Question

If the radius of the Earth contracted by 0.1%, it’s mass remains the same. Then weight of the body at Earth’s surface will increase by-

& (1).\,0.1\% \\\ & (2).\,0.2\% \\\ & (3)\,0.3\% \\\ & (4)\,remains\,the\,same \\\ \end{aligned}$$

Answer 1

Weight depends on acceleration due to gravity. Using Newton’s Law of gravitation determine the relation between $g$ and $R$ and use it to calculate the change in acceleration due to gravity. We can use that change to calculate the change in weight of the body on the surface of Earth in the given conditions.

Formula used:
$g=\dfrac{GM}{{{R}^{2}}}$
$W=mg$

Complete step by step solution:
The weight ($W$ ) of an object on the surface of the Earth is the gravitational force acting on it due to the Earth.
It is given by-
$W=mg$
Here, $m$ is the mass of the object
$g$ is the acceleration due to gravity.
Acceleration due to gravity depends on the radius as-
$g=\dfrac{GM}{{{R}^{2}}}$ ………….. (1)
Here, $G$ is the gravitational constant and its value is $6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$
$M$ is the mass of Earth and its value is $6\times {{10}^{24}}kg$
$R$ is the radius of Earth.
When radius contracts by $0.1$ , the new $g$ will be-
$g'=\dfrac{GM}{{{(R-0.1R)}^{2}}}$ ………………...(2)
From eq (1) and eq (2)

& \Delta g=\dfrac{GM}{{{\left( R-\dfrac{0.1}{100}R \right)}^{2}}}-\dfrac{GM}{{{R}^{2}}} \\\ & =GM(\dfrac{{{R}^{2}}-{{\left( R-\dfrac{0.1}{100}R \right)}^{2}}}{{{R}^{2}}\cdot {{\left( R-\dfrac{0.1}{100}R \right)}^{2}}})\, \\\ & \Delta g=\dfrac{GM}{{{R}^{2}}}\left( 1-{{\left( 1-\dfrac{0.1}{100} \right)}^{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[{{\left( R-\dfrac{0.1}{100}R \right)}^{2}}\,is\,neglected] \\\ & \Delta g=\dfrac{GM}{{{R}^{2}}}\left( 1-\left( 1-2\dfrac{0.1}{100} \right) \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[By\,rules\,of\,\exponents] \\\ & \Rightarrow \Delta g=\dfrac{0.2g}{100} \\\ & \therefore \Delta g=0.2\%\,of\,g\,\,\,\, \\\ \end{aligned}$$ $$g$$ changes by $$0.2%$$ . So the weight will change by- $$\begin{aligned} & \Delta W=W'-W=mg'-mg \\\ & \Delta W=m(g'-g)=m\Delta g \\\ & \Delta W=m\left( \dfrac{0.2g}{100} \right)=\dfrac{0.2W}{100} \\\ & \\\ & \\\ \end{aligned}$$ Therefore, the weight change is $$0.2%\,of\,W$$ , **So, the correct answer is “Option (2)”.** **Note:** When the mass is constant, the acceleration of gravity for a body is inversely proportional to the square of the radius of the body. But when the radius is constant, acceleration due to gravity will be directly proportional to the mass of the body. Weight will be constant as long as $$g$$ is constant.