Question

If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth \(\alpha h \ (h \ll Re)\) from the earth surface. The value of α will be ________.
(Use R_e=6400 km)

Answer 1

Correct Answer: 2

Let \( g_0 \) be the acceleration due to gravity at the earth's surface. The acceleration due to gravity at a height \( h \) above the surface is given by: 

\( g_h = \frac{g_0}{(1 + \frac{h}{R_e})^2} \)

The acceleration due to gravity at a depth \( d = \alpha h \) below the surface is given by:

\( g_d = g_0 \left(1 - \frac{\alpha h}{R_e}\right) \)

According to the problem, \( g_h = g_d \). Therefore:

\[\frac{g_0}{(1 + \frac{h}{R_e})^2} = g_0 \left(1 - \frac{\alpha h}{R_e}\right)\]

We can cancel \( g_0 \) from both sides and solve the resulting equation:

\[\frac{1}{(1 + \frac{h}{R_e})^2} = 1 - \frac{\alpha h}{R_e}\]

For simplicity, let's approximate \((1+\frac{h}{R_e})^2\) using binomial expansion, considering \( h \ll R_e \):

\((1 + \frac{h}{R_e})^2 \approx 1 + \frac{2h}{R_e}\)

Thus, the equation simplifies to:

\[\frac{1}{1 + \frac{2h}{R_e}} \approx 1 - \frac{\alpha h}{R_e}\]

Expanding the left side using binomial approximation again:

\[1 - \frac{2h}{R_e} \approx 1 - \frac{\alpha h}{R_e}\]

Equating coefficients of \(\frac{h}{R_e}\):

\[2 = \alpha\]

Thus, the value of \(\alpha\) is \(2\), which fits within the given range of \(2,2\).

Answer 2

\(g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})\)
\(⇒ 2h = d\)
\(⇒ α = 2\)
So, the answer is 2.