Question

The acceleration due to gravity at a height of 7 km above the earth is the same as at a depth \( d \) below the surface of the earth. Then \( d \) is:

• A. 7 km
• B. 2 km
• C. 3.5 km
• D. 14 km

Hint:

When comparing gravity at different altitudes, remember to apply the formula for gravity at height and depth. Gravity decreases with height and increases with depth, and these effects are governed by the inverse square law and linear dependence, respectively.

Answer 1

The Correct Option is D

The acceleration due to gravity \( g \) at a height \( h \) above the surface of the earth and at a depth \( d \) below the surface can be given by the following relationships: At height \( h \), the acceleration due to gravity \( g_h \) is: \[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \] where \( R \) is the radius of the earth. At depth \( d \), the acceleration due to gravity \( g_d \) is: \[ g_d = g \left( 1 - \frac{d}{R} \right) \] where \( g \) is the acceleration due to gravity at the surface. Given that the gravity at the height \( h = 7 \, \text{km} \) is the same as at the depth \( d \), we can equate \( g_h = g_d \): \[ \frac{g}{(1 + \frac{7}{R})^2} = g \left( 1 - \frac{d}{R} \right) \] Simplifying: \[ \frac{1}{(1 + \frac{7}{R})^2} = 1 - \frac{d}{R} \] Now solving for \( d \). Let \( R = 6400 \, \text{km} \) (radius of Earth). First calculate \( 1 + \frac{7}{R} \) and square it: \[ 1 + \frac{7}{6400} = 1.00109375 \] \[ (1.00109375)^2 = 1.0021875 \] Now substitute this value back into the equation: \[ \frac{1}{1.0021875} = 1 - \frac{d}{6400} \] \[ 0.9978125 = 1 - \frac{d}{6400} \] \[ \frac{d}{6400} = 1 - 0.9978125 = 0.0021875 \] \[ d = 6400 \times 0.0021875 \approx 14 \, \text{km} \] Thus, the depth \( d \) is approximately 14 km.