Gravitation

Practice Questions:

Question 61: moderate

The radius of earth is about $6400\text{ km}$ and that of planet mars is $3200\text{ km}$. The mass of the earth is about $10$ times mass of planet mars. An object weighs $200\text{ N}$ on the surface of earth. Its weight on the surface of planet mars will be:

(1994)

1. $20\text{ N}$

2. $8\text{ N}$

3. $80\text{ N}$

4. $40\text{ N}$

View Answer

Gravity $g \propto \frac{M}{R^2}$. The ratio of weights is $W_m/W_e = (M_m/M_e) \times (R_e/R_m)^2 = (1/10) \times (6400/3200)^2 = 0.4$. Thus, the weight on Mars is $W_m = 0.4 \times 200 = 80\text{ N}$.

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Question 62: moderate

A body weighs $200 \text{ N}$ on the surface of the earth. How much will it weigh half way down to the centre of the earth?

(2019)

1. $150 \text{ N}$

2. $200 \text{ N}$

3. $250 \text{ N}$

4. $100 \text{ N}$

View Answer

The weight at depth $d$ is $W_d = W(1 - \frac{d}{R})$.nGiven $d = \frac{R}{2}$ (halfway to the center), we have $$W_d = 200(1 - \frac{1}{2})$.n$W_d = 200 \times \frac{1}{2} = 100 \text{ N}$$.

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Question 63: 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. $4R$

View Answer

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 64: 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. $50 \text{ N}$

View Answer

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 65: 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. $20 \text{ N/kg}$

View Answer

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 66: 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. $\frac{1}{R}$

View Answer

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

View Answer

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 68: 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 69: moderate

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)

(2020-Covid)

1. $\frac{R(n-1)}{n}$

2. $\frac{Rn}{(n-1)}$

3. $\frac{R}{n}$

4. $\frac{R}{n^2}$

View Answer

The acceleration due to gravity at depth $d$ is $g_d = g(1 - \frac{d}{R})$.nGiven $g_d = \frac{g}{n}$, we have $\frac{g}{n} = g(1 - \frac{d}{R})$. Solving for $d$: $$1 - \frac{d}{R} = \frac{1}{n} \Rightarrow \frac{d}{R} = \frac{n-1}{n} \Rightarrow d = \frac{R(n-1)}{n}$$.

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Question 70: 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. $\frac{3}{2}mgR$

View Answer

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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