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NEET PHYSICSEasy

A body weighs 72 N72 \text{ 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?

A

32 N32 \text{ N}

B

30 N30 \text{ N}

C

24 N24 \text{ N}

D

48 N48 \text{ N}

Step-by-Step Solution

  1. Formula: The gravitational force (weight) WhW_h acting on a body of mass mm at a height hh above the surface of the Earth is given by: Wh=mgh=mg(RR+h)2W_h = m g_h = m g \left( \frac{R}{R+h} \right)^2 where W=mgW = mg is the weight on the surface and RR is the radius of the Earth [Eq. 7.13, 7.14].
  2. Given Data:
  • Weight on surface, W=72 NW = 72 \text{ N}.
  • Height, h=R2h = \frac{R}{2}.
  1. Calculation: Substitute h=R/2h = R/2 into the formula: Wh=72(RR+R2)2W_h = 72 \left( \frac{R}{R + \frac{R}{2}} \right)^2 Wh=72(R3R2)2W_h = 72 \left( \frac{R}{\frac{3R}{2}} \right)^2 Wh=72(23)2W_h = 72 \left( \frac{2}{3} \right)^2 Wh=72(49)W_h = 72 \left( \frac{4}{9} \right) Wh=8×4=32 NW_h = 8 \times 4 = 32 \text{ N}
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