JEE Main · 2005 · 08 May 2005 · Offline · Gravitation

The change in the value of

g

at a height '

h

' above the surface of the earth is the same as at a depth 'd' below the surface of earth. When both 'd' and '

h

' are much smaller than the radius of earth, then which one of the following is correct?

(A) $d = \frac{h}{2}$
(B) $d = \frac{3 h}{2}$
(C) $d = 2 h$ ✓ Correct
(D) $d = h$

Answer: (C)

Solution

Above the surface

g

falls as

g_{h} = g (1 - \frac{2 h}{R})

for small

h

, a change of

\frac{2 g h}{R}

; below the surface it falls as

g_{d} = g (1 - \frac{d}{R})

, a change of

\frac{g d}{R}

. Setting the two changes equal,

\frac{2 g h}{R} = \frac{g d}{R}

, so

d = 2 h

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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