Question

Find the height at which the weight will be the same as at the same depth from the surface of the earth.

& A.\,\dfrac{R}{2} \\\ & B.\,\sqrt{5}R-R \\\ & C.\,\dfrac{\sqrt{5}R-R}{2} \\\ & D.\,\dfrac{\sqrt{3}R-R}{2} \\\ \end{aligned}$$

Answer 1

We have to equate the equation of the acceleration due to the gravity of a body at some height on the surface of the earth with the acceleration due to the gravity of a body at some depth below the surface of the earth. By solving this equation, we will get the expression for the height.
Formula used:
${{g}_{h}}=\dfrac{g{{R}^{2}}}{{{(R+h)}^{2}}}$
${{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)$

Complete answer:
From the given information, we have the data as follows.
At some height, the weight will be the same as at the same depth from the surface of the earth.
The weight of a body can be represented as the product of the mass of the body and the acceleration due to gravity.
$W=mg$
The mass of the body remains constant irrespective of the place. Thus, the value of the acceleration due to gravity should remain the same as the weight of the body is supposed to remain the same.

The acceleration due to the gravity of a body at some height on the surface of the earth is given by the formula as follows.
${{g}_{h}}=\dfrac{g{{R}^{2}}}{{{(R+h)}^{2}}}$
The acceleration due to the gravity of a body at some depth below the surface of the earth, that is, depth is given by the formula as follows.
${{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)$
As the weight of the body equals at some height and the depth, so, we have,
${{g}_{d}}=g\left( 1-\dfrac{h}{R} \right)$
As the value of the acceleration due to gravity should remain the same as the weight of the body is supposed to remain the same.
${{g}_{h}}={{g}_{d}}$
Substitute the equations in the above formula
$\dfrac{g{{R}^{2}}}{{{(R+h)}^{2}}}=g\left( 1-\dfrac{h}{R} \right)$
Continue further computation.

& \dfrac{{{R}^{2}}}{{{(R+h)}^{2}}}=\left( 1-\dfrac{h}{R} \right) \\\ & \Rightarrow \dfrac{1}{{{R}^{2}}}{{(R+h)}^{2}}\left( 1-\dfrac{h}{R} \right)=1 \\\ & \therefore \left( \dfrac{{{R}^{2}}+{{h}^{2}}+2hR}{{{R}^{2}}} \right)\left( 1-\dfrac{h}{R} \right)=1 \\\ \end{aligned}$$ Cancel out common terms. $$\begin{aligned} & \left( \dfrac{{{R}^{2}}}{{{R}^{2}}}+\dfrac{{{h}^{2}}}{{{R}^{2}}}+\dfrac{2hR}{{{R}^{2}}} \right)\left( 1-\dfrac{h}{R} \right)=1 \\\ & \therefore \left( 1+\dfrac{{{h}^{2}}}{{{R}^{2}}}+\dfrac{2h}{R} \right)\left( 1-\dfrac{h}{R} \right)=1 \\\ \end{aligned}$$ Take out the brackets. $$\begin{aligned} & 1+\dfrac{{{h}^{2}}}{{{R}^{2}}}+\dfrac{2h}{R}-\dfrac{h}{R}-\dfrac{{{h}^{3}}}{{{R}^{3}}}-\dfrac{2{{h}^{2}}}{{{R}^{2}}}=1 \\\ & \therefore -\dfrac{{{h}^{2}}}{{{R}^{2}}}-\dfrac{{{h}^{3}}}{{{R}^{3}}}+\dfrac{h}{R}=0 \\\ \end{aligned}$$ Take out the common terms. $$\begin{aligned} & \left( -\dfrac{h}{R} \right)\left( \dfrac{{{h}^{2}}}{{{R}^{2}}}+\dfrac{h}{R}-1 \right)=0 \\\ & \Rightarrow \left( \dfrac{{{h}^{2}}}{{{R}^{2}}}+\dfrac{h}{R}-1 \right)=0 \\\ & \therefore {{h}^{2}}+hR-{{R}^{2}}=0 \\\ \end{aligned}$$ Solve the above quadratic equation. $$\begin{aligned} & {{h}^{2}}+hR-{{R}^{2}}=0 \\\ & \Rightarrow h=\dfrac{-R\pm \sqrt{{{R}^{2}}+4{{R}^{2}}}}{2\times 1} \\\ & \therefore h=\dfrac{\sqrt{5}R-R}{2} \\\ \end{aligned}$$ $$\therefore $$ The height at which the weight will be the same as at the same depth from the surface of the earth is $$\dfrac{\sqrt{5}R-R}{2}$$. **Thus, option (C) is correct.** **Note:** As the options contain the parameter radius, thus, we need to obtain the expression for the height above the surface of the earth at which the weight equals the weight at some depth below the surface of the earth.