Gravitation

Practice Questions:

Question 71: moderate

Infinite number of bodies, each of mass $2 \text{ kg}$ are situated on x-axis at distances $1 \text{ m}$, $2 \text{ m}$, $4 \text{ m}$, $8 \text{ m}$, ….. respectively, from the origin. The resulting gravitational potential due to this system at the origin will be:

(2013)

1. $-4G$

2. $-G$

3. $-\frac{8}{3}G$

4. $-\frac{4}{3}G$

View Answer

Total potential $V = -GM \sum \frac{m}{r} = -2G (1 + \frac{1}{2} + \frac{1}{4} + ...) = -2G \times \frac{1}{1 - 0.5} = -4G$.

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Question 72: moderate

A body of mass ‘$m$’ taken from the earth’s surface to the height equal to twice the radius ($R$) of the earth. The change in potential energy of body will be:

(2013)

1. $\frac{1}{3}mgR$

2. $2 mgR$

3. $\frac{2}{3}mgR$

4. $3 mgR$

View Answer

Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. For $h = 2R$, $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$.

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Question 73: moderate

A particle of mass $M$ is situated at the center of a spherical shell of same mass and radius $a$. The magnitude of the gravitational potential at a point situated at $\frac{a}{2}$ distance from the center, will be:

(2011 Mains)

1. $\frac{2GM}{a}$

2. $\frac{3GM}{a}$

3. $\frac{4GM}{a}$

4. $\frac{GM}{a}$

View Answer

Potential at distance $a/2$ is $V = V_{\text{shell}} + V_{\text{particle}} = -\frac{GM}{a} - \frac{GM}{a/2} = -\frac{3GM}{a}$. Magnitude is $\frac{3GM}{a}$.

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Question 74: moderate

The escape velocity from the Earth’s surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and same mass density is:

(2021)

1. $2v$

2. $3v$

3. $4v$

4. $v$

View Answer

Escape velocity $v = R \sqrt{\frac{8}{3} \pi G \rho}$. Since $v \propto R$ for constant density, a planet with 4 times the radius will have $v' = 4v$.

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Question 75: moderate

A satellite of mass $m$ is orbiting the earth (of radius $R$) at a height $h$ from its surface. The total energy of the satellite in terms of $g_0$, the value of acceleration due to gravity at the earth’s surface, is:

(2016 – II)

1. $\frac{2mg_0 R^2}{R+h}$

2. $-\frac{2mg_0 R^2}{R+h}$

3. $\frac{mg_0 R^2}{2(R+h)}$

4. $-\frac{mg_0 R^2}{2(R+h)}$

View Answer

Total energy of a satellite is the sum of kinetic and potential energy, given by $E = -\frac{GMm}{2(R+h)}$. Substituting $GM = g_0 R^2$, we obtain $E = -\frac{mg_0 R^2}{2(R+h)}$. Thus, option D is correct.

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Question 76: moderate

A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass $= 5.98 \times 10^{24}\text{ kg}$) have to be compressed to be a black hole?

(2014)

1. $10^{-2}\text{ m}$

2. $10^{-6}\text{ m}$

3. $10\text{ m}$

4. $100\text{ m}$

View Answer

For a black hole, the escape velocity equals the speed of light $c$, leading to the radius formula $R = \frac{2GM}{c^2}$. Substituting the gravitational constant, mass of earth, and speed of light gives $R \approx 10^{-2}\text{ m}$. Thus, option A is correct.

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Question 77: moderate

The radii of circular orbits of two satellites A and B of the earth, are $4R$ and $R$, respectively. If the speed of satellite A is $3V$, then the speed of satellite B will be:

(2010 Pre)

1. $\frac{3V}{2}$

2. $\frac{3V}{4}$

3. $6V$

4. $12V$

View Answer

Orbital speed is inversely proportional to the square root of the radius ($v \propto \frac{1}{\sqrt{r}}$). Thus, $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}} = \sqrt{\frac{4R}{R}} = 2$, which means $v_B = 2 \times 3V = 6V$. Therefore, option C is correct.

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Question 78: moderate

For a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth, Then escape velocity for this planet will be:

(2000)

1. $11.2\text{ km/s}$

2. $22.4\text{ km/s}$

3. $5.6\text{ km/s}$

4. $44.8\text{ km/s}$

View Answer

Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. Since mass is constant and radius becomes one-fourth, $v_e$ increases by a factor of $\sqrt{4} = 2$. Thus, $v_p = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$. Option B is correct.

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Question 79: moderate

The escape velocity of a body on the surface of the earth is $11.2\text{ km/s}$. If the earth’s mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes:

(1997)

1. $22.4\text{ km/s}$

2. $44.8\text{ km/s}$

3. $5.6\text{ km/s}$

4. $11.2\text{ km/s}$

View Answer

Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. When mass becomes $2M$ and radius becomes $R/2$, $v'_e = \sqrt{\frac{2G(2M)}{R/2}} = 2v_e$. Thus, $v'_e = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$.

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