Question

The height at which the weight of a person becomes 25% of his weight on the surface of the earth is: (R$_E$ is the radius of the earth)

• A. $\frac{R_E}{2}$
• B. $\frac{R_E}{4}$
• C. $\frac{R_E}{8}$
• D. $2R_E$

Hint:

Gravitational force decreases with distance as $\frac{1}{r^2}$. When weight changes, set up the ratio and solve for the height above the surface.

Answer 1

The Correct Option is D

The weight of a person is due to gravitational force, $W = \frac{G M m}{r^2}$, where $r$ is the distance from the center of the Earth.
On the surface, $r = R_E$, so $W_{{surface}} = \frac{G M m}{R_E^2}$.
At height $h$, $r = R_E + h$, and the weight is 25% of the surface weight: $\frac{G M m}{(R_E + h)^2} = 0.25 \times \frac{G M m}{R_E^2}$.
Simplifying, $\frac{1}{(R_E + h)^2} = \frac{0.25}{R_E^2}$, so $(R_E + h)^2 = 4 R_E^2$.
Taking the square root, $R_E + h = 2 R_E$, thus $h = R_E$.
The distance from the center is $R_E + h = 2 R_E$, so the height $h = R_E$, which corresponds to $2R_E$ from the center.