A body weighs $72 ext{ N}$ on the surface of the earth. What is the gravitational force on it at a height eq

Question

A body weighs $72 	ext{ N}$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth?

Accepted Answer

1. **Formula:** The gravitational force (weight) $W_h$ acting on a body of mass $m$ at a height $h$ above the surface of the Earth is given by:
   $$ W_h = m g_h = m g \left( \frac{R}{R+h} ight)^2 $$
   where $W = mg$ is the weight on the surface and $R$ is the radius of the Earth [Eq. 7.13, 7.14].
2. **Given Data:**
   - Weight on surface, $W = 72 	ext{ N}$.
   - Height, $h = \frac{R}{2}$.
3. **Calculation:**
   Substitute $h = R/2$ into the formula:
   $$ W_h = 72 \left( \frac{R}{R + \frac{R}{2}} ight)^2 $$
   $$ W_h = 72 \left( \frac{R}{\frac{3R}{2}} ight)^2 $$
   $$ W_h = 72 \left( \frac{2}{3} ight)^2 $$
   $$ W_h = 72 \left( \frac{4}{9} ight) $$
   $$ W_h = 8 	imes 4 = 32 	ext{ N} $$

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