First slide

Variation of g

Question

Let A and B be the points respectively above and below the earth's surface each at a distance equal to half the radius of the earth. If the acceleration due to gravity at these points be g_A and g_A respectively, then g_B : g_A is :

Easy

A

1:1
B

9:8
C

8:9
D

zero

Solution

$\begin{array}{l} g_{A} = \frac{GM}{(R + R / 2)^{2}} = \frac{4 GM}{9 R^{2}} = \frac{4}{9} g \\ g_{B} = g [1 - \frac{d}{R}] here d = \frac{R}{2}; g_{B} = \frac{g}{2} ∴ \frac{g_{B}}{g_{A}} = \frac{9}{8} \end{array}$

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