Question

The change in the value of acceleration due to gravity (g) at a height h above the surface of the earth is same as that at a depth d below the surface. If both h and d are much smaller than the radius of the earth (R), then

• A. $d=h$
• B. $d=2h$
• C. $h=2d$
• D. $d=h/2$
• E. $d=h^{2}$

Hint:

Gravity decreases twice as fast as you go up into the air compared to as you go down into a mine. Thus, you only have to go half the distance into the Earth to see the same weight loss as you would going up into the atmosphere.

Answer 1

The Correct Option is B

Concept:
Physics - Variation of g with Altitude and Depth.
Step 1: State the formula for variation with height.
For $h \ll R$, the acceleration due to gravity at height $h$ ($g_h$) is: $$ g_h = g \left( 1 - \frac{2h}{R} \right) $$ The change in gravity is $\Delta g_h = g - g_h = \frac{2gh}{R}$.
Step 2: State the formula for variation with depth.
The acceleration due to gravity at depth $d$ ($g_d$) is: $$ g_d = g \left( 1 - \frac{d}{R} \right) $$ The change in gravity is $\Delta g_d = g - g_d = \frac{gd}{R}$.
Step 3: Equate the changes.
The problem states the changes are the same ($\Delta g_h = \Delta g_d$): $$ \frac{2gh}{R} = \frac{gd}{R} $$
Step 4: Simplify and solve.
Cancel $g$ and $R$ from both sides: $$ 2h = d \implies d = 2h $$