First slide

Variation of g

Question

Weight of a body at a height R/2 (Radius of earth = R) above the surface of earth is 80 N. What will be its weight at a depth R/2 below te surface of earth?

Moderate

A

75 N
B

25 N
C

90 N
D

120 N

Solution

$\begin{array}{l} W_{1} = m g_{1} = m \frac{G M}{r_{1}^{2}} \\ W_{2} = m g_{2} = m . g_{s} \frac{r_{2}}{R} = m \frac{G M}{R^{3}} r_{2} \end{array}$
$∴ \frac{W_{2}}{W_{1}} = \frac{r_{2}}{R^{3}} \times r_{1}^{2} = \frac{\frac{R}{2} \times {(R + \frac{R}{2})}^{2}}{R^{3}} = \frac{9}{8}$
$\Rightarrow W_{2} = 80 \times \frac{9}{8} N = 90 N$

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