2.3 \times 10^8 - x)

km. Setting the forces equal:

\frac{GM_E}{x^2} = \frac{GM_M}{(2.3 \times 10^8 - x)^2}

Since

M_M = 0.11M_E

, this becomes:

\frac{1}{x^2} = \frac{0.11}{(2.3 \times 10^8 - x)^2}

Taking the square root of both sides:

\frac{1}{x} = \frac{\sqrt{0.11}}{2.3 \times 10^8 - x}

Cross-multiplying and solving:

2.3 \times 10^8 - x = x\sqrt{0.11}

Since

\sqrt{0.11} \approx 0.33

, we get:

2.3 \times 10^8 = x(1 + 0.33) = 1.33x

Therefore:

x = \frac{2.3 \times 10^8}{1.33} \approx 1.8 \times 10^8 \text{ km}

This confirms answer A is correct. Answer B (

1.2 \times 10^8

) would place the point too close to Earth, underestimating Earth's stronger gravity. Answer C (

1.5 \times 10^8

) represents a common error of assuming the midpoint. Answer D (

2.0 \times 10^8

) places the point too far from Earth. Remember: gravitational equilibrium points are always closer to the less massive object, since gravity weakens with distance squared.

Question 19

A binary star system consists of two stars, each of mass $M$ , orbiting their common center of mass in circular orbits of radius $R$ . If one star suddenly loses half its mass (while maintaining its orbital speed), what happens to the gravitational force between the stars immediately after the mass loss?

The force remains the same because distance is unchanged
The force decreases to one-quarter its original value
The force decreases to half its original value (correct answer)
The force decreases to $\frac{\sqrt{2}}{2}$ times its original value

Explanation: When analyzing gravitational force problems, always start with Newton's law of universal gravitation:

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

. This tells you that gravitational force depends on both masses and the distance between them. Initially, the force between the two stars is

F_0 = G\frac{M \cdot M}{(2R)^2} = G\frac{M^2}{4R^2}

. Note that the distance between stars is

2R

since each orbits at radius

R

from their center of mass. When one star loses half its mass, it becomes mass

\frac{M}{2}

while the other remains mass

M

. The problem states the orbital speed is maintained, so immediately after mass loss, the distance is still

2R

. The new force becomes:

F_{new} = G\frac{M \cdot \frac{M}{2}}{(2R)^2} = G\frac{M^2/2}{4R^2} = \frac{1}{2} \cdot G\frac{M^2}{4R^2} = \frac{F_0}{2}

The force decreases to half its original value. Choice A incorrectly assumes force depends only on distance. While distance is unchanged, the reduced mass product affects the force. Choice B applies the mass reduction twice (

\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

), which would only be correct if both masses were halved. Choice D incorrectly applies energy or momentum considerations rather than the direct force calculation. Study tip: In gravitational force problems, always identify what changes (masses, distances) and what stays constant. Apply Newton's law of gravitation directly, being careful about the distance between centers of mass, not orbital radii.

Question 20

A spherical planet has uniform density $\rho$ and radius $R$ . A small object is placed at a distance $\frac{R}{2}$ below the planet's surface (i.e., at distance $\frac{R}{2}$ from the center). What is the gravitational force on the object compared to the force it would experience on the planet's surface?

The force is $\frac{1}{8}$ of the surface value
The force is $\frac{1}{4}$ of the surface value
The force is $\frac{1}{2}$ of the surface value (correct answer)
The force is $2$ times the surface value

Explanation: When analyzing gravitational force inside a uniform sphere, you need to apply a key principle: only the mass within your radius contributes to the gravitational force you experience. The mass in the spherical shell outside your position has zero net gravitational effect. At distance

\frac{R}{2}

from the center, only the mass within a sphere of radius

\frac{R}{2}

affects the object. Since the planet has uniform density

\rho

, this inner mass is:

M_{inner} = \frac{4}{3}\pi\left(\frac{R}{2}\right)^3\rho = \frac{1}{8} \cdot \frac{4}{3}\pi R^3\rho = \frac{1}{8}M_{total}

The gravitational force is

F = \frac{GM_{inner}m}{r^2}

, where

r = \frac{R}{2}

. Substituting:

F_{inside} = \frac{G \cdot \frac{1}{8}M_{total} \cdot m}{\left(\frac{R}{2}\right)^2} = \frac{G \cdot \frac{1}{8}M_{total} \cdot m}{\frac{1}{4}R^2} = \frac{1}{2} \cdot \frac{GM_{total}m}{R^2}

Since the surface force is

F_{surface} = \frac{GM_{total}m}{R^2}

, the force inside is

\frac{1}{2}

the surface value. Answer A (

\frac{1}{8}

) incorrectly uses only the mass ratio without accounting for the distance change. Answer B (

\frac{1}{4}

) likely confuses the radius ratio with the force ratio. Answer D (2 times) incorrectly assumes force increases as you go deeper. Study tip: For uniform spheres, gravitational force increases linearly from the center to the surface, reaching maximum at the surface. Remember that

F \propto \frac{M_{enclosed}}{r^2}

, where both mass and distance matter.