Question

What is the height from the surface of earth, where acceleration due to gravity will be \( \frac{1}{4} \) of that of the earth? (Re = 6400 km)

• A. \(6400 \, {km}\)
• B. \(3200 \, {km}\)
• C. \(1600 \, {km}\)
• D. \(640 \, {km}\)

Hint:

To find where gravity is a fraction \( f \) of its surface value, solve the equation \( \left(\frac{R_e}{R_e + h}\right)^2 = f \) for \( h \), which typically involves simple algebraic manipulation.

Answer 1

The Correct Option is B

To find the height \( h \) at which the acceleration due to gravity is \( \frac{1}{4} \) of its surface value \( g \), we use the formula for gravitational acceleration at height \( h \) above the Earth: \[ g_h = g \left(\frac{R_e}{R_e + h}\right)^2 \] Where: - \( g_h \) is the gravitational acceleration at height \( h \), - \( R_e \) is the radius of the Earth. Given that \( g_h = \frac{g}{4} \), we substitute this into the equation: \[ \frac{g}{4} = g \left(\frac{R_e}{R_e + h}\right)^2 \] Now, we cancel \( g \) from both sides and simplify: \[ \frac{1}{4} = \left(\frac{R_e}{R_e + h}\right)^2 \] Taking the square root of both sides: \[ \frac{1}{2} = \frac{R_e}{R_e + h} \] Now, cross-multiply to solve for \( h \): \[ R_e = 2(R_e + h) \] \[ R_e = 2R_e + 2h \] \[ 2h = R_e \] \[ h = \frac{R_e}{2} = 3200 \, {km} \] Thus, the height where the acceleration due to gravity is \( \frac{1}{4} \) of its surface value is \( \boxed{3200 \, {km}} \).