Question

A body weighs 45 N on the surface of the earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be ______.

• A. 20 N
• B. 22.5 N
• C. 30 N
• D. 36 N

Hint:

Never use the approximation $g' = g(1 - 2h/R)$ unless $h \ll R$ (less than 5% of the radius). For large heights like $h = R/2$, you MUST use the exact formula $g' = g [R / (R+h)]^2$ to get the correct answer!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the weight of a body on the Earth's surface and need to calculate its new weight when it is elevated to a height $h = R/2$, where $R$ is the Earth's radius.

Step 2: Detailed Explanation:
The weight of a body is the gravitational force acting on it, $W = mg$.
On the surface of the Earth, acceleration due to gravity is $g$.
$W_{\text{surface}} = mg = 45 \text{ N}$.
The acceleration due to gravity at a significant height $h$ above the Earth's surface is given by the formula:
$g' = g \left( \frac{R}{R + h} \right)^2$
We are given $h = \frac{R}{2}$. Substitute this into the formula:
$g' = g \left( \frac{R}{R + R/2} \right)^2$
$g' = g \left( \frac{R}{\frac{3R}{2}} \right)^2$
The $R$ terms cancel out perfectly:
$g' = g \left( \frac{1}{\frac{3}{2}} \right)^2$
$g' = g \left( \frac{2}{3} \right)^2$
$g' = \frac{4}{9} g$
The new weight ($W_{\text{height}}$) is the mass multiplied by the new gravity $g'$:
$W_{\text{height}} = m g' = m \left( \frac{4}{9} g \right)$
$W_{\text{height}} = \frac{4}{9} (mg)$
Substitute the original surface weight ($mg = 45 \text{ N}$):
$W_{\text{height}} = \frac{4}{9} \times 45$
$W_{\text{height}} = 4 \times 5 = 20 \text{ N}$

Step 3: Final Answer:
The gravitational force will be 20 N, matching option (a).