Gravitation MCQs(Quiz) with Answers and Explanations

At what distance from the center of Earth will the gravitational acceleration become one-fourth of its value at Earth’s surface? (Assume Earth is a uniform sphere and neglect rotation.)

1. \(6R\)
2. \(8R\)
3. \(2R\)
4. \(4R\)

Explanation

Outside Earth,

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

We require

\[
\frac{GM}{r^2}
=
\frac{g_0}{4}
=
\frac{GM}{4R^2}
\]

Thus,

\[
r^2=4R^2
\]

\[
r=2R
\]

Notice that this distance is measured from Earth’s center, not from the surface.

A particle is placed at the exact center of a thin uniform spherical shell. What is the gravitational potential energy of the particle due to the shell?

1. Zero
2. \(-\dfrac{GMm}{R}\)
3. \(-\dfrac{GMm}{2R}\)
4. \(\dfrac{GMm}{R}\)

Explanation

Inside a uniform spherical shell, the gravitational field is zero everywhere.

However, gravitational potential is not zero.

The potential at every interior point is

\[
V=-\frac{GM}{R}
\]

Therefore,

\[
U=mV
=
-\frac{GMm}{R}
\]

This is a classic distinction between gravitational field and gravitational potential.

Two identical planets have the same mass, but Planet X has twice the radius of Planet Y. What is the ratio \(\dfrac{g_x}{g_y}\)?

1. \(\dfrac{1}{2}\)
2. \(2\)
3. \(\dfrac{1}{8}\)
4. \(\dfrac{1}{4}\)

Explanation

Surface gravity is

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

Since masses are equal,

\[
\frac{g_X}{g_Y}
=
\frac{R_Y^2}{R_X^2}
\]

Given

\[
R_X=2R_Y
\]

Thus,

\[
\frac{g_X}{g_Y}
=
\frac{1}{4}
\]

A satellite is orbiting Earth in a circular orbit. Which quantity remains constant during the motion?

1. Its total mechanical energy
2. Its gravitational force vector
3. Its acceleration vector
4. Its linear momentum vector

Explanation

The magnitude of force remains constant, but its direction changes continuously.

Similarly, acceleration and momentum vectors continuously change direction.

However, in a stable circular orbit,

\[
E=K+U
=
-\frac{GMm}{2r}
\]

which remains constant.

A particle is moved from infinity to a point where the gravitational potential is \(-8\times10^7\) J/kg. How much work is done by an external agent per unit mass if the particle is brought slowly?

1. \(8\times10^7\) J/kg
2. \(-8\times10^7\) J/kg
3. \(4\times10^7\) J/kg
4. \(-8\times10^7\) J/kg

Explanation

For a slow transfer,

\[
W_{ext}
=
\Delta U
\]

Per unit mass,

\[
\Delta V
=
-8\times10^7-0
\]

\[
=
-8\times10^7\;\text{J/kg}
\]

The external agent does negative work because gravity assists the motion.

At which location is the gravitational field zero while the gravitational potential is non-zero?

1. At the center of a uniform spherical shell
2. At infinity
3. On Earth’s surface
4. At the center of a solid uniform sphere

Explanation

Inside a spherical shell,

\[
g=0
\]

everywhere.

However,

\[
V=-\frac{GM}{R}
\]

which is non-zero.

This is one of the most important shell theorem results.

The escape speed from a planet is 12 km/s. Ignoring atmospheric effects, what minimum speed is required for a spacecraft to reach infinity with a residual speed of 5 km/s?

1. 13 km/s
2. 17 km/s
3. \(\sqrt{119}\) km/s
4. 13 km/s

Explanation

Using energy conservation,

\[
\frac{1}{2}v^2
=
\frac{1}{2}v_e^2
+
\frac{1}{2}(5)^2
\]

\[
v^2
=
12^2+5^2
\]

\[
=169
\]

\[
v=13\;\text{km/s}
\]

Which statement best explains why astronauts appear weightless inside an orbiting spacecraft?

1. The gravitational force is zero in orbit
2. The spacecraft is outside Earth’s gravitational field
3. The astronauts and spacecraft are in continuous free fall together
4. There is no mass in space

Explanation

At typical orbital altitudes, gravity is still strong.

The apparent weightlessness occurs because both the spacecraft and astronauts accelerate toward Earth at the same rate.

Thus, no supporting normal force acts on the astronauts.

A planet has the same density as Earth but twice Earth’s radius. What is the ratio of its escape velocity to Earth’s escape velocity?

1. \(\sqrt2\)
2. 2
3. \(2\sqrt2\)
4. \(\dfrac{1}{\sqrt2}\)

Explanation

Since density is constant,

\[
M\propto R^3
\]

Escape velocity is

\[
v_e=\sqrt{\frac{2GM}{R}}
\]

Therefore,

\[
v_e\propto R
\]

Doubling the radius doubles the escape velocity.

Hence,

\[
\frac{v_{e,new}}{v_{e,Earth}}
=2
\]

Three identical masses are placed at the vertices of an equilateral triangle. What is the gravitational field at the centroid due to the three masses?

1. \(\dfrac{GM}{a^2}\)
2. \(\dfrac{3GM}{a^2}\)
3. Zero
4. \(\dfrac{\sqrt3 GM}{a^2}\)

Explanation

Each mass produces a gravitational field at the centroid.

The three field vectors have equal magnitudes and are separated by \(120^\circ\).

The vector sum of three equal vectors separated by \(120^\circ\) is

\[
\vec g_{net}=0
\]

This result follows purely from symmetry.

A satellite is orbiting Earth in a circular orbit of radius \(R\). Its orbital speed is \(v\). If the orbital radius becomes \(4R\), what will be the new orbital speed?

1. \(\dfrac{v}{4}\)
2. \(\dfrac{v}{6}\)
3. \(\dfrac{v}{2}\)
4. \(2v\)

Explanation

For a circular orbit,

\[
v=\sqrt{\frac{GM}{r}}
\]

Therefore,

\[
\frac{v’}{v}
=
\sqrt{\frac{R}{4R}}
=
\frac{1}{2}
\]

Hence,

\[
v’=\frac{v}{2}
\]

The total mechanical energy of a satellite in a circular orbit is \(-200\) MJ. What is its kinetic energy?

1. 100 MJ
2. 200 MJ
3. 400 MJ
4. 50 MJ

Explanation

For a circular orbit,

\[
E=-K
\]

since

\[
K=\frac{GMm}{2r}
\]

and

\[
E=-\frac{GMm}{2r}
\]

Thus,

\[
K=200\ \text{MJ}
\]

A satellite is moved from a circular orbit of radius \(R\) to another circular orbit of radius \(2R\). The ratio of the orbital periods is:

1. \(\sqrt{2}\)
2. \(2\)
3. \(2\sqrt{2}\)
4. \(4\sqrt{2}\)

Explanation

Kepler’s third law gives

\[
T^2\propto r^3
\]

Therefore,

\[
\frac{T_2}{T_1}
=
\left(\frac{2R}{R}\right)^{3/2}
=
2\sqrt2
\]

A geostationary satellite must orbit in:

1. A polar orbit
2. The equatorial plane with a period of 24 hours
3. An orbit inclined at \(45^\circ\)
4. Any circular orbit with constant speed

Explanation

A geostationary satellite must remain above the same point on Earth.

Therefore it must:

– Move in the equatorial plane
– Have the same angular velocity as Earth
– Have an orbital period of approximately 24 hours

The escape velocity from Earth is \(v_e\). A particle is projected vertically upward with speed \(\frac{v_e}{2}\). Neglect air resistance. Which statement is correct?

1. The particle will return to Earth
2. The particle will reach infinity with finite speed
3. The particle will just escape Earth’s gravity
4. The particle will move in a circular orbit

Explanation

Escape requires

\[
v\ge v_e
\]

Since

\[
v=\frac{v_e}{2}
\]

the total mechanical energy remains negative.

The particle cannot escape and must eventually return.

A planet has mass \(4M\) and radius \(2R\). If Earth’s surface gravity is \(g\), what is the surface gravity of the planet?

1. \(\dfrac{g}{2}\)
2. \(4g\)
3. \(g\)
4. \(2g\)

Explanation

Surface gravity:

\[
g’=\frac{G(4M)}{(2R)^2}
\]

\[
g’=\frac{4GM}{4R^2}
\]

\[
g’=g
\]

A tunnel is drilled through the center of a uniform Earth. Neglecting rotation and air resistance, the motion of a particle released into the tunnel is:

1. Uniform circular motion
2. Projectile motion
3. Simple harmonic motion
4. Uniform acceleration motion

Explanation

Inside a uniform Earth,

\[
g\propto r
\]

Therefore,

\[
F=-kr
\]

which is the restoring-force equation of SHM.

Hence the particle executes simple harmonic motion.

At what altitude above Earth’s surface does the gravitational acceleration become \(\frac{g}{9}\)? Let Earth’s radius be \(R\).

1. \(R\)
2. \(2R\)
3. \(3R\)
4. \(8R\)

Explanation

At altitude \(h\),

\[
g’=
g\left(\frac{R}{R+h}\right)^2
\]

Given,

\[
g’=\frac{g}{9}
\]

Thus,

\[
\left(\frac{R}{R+h}\right)^2
=
\frac{1}{9}
\]

\[
R+h=3R
\]

\[
h=2R
\]

Two identical masses are separated by a distance \(d\). At what point on the line joining them is the net gravitational field zero?

1. At a distance \(\dfrac{d}{4}\) from either mass
2. Nowhere
3. At the midpoint
4. At infinity only

Explanation

At the midpoint, the gravitational fields due to both masses have equal magnitude and opposite directions.

Therefore,

\[
g_{net}=0
\]

by symmetry.

A satellite is orbiting Earth in a circular orbit. If its speed is suddenly increased without changing its position, what will happen immediately afterward?

1. It will fall directly toward Earth
2. It will move into an elliptical orbit
3. It will instantly escape Earth
4. It will remain in the same circular orbit

Explanation

The original speed was exactly the circular-orbit speed.

A sudden increase in speed changes the satellite’s total energy.

The new trajectory becomes an ellipse, provided the speed remains below escape velocity.

This is the basic idea behind orbital transfer maneuvers.

A uniform solid sphere of radius R and mass M has a narrow tunnel drilled through its center. A particle is released from rest at a distance \(\frac{R}{2}\) from the center inside the tunnel. What is its maximum speed during the motion?

1. \(\sqrt{\dfrac{GM}{2R}}\)
2. \(\sqrt{\dfrac{GM}{R}}\)
3. \(\sqrt{\dfrac{GM}{4R}}\)
4. \(\sqrt{\dfrac{2GM}{R}}\)

Explanation

Inside a uniform sphere,

\[
g(r)=\frac{GM}{R^3}r
\]

The motion is SHM with

\[
\omega=\sqrt{\frac{GM}{R^3}}
\]

Amplitude:

\[
A=\frac{R}{2}
\]

Maximum speed:

\[
v_{max}=\omega A
\]

\[
=\sqrt{\frac{GM}{R^3}}\cdot\frac{R}{2}
\]

\[
=\sqrt{\frac{GM}{4R}}
\]

Four identical masses M are placed at the corners of a square of side a. What is the gravitational potential at the center?

1. \(-\dfrac{4\sqrt2GM}{a}\)
2. \(-\dfrac{2\sqrt2GM}{a}\)
3. \(-\dfrac{8GM}{a}\)
4. \(-\dfrac{GM}{a}\)

Explanation

Distance from center to each corner:

\[
r=\frac{a}{\sqrt2}
\]

Potential due to one mass:

\[
V_1=-\frac{GM}{r}
\]

\[
=-\frac{\sqrt2GM}{a}
\]

For four masses,

\[
V=4V_1
\]

\[
=-\frac{4\sqrt2GM}{a}
\]

A satellite moves from a circular orbit of radius R into an elliptical orbit whose perigee is R and apogee is 3R. What is the semi-major axis of the new orbit?

1. \(R\)
2. \(\dfrac{3R}{2}\)
3. \(2R\)
4. \(3R\)

Explanation

For an ellipse,

\[
a=\frac{r_p+r_a}{2}
\]

Thus,

\[
a=\frac{R+3R}{2}
\]

\[
=2R
\]

Two planets have the same radius. Planet A has four times the density of Planet B. What is the ratio of their surface gravities \(\frac{g_A}{g_B}\)?

1. \(2\)
2. \(4\)
3. \(8\)
4. \(16\)

Explanation

Since

\[
M=\frac{4}{3}\pi R^3\rho
\]

and radius is constant,

\[
M\propto\rho
\]

Surface gravity:

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

Therefore,

\[
g\propto\rho
\]

Hence,

\[
\frac{g_A}{g_B}=4
\]

A particle is located midway between two masses M and 9M separated by distance d. What is the direction of the net gravitational field at the midpoint?

1. Toward M
2. Toward 9M
3. Zero
4. Perpendicular to the line joining them

Explanation

At the midpoint,

\[
g_M=\frac{GM}{(d/2)^2}
\]

\[
g_{9M}=\frac{9GM}{(d/2)^2}
\]

The stronger field is produced by 9M.

Therefore the net field points toward 9M.

A particle is projected vertically upward from Earth’s surface with speed equal to \(\frac{1}{\sqrt2}\) times the escape speed. Neglect air resistance. What is the maximum distance from Earth’s center reached by the particle?

1. \(R\)
2. \(\dfrac{3R}{2}\)
3. \(2R\)
4. \(4R\)

Explanation

Initial energy:

\[
E=\frac12m\left(\frac{v_e}{\sqrt2}\right)^2-\frac{GMm}{R}
\]

Since

\[
v_e^2=\frac{2GM}{R}
\]

\[
E=-\frac{GMm}{2R}
\]

At maximum distance,

\[
K=0
\]

\[
E=-\frac{GMm}{r_{max}}
\]

Thus,

\[
r_{max}=2R
\]

A planet has radius R. At what depth below the surface will the gravitational acceleration become half its surface value? Assume uniform density.

1. \(\dfrac{R}{4}\)
2. \(\dfrac{R}{2}\)
3. \(\dfrac{2R}{3}\)
4. \(\dfrac{3R}{4}\)

Explanation

Inside a uniform planet,

\[
g’=g\left(1-\frac{d}{R}\right)
\]

Given,

\[
g’=\frac{g}{2}
\]

Therefore,

\[
1-\frac{d}{R}=\frac12
\]

\[
d=\frac{R}{2}
\]

Three identical masses M are located at coordinates \((a,0)\), \((-a,0)\), and \((0,a)\). What is the direction of the net gravitational field at the origin?

1. Positive x-axis
2. Negative x-axis
3. Positive y-axis
4. Negative y-axis

Explanation

The fields due to masses at \((a,0)\) and \((-a,0)\) cancel each other.

Only the field due to the mass at \((0,a)\) remains.

A gravitational field always points toward the mass.

Therefore the net field points along the positive y-axis.

The total energy of a satellite in an elliptical orbit depends primarily on:

1. The eccentricity only
2. The orbital speed at perigee
3. The orbital speed at apogee
4. The semi-major axis of the orbit

Explanation

For any bound Kepler orbit,

\[
E=-\frac{GMm}{2a}
\]

where \(a\) is the semi-major axis.

The energy does not depend directly on eccentricity.

A thin spherical shell of radius R has mass M. A point mass m is located at a distance \(\frac{R}{2}\) from the center inside the shell. What is the gravitational force exerted by the shell on the mass?

1. \(\dfrac{GMm}{R^2}\)
2. \(\dfrac{4GMm}{R^2}\)
3. \(\dfrac{GMm}{4R^2}\)
4. 0

Explanation

According to the shell theorem:

– The gravitational field is zero everywhere inside a uniform spherical shell.
– Therefore the net force on any mass inside the shell is zero.

Thus,

\[
F=0
\]

even though the gravitational potential is not zero.

This is one of the most important and frequently misunderstood results in gravitation.