IB Physics Quiz: Apply Gravitational Fields

20 questions · exam conditions

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Apply Gravitational FieldsQuestion 1 of 20

Planet X has mass M and radius R. Planet Y has mass 2M and radius 2R. An object has weight W when on the surface of Planet X. What is the weight of the same object on the surface of Planet Y?

W/4

W/2

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IB Physics Quiz

IB Physics Quiz: Apply Gravitational Fields

Practice Apply 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 Apply 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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Question 1

Planet X has mass M and radius R. Planet Y has mass 2M and radius 2R. An object has weight W when on the surface of Planet X. What is the weight of the same object on the surface of Planet Y?

W/4
W/2 (correct answer)
W
2W

Explanation: Weight is the gravitational force, which is proportional to M/R². For Planet X, W ∝ M/R². For Planet Y, the new weight W' ∝ (2M)/(2R)² = 2M/(4R²) = (1/2) * (M/R²). Therefore, the new weight W' is W/2.

Question 2

The total energy of a satellite of mass m in a circular orbit of radius r around a planet of mass M is E = -GMm/(2r). What is the minimum work that must be done by an external agent to move this satellite to an orbit of radius 2r?

GMm/(4r) (correct answer)
GMm/(2r)
3GMm/(4r)
GMm/r

Explanation: Work done is the change in total energy, W = ΔE = E_final - E_initial. The initial energy is E_initial = -GMm/(2r). The final energy in the orbit of radius 2r is E_final = -GMm/(2(2r)) = -GMm/(4r). The work done is W = (-GMm/(4r)) - (-GMm/(2r)) = -GMm/(4r) + 2GMm/(4r) = GMm/(4r).

Question 3

Assuming a planet has uniform density ρ and radius R, how does the gravitational field strength g inside the planet vary with distance r from its center (for r < R)?

g is proportional to r. (correct answer)
g is proportional to 1/r².
g is constant.
g is proportional to 1/r.

Explanation: For a point inside the planet at radius r, the gravitational force is due only to the mass enclosed within that radius. Mass enclosed M_enc = ρ × V = ρ × (4/3)πr³. The field strength is g = GM_enc/r² = G[ρ(4/3)πr³]/r² = (4/3)Gρπ * r. Since G, ρ, and π are constants, g is directly proportional to r.

Question 4

Two spherical masses, m₁ and m₂, are separated by a distance r. The magnitude of the gravitational force between them is F. If the distance is doubled to 2r and the mass m₁ is also doubled to 2m₁, what is the new magnitude of the gravitational force?

F/4
F/2 (correct answer)
F
2F

Explanation: The original force is F = Gm₁m₂/r². The new force F' = G(2m₁)m₂/(2r)² = G(2m₁m₂)/(4r²) = (2/4) * (Gm₁m₂/r²) = (1/2)F. The new force is F/2.

Question 5

The gravitational potential on the surface of a planet is -6.0 x 10⁷ J kg⁻¹. A 10 kg satellite is launched from the surface and placed in a high orbit where the gravitational potential is -2.0 x 10⁷ J kg⁻¹. What is the minimum work done by the rocket engines for this change in position?

4.0 x 10⁷ J
8.0 x 10⁷ J
4.0 x 10⁸ J (correct answer)
8.0 x 10⁸ J

Explanation: The work done against the gravitational field is equal to the change in gravitational potential energy, W = ΔE_p. The change in potential energy is mΔV, where ΔV is the change in gravitational potential. ΔV = V_final - V_initial = (-2.0 x 10⁷) - (-6.0 x 10⁷) = 4.0 x 10⁷ J kg⁻¹. The work done is W = mΔV = 10 kg × (4.0 x 10⁷ J kg⁻¹) = 4.0 x 10⁸ J. This ignores the change in kinetic energy, giving the minimum work required for the position change.

Question 6

Consider a hypothetical situation where the gravitational force is an inverse cube law, F ∝ 1/r³. For a planet in a stable circular orbit of radius r around a star of mass M, how would its orbital speed v depend on r?

v ∝ 1/r (correct answer)
v ∝ 1/√r
v ∝ r
v is independent of r

Explanation: The centripetal force is mv²/r. The gravitational force is F = k/r³ for some constant k. For a stable orbit, mv²/r = k/r³. Rearranging for v² gives v² = k/(mr²). Taking the square root gives v = √(k/m) * (1/r). Since k and m are constant for the planet, v is proportional to 1/r.

Question 7

The escape speed from the surface of a planet of mass M and radius R is v_esc. What is the escape speed from a planet with mass 4M and radius R/2?

√2 v_esc
2 v_esc
2√2 v_esc (correct answer)
8 v_esc

Explanation: The escape speed is given by the formula v_esc = √(2GM/R). For the new planet, the mass M' = 4M and the radius R' = R/2. The new escape speed v'_esc = √(2G(4M)/(R/2)) = √(16GM/R) = √8 * √(2GM/R) = 2√2 v_esc.

Question 8

A spacecraft is at a point in space where the gravitational fields from the Earth and the Moon cancel out. Let M_E be the mass of the Earth, M_M the mass of the Moon, and D the distance between their centers. How far is this point from the center of the Earth?

D / (1 + √(M_M/M_E)) (correct answer)
D * M_E / (M_E + M_M)
D * √(M_E/M_M)
D / (1 - √(M_M/M_E))

Explanation: Let the distance from Earth be x. The distance from the Moon is D-x. The gravitational fields must be equal in magnitude: GM_E/x² = GM_M/(D-x)². This simplifies to M_E/x² = M_M/(D-x)². Rearranging gives (D-x)²/x² = M_M/M_E. Taking the square root: (D-x)/x = √(M_M/M_E). Then D/x - 1 = √(M_M/M_E), so D/x = 1 + √(M_M/M_E). Finally, x = D / (1 + √(M_M/M_E)).

Question 9

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 strength zero?

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

Explanation: Let the point be at distance

x

from mass

M

and

(d-x)

from mass

2M

. For zero field:

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

. This gives

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

, so

(d-x) = x\sqrt{2}

. Solving:

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

, therefore

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

. Option A assumes equal division by mass ratio, B assumes midpoint, C involves incorrect square root manipulation.

Question 10

Two identical satellites orbit a planet at different altitudes. Satellite A has an orbital period of $4.0 \text{ hours}$ and experiences a gravitational field strength of $6.0 \text{ N kg}^{-1}$ . Satellite B has an orbital period of $8.0 \text{ hours}$ . What is the gravitational field strength experienced by satellite B?

$1.5 \text{ N kg}^{-1}$
$2.4 \text{ N kg}^{-1}$ (correct answer)
$3.0 \text{ N kg}^{-1}$
$4.8 \text{ N kg}^{-1}$

Explanation: From Kepler's third law,

\frac{T_B^2}{T_A^2} = \frac{r_B^3}{r_A^3}

, so

\frac{64}{16} = \frac{r_B^3}{r_A^3}

, giving

\frac{r_B}{r_A} = 4^{1/3} = 1.587

. Since gravitational field strength

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

\frac{g_B}{g_A} = \frac{r_A^2}{r_B^2} = \frac{1}{(1.587)^2} = \frac{1}{2.52} = 0.397

. Therefore

g_B = 6.0 \times 0.397 = 2.4 \text{ N kg}^{-1}

. Option A uses

\frac{1}{4}

ratio incorrectly, C uses

\frac{1}{2}

, and D uses incorrect proportionality.

Question 11

A comet follows an elliptical orbit around the Sun. At its closest approach (perihelion), it is $1.0 \times 10^{11} \text{ m}$ from the Sun and experiences a gravitational field strength of $5.9 \times 10^{-3} \text{ N kg}^{-1}$ . At its farthest point (aphelion), it is $4.0 \times 10^{11} \text{ m}$ from the Sun. What is the ratio of the comet's speeds $\frac{v_{perihelion}}{v_{aphelion}}$ ?

$2.0$
$4.0$ (correct answer)
$6.9$
$16$

Explanation: Using conservation of angular momentum for the elliptical orbit:

L = mr_p v_p = mr_a v_a

, so

\frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{4.0 \times 10^{11}}{1.0 \times 10^{11}} = 4.0

. The gravitational field information is provided as a distractor. Option A uses

\sqrt{4} = 2

, option C might come from incorrectly using the field strength data, and option D uses

(r_a/r_p)^2 = 16

Question 12

A spherical planet has a uniform density $\rho$ and radius $R$ . A tunnel is drilled from the surface to the center. What is the gravitational field strength at a distance $\frac{R}{2}$ from the center along this tunnel?

$\frac{2\pi G\rho R}{3}$ (correct answer)
$\frac{4\pi G\rho R}{3}$
$\frac{2\pi G\rho R}{9}$
$\frac{\pi G\rho R}{6}$

Explanation: For a uniform sphere, the gravitational field at radius

r

from the center (where

r < R

) is due only to the mass within radius

r

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

. The mass within radius

r

M(r) = \rho \times \frac{4}{3}\pi r^3

. At

r = \frac{R}{2}

g = \frac{G \times \rho \times \frac{4}{3}\pi \times (\frac{R}{2})^3}{(\frac{R}{2})^2} = \frac{G\rho \times \frac{4}{3}\pi \times \frac{R^3}{8}}{\frac{R^2}{4}} = \frac{G\rho \times \frac{\pi R^3}{6}}{\frac{R^2}{4}} = \frac{2\pi G\rho R}{3}

. Options B, C, and D represent common algebraic errors in the calculation.

Question 13

A spacecraft travels from Earth's surface to the Moon's surface. Given that Earth's mass is $81$ times the Moon's mass and Earth's radius is $3.7$ times the Moon's radius, what is the ratio of the escape speeds $\frac{v_{Earth}}{v_{Moon}}$ ?

$4.7$ (correct answer)
$6.5$
$8.1$
$21.9$

Explanation: Escape speed is

v_{escape} = \sqrt{\frac{2GM}{R}}

. Therefore

\frac{v_{Earth}}{v_{Moon}} = \sqrt{\frac{M_{Earth}}{M_{Moon}} \times \frac{R_{Moon}}{R_{Earth}}} = \sqrt{81 \times \frac{1}{3.7}} = \sqrt{21.9} = 4.7

. Option B might result from using

\sqrt{81/3.7}

incorrectly, option C uses the mass ratio directly, and option D uses

81/3.7

without taking the square root.

Question 14

The gravitational field lines around a planet show the direction of the force on a test mass. Which statement correctly describes the equipotential surfaces associated with this field?

Equipotential surfaces are parallel to the gravitational field lines.
Equipotential surfaces are spaced further apart where the gravitational field is weaker. (correct answer)
The work done moving a mass along an equipotential surface is positive.
The gravitational potential increases as the distance from the planet increases.

Explanation: Equipotential surfaces are surfaces of constant potential. The gravitational field is stronger where these surfaces are closer together. Conversely, they are spaced further apart where the field is weaker. Field lines are always perpendicular to equipotential surfaces. No work is done moving a mass along an equipotential surface (W = mΔV = 0). Gravitational potential V = -GM/r, which becomes less negative (increases) as r increases.

Question 15

Three identical masses $m$ are arranged at the vertices of an equilateral triangle with side length $a$ . What is the magnitude of the gravitational field at the center of the triangle due to all three masses?

$\frac{3Gm}{a^2}$
$\frac{3\sqrt{3}Gm}{a^2}$
$\frac{9Gm}{a^2}$
$0$ (correct answer)

Explanation: The center of an equilateral triangle is equidistant from all three vertices. Each mass contributes a gravitational field of magnitude

\frac{Gm}{r^2}

toward itself, where

r = \frac{a}{\sqrt{3}}

is the distance from center to vertex. However, by symmetry, the three field vectors are separated by

120°

and their vector sum is zero. This is a classic result that students often miss by trying to add magnitudes instead of vectors. Options A, B, and C represent attempts to add the field magnitudes rather than the field vectors.

Question 16

A space station orbits Earth at an altitude where the gravitational field strength is $4.9 \text{ N kg}^{-1}$ . An astronaut in the station drops a tool, which appears to float beside her. From the perspective of an observer on Earth's surface (where $g = 9.8 \text{ N kg}^{-1}$ ), what is the acceleration of the tool?

$0 \text{ m s}^{-2}$
$9.8 \text{ m s}^{-2}$ toward Earth's center
$4.9 \text{ m s}^{-2}$ toward Earth's center (correct answer)
$4.9 \text{ m s}^{-2}$ tangential to the orbit

Explanation: This question tests your understanding of reference frames and gravitational acceleration in orbital mechanics. The key insight is recognizing that different observers measure different accelerations for the same object. From an Earth-based observer's perspective, the tool experiences gravitational acceleration equal to the local gravitational field strength at the space station's altitude. Since the gravitational field strength is

4.9 \text{ N kg}^{-1}

at that altitude, the tool accelerates at

4.9 \text{ m s}^{-2}

toward Earth's center. This acceleration keeps the tool in orbit alongside the space station. Looking at the incorrect options: Option A (

0 \text{ m s}^{-2}

) reflects what the astronaut observes in her reference frame—the tool appears to float because both she and the tool are in free fall together. However, the question specifically asks for the Earth observer's perspective. Option B (

9.8 \text{ m s}^{-2}

) incorrectly uses Earth's surface gravity, but gravitational acceleration decreases with distance from Earth's center. At the space station's altitude, gravity is weaker than at Earth's surface. Option D suggests tangential acceleration, but gravitational force always acts radially toward Earth's center, not tangentially. The tool appears weightless to the astronaut because both are accelerating at the same rate due to gravity—this is the essence of free fall in orbit. Study tip: When analyzing orbital motion problems, always clarify which reference frame you're using. Objects in orbit are simultaneously accelerating (from an external observer's view) and weightless (from their own reference frame).

Question 17

An object is moved from the surface of the Earth (radius R) to an altitude of R above the surface. What is the change in its gravitational potential energy?

mgR/2 (correct answer)
mgR
2mgR
GMm/(2R)

Explanation: Gravitational potential energy is E_p = -GMm/r. At the surface, r₁ = R, so E_p1 = -GMm/R. At altitude R, the new distance from the center is r₂ = 2R, so E_p2 = -GMm/(2R). The change is ΔE_p = E_p2 - E_p1 = (-GMm/(2R)) - (-GMm/R) = GMm/R - GMm/(2R) = GMm/(2R). Since g = GM/R² on the surface, we can write GM = gR². Substituting this gives ΔE_p = (gR²m)/(2R) = mgR/2.

Question 18

A satellite is in a stable circular orbit of radius R around a planet. Its orbital speed is v. If the satellite is moved to a new stable circular orbit of radius 4R, what is its new orbital speed?

v/4
v/2 (correct answer)
2v
4v

Explanation: For a circular orbit, the gravitational force provides the centripetal force: GMm/R² = mv²/R. This simplifies to v² = GM/R, so v = √(GM/R). The speed is inversely proportional to the square root of the radius (v ∝ 1/√R). If the radius R is multiplied by 4, the new speed will be multiplied by 1/√4 = 1/2. The new speed is v/2.

Question 19

Two planets, P1 and P2, orbit the same star. The orbital period of P1 is T and its mean orbital radius is R. The mean orbital radius of P2 is 4R. What is the orbital period of P2?

2T
4T
8T (correct answer)
16T

Explanation: According to Kepler's Third Law, the square of the orbital period is proportional to the cube of the mean orbital radius (T² ∝ R³). So, (T₂/T₁)² = (R₂/R₁)³. Given R₂ = 4R₁, we have (T₂/T)² = (4R/R)³ = 4³ = 64. Therefore, T₂² = 64T², and T₂ = √64 * T = 8T.

Question 20

The gravitational potential at a point P is V. What is the definition of this statement?

V is the force per unit mass exerted on a small test mass placed at P.
V is the work done per unit mass to move a small test mass from infinity to P. (correct answer)
V is the gravitational potential energy of a small test mass placed at P.
V is the work done by the gravitational field to move a unit mass from P to infinity.

Explanation: Gravitational potential (V) at a point is defined as the work done per unit mass by an external agent in bringing a small test mass from a point of zero potential (infinity) to that point. Work done by the field would have the opposite sign. Force per unit mass is gravitational field strength (g). Gravitational potential energy (Ep) depends on the test mass (Ep = mV).