Question:

A 90 kg body placed at 2R 2R distance from the surface of the earth experiences gravitational pull of:
(R R = Radius of Earth, g=10ms2 g = 10 \, \text{ms}^{-2} ).

Updated On: Mar 22, 2025
  • 300 N
  • 225 N
  • 120 N
  • 100 N
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The Correct Option is D

Solution and Explanation

1. Calculate Gravitational Acceleration at 2R from the Earth’s Surface:
The gravitational acceleration g g at a height h h from the Earth’s surface is given by:
g=gs(1+hR)2, g = g_s \left(1 + \frac{h}{R}\right)^{-2}, where gs g_s is the gravitational acceleration at the Earth’s surface and R R is the Earth’s radius.
2. Substitute h=2R h = 2R :
g=gs(1+2RR)2=gs(3)2=gs9. g = g_s \left(1 + \frac{2R}{R}\right)^{-2} = g_s (3)^{-2} = \frac{g_s}{9}. Here, gs=10m/s2 g_s = 10 \, \text{m/s}^2 .
3. Calculate the Gravitational Force:
The gravitational force F F acting on the 90 kg body is:
F=mg=90×gs9=90×109=100N. F = mg = 90 \times \frac{g_s}{9} = 90 \times \frac{10}{9} = 100 \, \text{N}.

Answer: 100 N

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