Acceleration Due to Gravity and its variation

Practice Questions:

Question 11: moderate

For moon, its mass is $\frac{1}{81}$ of earth mass and its diameter is $\frac{1}{3.7}$ of earth diameter. If acceleration due to gravity at earth surface is $9.8\text{ m/s}^2$ then at moon its value is:

(1999)

1. $2.86\text{ m/s}^2$

2. $1.65\text{ m/s}^2$

3. $8.65\text{ m/s}^2$

4. $5.16\text{ m/s}^2$

View Answer

Using $g \propto \frac{M}{R^2}$, we have $g_m = g_e \times (\frac{M_m}{M_e}) \times (\frac{R_e}{R_m})^2$. Substituting the values: $g_m = 9.8 \times \frac{1}{81} \times (3.7)^2 \approx 1.65\text{ m/s}^2$.

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Question 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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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