Rankers Physics
Topic: Gravitation
Subtopic: Acceleration Due to Gravity and its variation

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 earth when both d and h are much smaller than the radius of the earth, which one of the following is correct ?
d = h/2
d = 3h/2
d = 2h
d = h

Solution:

The change in the value of \( g \) at height \( h \) above the surface of the Earth and at depth \( d \) below the surface can be expressed using the following formulas:

1. **At height \( h \)**:
\[
g_h = g \left(1 - \frac{2h}{R}\right) \quad \text{(for small } h\text{)}
\]

The change in \( g \) is:
\[
\Delta g_h = g - g_h = g \frac{2h}{R} = \frac{2gh}{R}.
\]

2. **At depth \( d \)**:
\[
g_d = g \left(1 - \frac{d}{R}\right) \quad \text{(for small } d\text{)}
\]

The change in \( g \) is:
\[
\Delta g_d = g - g_d = g \frac{d}{R}.
\]

Setting the changes equal gives:

\[
\frac{2gh}{R} = \frac{g d}{R}.
\]

Cancelling \( g \) and \( R \) (assuming they are non-zero):

\[
2h = d.
\]

Thus, the relation between depth \( d \) and height \( h \) is:

\[
d = 2h.
\]

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