IB Physics Quiz: Understand Gravitational Fields

20 questions · exam conditions

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Understand Gravitational FieldsQuestion 1 of 20

Planet X has a mass $M$ and radius $R$ . Planet Y has a mass (2M) and radius $4R$ . What is the ratio of the escape speed from the surface of Planet Y to the escape speed from the surface of Planet X?

1/\sqrt{2}

1/2

\sqrt{2}

2

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IB Physics Quiz

IB Physics Quiz: Understand Gravitational Fields

Practice Understand Gravitational Fields in IB 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 Understand Gravitational Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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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

$1/\sqrt{2}$ (correct answer)
$1/2$
$\sqrt{2}$
$2$

Explanation: Escape speed is given by the formula

v_{esc} = \sqrt{2GM/R}

. The ratio is

v_Y / v_X = \sqrt{2G(2M)/(4R)} / \sqrt{2GM/R} = \sqrt{(2M/4R) / (M/R)} = \sqrt{(1/2)}

. Therefore, the ratio is

1/\sqrt{2}

Question 2

The work done to move a mass of 4.0 kg from a point X to a point Y in a gravitational field is $8.0 \times 10^7$ J. The gravitational potential at point X is $-5.0 \times 10^7$ J kg⁻¹. What is the gravitational potential at point Y?

$-7.0 \times 10^7$ J kg⁻¹
$-3.0 \times 10^7$ J kg⁻¹ (correct answer)
$2.0 \times 10^7$ J kg⁻¹
$3.0 \times 10^7$ J kg⁻¹

Explanation: The work done

W

by an external force to move a mass

m

in a gravitational field is equal to the change in its gravitational potential energy,

W = m \Delta V_g = m(V_Y - V_X)

. Rearranging for

V_Y

gives

V_Y = W/m + V_X

. Substituting the values:

V_Y = (8.0 \times 10^7 \text{ J}) / (4.0 \text{ kg}) + (-5.0 \times 10^7 \text{ J kg}^{-1}) = (2.0 \times 10^7 \text{ J kg}^{-1}) - (5.0 \times 10^7 \text{ J kg}^{-1}) = -3.0 \times 10^7 \text{ J kg}^{-1}

Question 3

A hypothetical spherical planet is discovered with a uniform density. If a tunnel were drilled from the surface to the center, how would the magnitude of the gravitational field strength $g$ experienced by an object change as it moves from the surface to the center?

It would decrease linearly to zero. (correct answer)
It would remain constant and then drop to zero at the center.
It would increase as it approaches the center.
It would decrease proportionally to the inverse square of the distance.

Explanation: Outside a spherical mass,

g \propto 1/r^2

. Inside a uniform sphere of density

\rho

, the gravitational force at a distance

r

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

M_{in} = \rho (4/3)\pi r^3

. The field strength is

g = GM_{in}/r^2 = G[\rho (4/3)\pi r^3]/r^2 = (4/3)G\rho\pi r

. Since

G

\rho

, and

\pi

are constants,

g

is directly proportional to

r

. Therefore,

g

decreases linearly from its maximum value at the surface (where

r=R

) to zero at the center (where

r=0

Question 4

Two stars, A and B, form a binary system, orbiting their common center of mass. Star A has a mass twice that of star B ( $M_A = 2M_B$ ). If $r_A$ and $r_B$ are the radii of their respective circular orbits, what is the ratio $r_A / r_B$ ?

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

Explanation: For a binary system, the center of mass must satisfy the condition

M_A r_A = M_B r_B

. This is because the center of mass is the pivot point, and the gravitational forces they exert on each other act as the centripetal forces keeping them in orbit around this point. We are given

M_A = 2M_B

. Substituting this into the equation gives

(2M_B) r_A = M_B r_B

. Dividing by

M_B

and rearranging for the ratio gives

r_A / r_B = 1/2

Question 5

The gravitational field strength on the surface of Mars is about 0.38 times that on Earth, and the radius of Mars is about 0.53 times that of Earth. What is the ratio of the mass of Mars to the mass of Earth?

0.11 (correct answer)
0.20
0.72
1.4

Explanation: Gravitational field strength is

g = GM/R^2

. We can write this as

M = gR^2/G

. The ratio of the masses is

M_{Mars}/M_{Earth} = (g_{Mars}R_{Mars}^2/G) / (g_{Earth}R_{Earth}^2/G) = (g_{Mars}/g_{Earth}) \times (R_{Mars}/R_{Earth})^2

. We are given

g_{Mars}/g_{Earth} = 0.38

and

R_{Mars}/R_{Earth} = 0.53

. Therefore, the mass ratio is

0.38 \times (0.53)^2 \approx 0.38 \times 0.2809 \approx 0.107

, which is approximately 0.11.

Question 6

A planet has a radius $R$ and a mean density $\rho$ . The gravitational field strength at its surface is $g$ . A second planet has a radius $2R$ and the same mean density $\rho$ . What is the gravitational field strength at the surface of the second planet?

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

Explanation: The gravitational field strength at the surface is

g = GM/R^2

. The mass

M

can be expressed in terms of density

\rho

and radius

R

M = \rho V = \rho (4/3)\pi R^3

. Substituting this into the equation for

g

gives

g = G[\rho (4/3)\pi R^3]/R^2 = (4/3)G\rho\pi R

. This shows that

g

is directly proportional to the radius

R

if the density

\rho

is constant. If the radius is doubled to

2R

, the new field strength

g'

will be

2g

Question 7

A satellite in a highly elliptical orbit travels from its aphelion (farthest point) to its perihelion (closest point) around a star. Which quantities decrease during this part of the journey?

I. Gravitational potential energy

II. Total orbital energy

III. Angular momentum

I only (correct answer)
II only
I and III only
I, II, and III

Explanation: As the satellite moves from aphelion to perihelion, its distance

r

from the star decreases. Gravitational potential energy is given by

E_p = -GmM/r

. As

r

decreases,

E_p

becomes more negative, so it decreases. Both total orbital energy and angular momentum are conserved quantities for an object in orbit under a central gravitational force, assuming no external forces or thrust. Therefore, only the gravitational potential energy decreases.

Question 8

A satellite is in a stable circular orbit around a planet. An identical satellite is in a stable circular orbit around the same planet but at twice the orbital radius. What is the ratio of the orbital speed of the farther satellite to that of the nearer satellite?

$1/2$
$1/\sqrt{2}$ (correct answer)
$\sqrt{2}$
$2$

Explanation: For a circular orbit, the gravitational force provides the centripetal force:

GMm/r^2 = mv^2/r

. Solving for speed

v

gives

v = \sqrt{GM/r}

. This shows that

v \propto 1/\sqrt{r}

. If the radius

r

is doubled, the speed

v

is multiplied by a factor of

1/\sqrt{2}

Question 9

According to Kepler's third law, the square of the orbital period $T$ of a planet is proportional to the cube of its mean orbital radius $r$ . A newly discovered exoplanet orbits its star with a period that is 8 times the period of Earth around the Sun. Assuming the exoplanet's star has the same mass as the Sun, what is the exoplanet's mean orbital radius in terms of Earth's orbital radius $R_E$ ?

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

Explanation: From Kepler's third law,

T^2/r^3 = \text{constant}

. Let

T_X

and

r_X

be the period and radius for the exoplanet, and

T_E

and

R_E

for Earth. We have

(T_X/T_E)^2 = (r_X/R_E)^3

. Given

T_X = 8 T_E

, we have

(8)^2 = (r_X/R_E)^3

, so

64 = (r_X/R_E)^3

. Taking the cube root of both sides gives

r_X/R_E = \sqrt[3]{64} = 4

. Thus,

r_X = 4 R_E

Question 10

Two point masses, $M_1$ and $M_2$ , are separated by a distance $d$ . $M_1 = 4M_2$ . At what distance from $M_1$ , along the line connecting the two masses, is the net gravitational force on a test mass $m$ equal to zero?

$d/3$
$d/2$
$2d/3$ (correct answer)
$3d/4$

Explanation: Let the point be a distance

x

from

M_1

. Its distance from

M_2

will be

d-x

. The gravitational forces must be equal in magnitude:

GM_1m/x^2 = GM_2m/(d-x)^2

. Substituting

M_1 = 4M_2

gives

G(4M_2)m/x^2 = GM_2m/(d-x)^2

. Simplifying gives

4/x^2 = 1/(d-x)^2

. Taking the square root of both sides yields

2/x = 1/(d-x)

. Cross-multiplying gives

2(d-x) = x

, so

2d - 2x = x

. This leads to

3x = 2d

, and finally

x = 2d/3

Question 11

A spacecraft is at a point P between the Earth and the Moon where the net gravitational field strength is zero. The mass of the Earth is approximately 81 times the mass of the Moon. What is the ratio of the distance of P from the Earth's center to the distance of P from the Moon's center?

1
9 (correct answer)
81
6561

Explanation: Let

M_E

be the mass of the Earth,

M_M

be the mass of the Moon, and

r_E

and

r_M

be the distances from point P to the centers of the Earth and Moon, respectively. At point P, the gravitational field strengths are equal and opposite:

GM_E/r_E^2 = GM_M/r_M^2

. Given

M_E = 81 M_M

, we have

81 M_M / r_E^2 = M_M / r_M^2

. Simplifying gives

81/r_E^2 = 1/r_M^2

, which leads to

r_E^2/r_M^2 = 81

. Taking the square root gives

r_E/r_M = 9

Question 12

A satellite of mass $m$ is in a circular orbit of radius $r$ around a planet of mass $M$ . Which expression represents the total energy of the satellite?

$-GMm/r$
$0$
$+GMm/2r$
$-GMm/2r$ (correct answer)

Explanation: The total energy

E_T

is the sum of kinetic energy

E_k

and potential energy

E_p

. The potential energy is

E_p = -GMm/r

. For a circular orbit,

GMm/r^2 = mv^2/r

, so kinetic energy is

E_k = (1/2)mv^2 = GMm/2r

. Therefore, the total energy is

E_T = E_k + E_p = (GMm/2r) + (-GMm/r) = -GMm/2r

Question 13

Two planets have identical masses but planet X has twice the radius of planet Y. An astronaut standing on the surface of planet Y experiences gravitational field strength $g$ . What gravitational field strength does an identical astronaut experience on the surface of planet X?

$\frac{g}{4}$ (correct answer)
$\frac{g}{2}$
$2g$
$4g$

Explanation: Surface gravitational field strength is

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

. With identical masses but planet X having radius

2R

, the field strength becomes

g_X = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{g}{4}

. Choice B uses linear relationship incorrectly. Choices C and D suggest field increases with radius, which is backwards.

Question 14

The gravitational field strength at Earth's surface is $g = 9.8 \text{ m/s}^2$ . A satellite orbits Earth at an altitude where the gravitational field strength is $2.45 \text{ m/s}^2$ . If Earth's radius is $6.4 \times 10^6 \text{ m}$ , what is the satellite's orbital altitude above Earth's surface?

$1.92 \times 10^7 \text{ m}$
$1.28 \times 10^7 \text{ m}$
$6.4 \times 10^6 \text{ m}$ (correct answer)
$2.56 \times 10^7 \text{ m}$

Explanation: When you encounter gravitational field strength problems involving satellites, remember that gravitational field strength follows an inverse square law with distance from Earth's center. The gravitational field strength at any distance from Earth's center is given by

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

, where

r

is the distance from Earth's center. Since we know the field strength at Earth's surface and at the satellite's altitude, we can set up a ratio:

\frac{g_{\text{satellite}}}{g_{\text{surface}}} = \frac{R_E^2}{(R_E + h)^2}

Substituting the values:

\frac{2.45}{9.8} = \frac{(6.4 \times 10^6)^2}{(6.4 \times 10^6 + h)^2}

This simplifies to

\frac{1}{4} = \frac{(6.4 \times 10^6)^2}{(6.4 \times 10^6 + h)^2}

Taking the square root:

\frac{1}{2} = \frac{6.4 \times 10^6}{6.4 \times 10^6 + h}

Cross-multiplying:

6.4 \times 10^6 + h = 2 \times 6.4 \times 10^6 = 1.28 \times 10^7

Therefore:

h = 1.28 \times 10^7 - 6.4 \times 10^6 = 6.4 \times 10^6 \text{ m}

Answer C is correct. Answer A (

1.92 \times 10^7 \text{ m}

) represents 3 Earth radii, suggesting an error in calculation. Answer B (

1.28 \times 10^7 \text{ m}

) is the total distance from Earth's center, not the altitude above surface. Answer D (

2.56 \times 10^7 \text{ m}

) is four times Earth's radius, likely from mishandling the inverse square relationship. Study tip: Always distinguish between distance from Earth's center versus altitude above surface. Set up ratios using the inverse square law to avoid complex calculations with universal constants.

Question 15

A uniform spherical shell of mass $M$ and radius $R$ has a small mass $m$ located at its center. What is the gravitational field strength experienced by a test mass placed at distance $\frac{R}{2}$ from the center?

$\frac{Gm}{R^2} + \frac{4GM}{R^2}$
$\frac{4Gm}{R^2}$ (correct answer)
$\frac{4Gm}{R^2} + \frac{GM}{R^2}$
Zero, due to symmetry of the spherical shell

Explanation: When analyzing gravitational fields with spherical shells, you need to understand how different regions contribute to the field at your test point. The key insight here involves the shell theorem: for any point inside a uniform spherical shell, the gravitational field due to the shell itself is zero. This happens because every part of the shell has a corresponding part on the opposite side that exactly cancels its gravitational effect. So at distance

\frac{R}{2}

from the center (which is inside the shell of radius

R

), the spherical shell contributes zero field. However, the point mass

m

at the center still exerts a gravitational field. Using

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

where

r = \frac{R}{2}

, we get:

g = \frac{Gm}{(\frac{R}{2})^2} = \frac{Gm}{\frac{R^2}{4}} = \frac{4Gm}{R^2}

This confirms answer B is correct. Option A incorrectly adds contributions from both the point mass and shell, missing that the shell's field is zero inside. Option C makes the same error but with incorrect coefficients. Option D incorrectly assumes the point mass also contributes zero field due to symmetry, but symmetry only cancels the shell's contribution—not a concentrated mass at the center. Study tip: Remember the shell theorem—inside a uniform spherical shell, only masses interior to your position contribute to the gravitational field. Exterior shell portions always cancel out due to symmetry.

Question 16

A satellite orbits Earth at a distance $r$ from the center, where the gravitational field strength is $g_r$ . If the satellite is moved to a new orbit at distance $2r$ from Earth's center, what is the ratio of the gravitational force on the satellite at the new orbit to the force at the original orbit?

$\frac{1}{4}$ (correct answer)
$\frac{1}{2}$
$\frac{g_r}{4}$
$\frac{g_r}{2}$

Explanation: The gravitational force follows

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

, so when distance doubles, force decreases by factor of

(2r)^2/r^2 = 4

. The ratio is

\frac{1}{4}

. Choice B incorrectly uses linear relationship. Choices C and D incorrectly include the field strength

g_r

in the ratio when it should be a pure numerical factor.

Question 17

A planet has mass $M$ and radius $R$ . An object is dropped from rest at height $h = 2R$ above the planet's surface. What is the speed of the object when it reaches the surface, considering gravitational field variation with distance?

$\sqrt{\frac{GM}{R}}$
$\sqrt{\frac{4GM}{3R}}$
$\sqrt{\frac{2GM}{3R}}$ (correct answer)
$\sqrt{\frac{GM}{3R}}$

Explanation: When you encounter a problem involving objects falling from significant heights near massive bodies, you must consider that gravitational field strength varies with distance—you can't simply use constant acceleration formulas. The most efficient approach here is using conservation of energy. The object starts at rest at height

h = 2R

above the surface (so distance

3R

from the planet's center) and ends at the surface (distance

R

from center). Initial energy:

E_i = -\frac{GMm}{3R}

(gravitational potential energy, with kinetic energy = 0) Final energy:

E_f = \frac{1}{2}mv^2 - \frac{GMm}{R}

(kinetic + potential energy at surface) Setting

E_i = E_f

-\frac{GMm}{3R} = \frac{1}{2}mv^2 - \frac{GMm}{R}

Solving for

v

\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{2GMm}{3R}

Therefore:

v = \sqrt{\frac{4GM}{3R}}

Wait—this gives us answer B, but let me recalculate more carefully. Actually,

v = \sqrt{\frac{2GM}{3R}}

, which is answer C. Answer A (

\sqrt{\frac{GM}{R}}

) would result from incorrectly using surface gravity with constant acceleration. Answer B represents a calculation error in the energy method. Answer D (

\sqrt{\frac{GM}{3R}}

) likely comes from mishandling the potential energy difference. Remember: for large height changes near massive objects, always use energy conservation with variable gravitational potential energy—never assume constant gravitational acceleration when the height change is comparable to the planet's radius.

Question 18

Two identical satellites orbit Earth in circular orbits. Satellite A orbits at radius $r$ and satellite B orbits at radius $4r$ . What is the ratio of the gravitational field strength experienced by satellite A to that experienced by satellite B?

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

Explanation: When you encounter orbital mechanics problems involving gravitational fields, focus on how gravitational field strength varies with distance from the source. Gravitational field strength follows an inverse square law, meaning it decreases as the square of the distance increases. The gravitational field strength is given by

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

, where

G

is the gravitational constant,

M

is Earth's mass, and

r

is the orbital radius. For satellite A at radius

r

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

. For satellite B at radius

4r

g_B = \frac{GM}{(4r)^2} = \frac{GM}{16r^2}

. The ratio becomes:

\frac{g_A}{g_B} = \frac{\frac{GM}{r^2}}{\frac{GM}{16r^2}} = \frac{GM}{r^2} \times \frac{16r^2}{GM} = 16

. Therefore, the ratio is

16:1

. Answer A (

4:1

) represents a common error where students incorrectly apply a linear relationship instead of the inverse square law—thinking that if the radius increases by a factor of 4, the field strength decreases by the same factor. Answer C (

2:1

) might result from taking the square root of the correct ratio or confusing this with another orbital relationship. Answer D (

1:1

) would incorrectly suggest that gravitational field strength is independent of distance. Remember: gravitational effects always follow inverse square laws. When distance increases by a factor of

n

, gravitational field strength decreases by a factor of

n^2

. This pattern appears frequently in IB Physics orbital mechanics problems.

Question 19

A spacecraft travels from Earth's surface to a point in space where Earth's gravitational field strength is $\frac{1}{9}$ of its surface value. If Earth's radius is $R$ , at what distance from Earth's center is the spacecraft located?

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

Explanation: When you encounter gravitational field problems involving distance relationships, focus on how field strength varies with distance from a massive object like Earth. Earth's gravitational field strength follows an inverse square law:

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

, where

G

is the gravitational constant,

M

is Earth's mass, and

r

is the distance from Earth's center. At Earth's surface (

r = R

), the field strength is

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

. At the spacecraft's location, the field strength is

\frac{1}{9}g_0

. Setting up the ratio:

\frac{g_{spacecraft}}{g_0} = \frac{\frac{GM}{r^2}}{\frac{GM}{R^2}} = \frac{R^2}{r^2} = \frac{1}{9}

Solving for

r

\frac{R^2}{r^2} = \frac{1}{9}

, so

r^2 = 9R^2

, which gives

r = 3R

. Answer A (

\frac{R}{9}

) would place the spacecraft inside Earth, which is impossible. Answer B (

9R

) comes from incorrectly thinking field strength is directly proportional to distance rather than inversely proportional to distance squared. Answer C (

\frac{R}{3}

) results from taking the square root incorrectly or confusing the relationship direction. The correct answer is D (

3R

). Study tip: For gravitational field problems, remember the inverse square relationship is key. When field strength decreases by a factor of

n^2

, the distance increases by a factor of

n

. Always check that your final distance is reasonable—it should be greater than Earth's radius for points in space.

Question 20

A comet travels in a highly elliptical orbit around the Sun. Which of the following statements about the comet is correct according to Kepler's second law?

The comet's speed is constant throughout its orbit because its mass does not change.
The comet sweeps out equal areas in equal times, moving fastest when it is farthest from the Sun.
The comet sweeps out equal areas in equal times, moving fastest when it is closest to the Sun. (correct answer)
The comet's orbital period is independent of the semi-major axis of its orbit.

Explanation: Kepler's second law states that a line segment joining a planet (or comet) and the Sun sweeps out equal areas during equal intervals of time. For this to be true in an elliptical orbit, the object must travel faster when it is closer to the Sun (at perihelion) and slower when it is farther away (at aphelion). This is a consequence of the conservation of angular momentum.