Question

The height $‘h’$ at which the weight of a body will be the same as that at the same depth $‘h’$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

• A. $\frac{\sqrt{3R}-R}{2}$
• B. $\frac{\sqrt{5}}{2}R-R$
• C. $\frac{\sqrt{5}R-R}{2}$
• D. $\frac{R}{2}$

Answer 1

The correct option is C

$\frac{\sqrt{5}R-R}{2}$

Solution:

Step 1: Given information:

The weight will be the same at a certain height as it will be at a certain depth below the earth's surface.
A body's weight can be calculated by multiplying its mass by the acceleration it experiences as a result of gravity.

$W=mg$

Step 2: Formula used and draw the required diagram:

Regardless of location, the body's mass is constant. As a result, since the body's weight should remain constant, the value of the acceleration caused by gravity should also remain constant.

Now the acceleration a body experiences as a result of gravity when it is elevated above earth's surface,

$g_h=\frac{g{R}^2}{{\left(R+h\right)}^2}...........................\left(1\right)$

And the acceleration of a body caused by gravity at a certain depth below the earth's surface, or depth,

$g_d=g\left(1-\frac{d}{R}\right)$

Note that the weight of the body equals some height and the depth, so

$g_d=g\left(1-\frac{h}{R}\right)..........................\left(2\right)$

Step 3: Formula used and substitute the values in the formula$g_d=g_h$

Since the body's weight should remain constant, the value of the acceleration caused by gravity should also remain constant.

Therefore from the equation $\left(1\right)$ and $\left(2\right)$.

$\begin{array}{rcl}\frac{g{R}^2}{{\left(R+h\right)}^2}& =& g\left(1-\frac{h}{R}\right)\\ \frac{R^2}{{\left(R+h\right)}^2}& =& \left(1-\frac{h}{R}\right)\\ \left(\frac{R^2+h^2+2hR}{R^2}\right)\left(1-\frac{h}{R}\right)& =& 1\end{array}$

Cancel out common terms and solve ,

$\begin{array}{rcl}\left(1+\frac{h^2}{R^2}+\frac{2h}{R}\right)\left(1-\frac{h}{R}\right)& =& 1\\ -\frac{h^2}{R^2}-\frac{h^3}{R^3}+\frac{h}{R}& =& 0\\ \left(-\frac{h}{R}\right)\left(\frac{h^2}{R^2}+\frac{h}{R}-1\right)& =& 0\\ h^2+hR-R^2& =& 0\end{array}$

Step 4: Solve the quadratic equation:

$\begin{array}{rcl}& \Rightarrow & h^2+hR-R^2=0\\ & \Rightarrow & h=\frac{-R\pm \sqrt{R^2+4R^2}}{2\times 1}\\ & \Rightarrow & h=\frac{\sqrt{5R}-R}{2}\end{array}$

Therefore, The height from the earth's surface where the weight will be equal to that at the same depth is$\frac{\sqrt{5}R-R}{2}$.

Hence, the correct option is (C).