Question

If mass of earth decreases by 25% and its radius increases by 50%, then acceleration due to gravity at its surface decreases by nearly
A. 89%
B. 67%
C. 33%
D. 11%

Answer 1

Here we will proceed by using the formula of acceleration due to gravity (g) i.e., $g = \dfrac{{GM}}{{{r^2}}}$. Then we will substitute the converted values of new mass and radius using the given decreased value of mass i.e. 25% and increased value of radius i.e. 50% to get the required answer.
Formula used: $g = \dfrac{{GM}}{{{r^2}}}$

Complete answer:
Firstly, we know the formula of acceleration due to gravity- value of g on earth.
Acceleration due to gravity is the acceleration gained by an object due to the gravitational force. Its SI unit of acceleration due to gravity is $\dfrac{m}{{{s^2}}}$. It has both magnitude and direction. It is a vector quantity.
The formula of acceleration due to gravity (g) is $g = \dfrac{{GM}}{{{r^2}}}$.
Where G = universal gravitational constant
M = mass of the earth
r = radius of the earth
Acceleration due to gravity depends on the mass and radius of the earth.
Here, we are given that
The mass of the earth decreases by 25%, then the remaining 75% of mass is
$M' = \dfrac{{75}}{{100}}M$
Also given- Radius of earth increase by 50%,
So, new radius $R' = 150\% {\text{ of }}R = \dfrac{{150}}{{100}}R$
Now, substituting the values of $M'$ and $R'$ in the formula of acceleration due to gravity,
We get, (here g’ is the rate at which the acceleration due to gravity is decreasing)
$\Rightarrow g' = \dfrac{{GM'}}{{{{\left( {R'} \right)}^2}}} = \dfrac{{G\left( {\dfrac{{75}}{{100}}M} \right)}}{{{{\left( {\dfrac{{150R}}{{100}}} \right)}^2}}}$
By cross-multiplying the above equation, we get

$$\Rightarrow g' = \dfrac{{GM}}{{{R^2}}}\left( {\dfrac{{75}}{{100}} \times \dfrac{{100 \times 100}}{{150 \times 150}}} \right) \\\ \Rightarrow g' = g\left( {\dfrac{5}{{15}}} \right){\text{ }}\left[ {\because g = \dfrac{{GM}}{{{R^2}}}} \right] \\\ \therefore g' = 0.33g \\\$$

The acceleration due to gravity decreases by
= g – 0.33g
= 0.67g
Hence acceleration due to gravity decreases by a percent of

$$= 0.67 \times 100\% \\\ = 67\% \\\$$

Therefore, the acceleration due to gravity at its surface decreases by nearly 67%

Hence, option B is correct.

Note:
While solving this question, we must know that the standard value of g on the surface of the earth at sea level is $9.8m/{s^2}$. Also, we must that the value of universal gravitational constant (g) is $\left( {6.67 \times {{10}^{ - 1}}N{M^2}/k{g^2}} \right)$.