Question: 32. A body weight 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:
(1) 16 N
(2) 27 N
(3) 32 N
(4) 36 N
Answer: Option (2)
Explanation:
The weight of the body on the surface of the Earth is due to gravitational force.
On the surface, the gravitational force is given as:
F = 48 \text{ N}The gravitational force at a distance r from
the center of the Earth varies inversely as the square of the distance:
F \propto \frac{1}{r^{2}}Let the radius of the Earth be R. On the surface,
the distance from the center is r = R.
The given height above the Earth’s surface is \frac{R}{3}.
Hence, the distance of the body from the center of the Earth becomes:
r = R + \frac{R}{3} = \frac{4R}{3}Using the inverse square law of gravitation,
the force at this height F^{\prime} is related to the force on the surface as:
\frac{F^{\prime}}{F} = \left(\frac{R}{\frac{4R}{3}}\right)^{2}Simplifying:
\frac{F^{\prime}}{48} = \left(\frac{3}{4}\right)^{2} = \frac{9}{16}So,
F^{\prime} = 48 \times \frac{9}{16} F^{\prime} = 27 \text{ N}Therefore, the gravitational force experienced by the body at
a height equal to one-third the radius of the Earth is 27 N.
Hence, the correct answer is Option (2).