Question

A 90 kg body placed at $2R$ distance from the surface of the earth experiences gravitational pull of:
($R$ = Radius of Earth, $g = 10 \, \text{ms}^{-2}$).

• A. 300 N
• B. 225 N
• C. 120 N
• D. 100 N

Answer 1

Answer: (D)

100 N

Answer 2

1. Calculate Gravitational Acceleration at 2R from the Earth’s Surface:
The gravitational acceleration $g$ at a height $h$ from the Earth’s surface is given by:
$g = g_s \left(1 + \frac{h}{R}\right)^{-2},$ where $g_s$ is the gravitational acceleration at the Earth’s surface and $R$ is the Earth’s radius.
2. Substitute $h = 2R$:
$g = g_s \left(1 + \frac{2R}{R}\right)^{-2} = g_s (3)^{-2} = \frac{g_s}{9}.$ Here, $g_s = 10 \, \text{m/s}^2$.
3. Calculate the Gravitational Force:
The gravitational force $F$ acting on the 90 kg body is:
$F = mg = 90 \times \frac{g_s}{9} = 90 \times \frac{10}{9} = 100 \, \text{N}.$

Answer: 100 N