Gravitation

Practice Questions:

Question 201: easy

The acceleration due to gravity at a height $1 \text{ km}$ above the earth is the same as at a depth d below the surface of earth. Then:

(2017-Delhi)

1. $d = 1 \text{ km}$

2. $d = \frac{3}{2} \text{ km}$

3. $d = 2 \text{ km}$

4. $d = \frac{1}{2} \text{ km}$

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For heights much smaller than the radius ($h \ll R$), $g_h \approx g(1 - \frac{2h}{R})$.nFor depth $d$, $g_d = g(1 - \frac{d}{R})$. Equating the two gives $1 - \frac{2h}{R} = 1 - \frac{d}{R}$.nThus $d = 2h$. Since $h = 1 \text{ km}$, $d = 2 \times 1 = 2 \text{ km}$.

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Question 202: moderate

The height at which the weight of a body becomes $1/16^{\text{th}}$, its weight on the surface of earth (radius R), is:

(2012 Pre)

1. $5R$

2. $15R$

3. $3R$

4. $4R$

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Weight at height $h$ is given by $W_h = \frac{W}{(1 + \frac{h}{R})^2}$. Given $W_h = \frac{W}{16}$, we equate: $\frac{1}{16} = \frac{1}{(1 + \frac{h}{R})^2}$. Taking the square root gives $$ 1 + \frac{h}{R} = 4 \Rightarrow \frac{h}{R} = 3 \Rightarrow h = 3R$$.

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Question 203: moderate

A body of weight $72 \text{ N}$ moves from the surface of earth to a height half of the radius of the earth, then gravitational force exerted on it will be:

(2000)

1. $36 \text{ N}$

2. $32 \text{ N}$

3. $144 \text{ N}$

4. $50 \text{ N}$

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Gravitational force (weight) at height $h$ is $F = \frac{W}{(1 + \frac{h}{R})^2}$.nSubstitute $h = \frac{R}{2}$ to get $F = \frac{72}{(1 + 0.5)^2}$.n$F = \frac{72}{2.25} = 32 \text{ N}$.

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Question 204: moderate

A body of mass $60 \text{ g}$ experiences a gravitational force of $3.0 \text{ N}$, when placed at a particular point. The magnitude of the gravitational field intensity at that point is:

(2022)

1. $180 \text{ N/kg}$

2. $0.05 \text{ N/kg}$

3. $50 \text{ N/kg}$

4. $20 \text{ N/kg}$

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Gravitational field intensity $E$ is given by $E = \frac{F}{m}$.nConvert mass to kg: $m = 60 \text{ g} = 0.06 \text{ kg}$.nSubstitute the values: $$E = \frac{3.0 \text{ N}}{0.06 \text{ kg}} = 50 \text{ N/kg}$$.

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Question 205: moderate

If the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to:

(1994, 89)

1. $R$

2. $R^0$ (independent of R)

3. $\frac{1}{R^2}$

4. $\frac{1}{R}$

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Centripetal force is provided by the given gravitational force: $$\frac{mv^2}{R} = \frac{k}{R}$$.

Solving for $v$, we get $v^2 = \frac{k}{m}$.nSince $k$ and $m$ are constants, $v$ is independent of $R$, meaning $v \propto R^0$.

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Question 206: moderate

The mean radius of earth is R, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth’s surface is g. What will be the radius of the orbit of a geostationary satellite?

(1992)

1. $(\frac{R^2g}{\omega^2})^{\frac{1}{3}}$

2. $(\frac{Rg}{\omega^2})^{\frac{1}{3}}$

3. $(\frac{R^2\omega^2}{g})^{\frac{1}{3}}$

4. $(\frac{R^2g}{\omega})^{\frac{1}{3}}$

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Gravitational force provides the centripetal force: $$\frac{GMm}{r^2} = m\omega^2r$$.nSince $$g = \frac{GM}{R^2}$$, we can write $$GM = gR^2$$.nSubstituting this gives $$\frac{gR^2}{r^2} = \omega^2r \Rightarrow r^3 = \frac{gR^2}{\omega^2} \Rightarrow r = (\frac{R^2g}{\omega^2})^{\frac{1}{3}}$$.

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Question 207: moderate

A body weighs $72 \text{ N}$ on the surface of the earth. What is the gravitation force on it, at a height equal to half the radius of the earth?

(2020)

1. $32 \text{ N}$

2. $30 \text{ N}$

3. $24 \text{ N}$

4. $48 \text{ N}$

View Answer

The weight at height $h$ is given by $$W_h = \frac{W}{(1 + \frac{h}{R})^2}$$.nSubstituting $h = \frac{R}{2}$, we get $$W_h = \frac{72}{(1 + \frac{1}{2})^2} = \frac{72}{(\frac{3}{2})^2}$.n$W_h = 72 \times \frac{4}{9} = 32 \text{ N}$$.

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Question 208: moderate

The work done to raise a mass $m$ from the surface of the earth to a height $h$, which is equal to the radius of the earth, is:

(2019)

1. $mgR$

2. $2mgR$

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

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

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Work done $W = \Delta U = \frac{mgh}{1+h/R}$. Substituting $h = R$, we get $W = \frac{mgR}{1+1} = \frac{1}{2}mgR$.

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Question 209: easy

At what height from the surface of earth the gravitation potential and the value of $g$ are $-5.4 \times 10^{7} \text{ J kg}^{-1}$ and $6.0 \text{ ms}^{-2}$ respectively. Take the radius of earth as $6400 \text{ km}$:

(2016 – I)

1. $2600 \text{ km}$

2. $1600 \text{ km}$

3. $1400 \text{ km}$

4. $2000 \text{ km}$

View Answer

Potential $V = -\frac{GM}{r} = -5.4 \times 10^{7}$ and $g = \frac{GM}{r^{2}} = 6.0$. Dividing $|V|$ by $g$ gives $r = 9000 \text{ km}$. Height $h = r - R = 9000 - 6400 = 2600 \text{ km}$.

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Question 210: 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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