Question 310: Gravitation

Question

The height at which the weight of the body becomes $\frac{1^{	ext {th }}}{16}$ of its weight on the surface of the earth of radius ' $R$ ' is

Accepted Answer

To determine the height at which the weight of a body becomes $\frac{1}{16}$ of its weight on the surface of the Earth, we start by using the formula for gravitational force:
$  F = \frac{G M m}{r^2}  $
Where:
$ F $ is the gravitational force (weight),
$ G $ is the gravitational constant,
$ M $ is the mass of the Earth,
$ m $ is the mass of the body,
$ r $ is the distance from the center of the Earth.
On the surface of the Earth, the weight $ W_0 $ is given by:
$  W_0 = \frac{G M m}{R^2}  $
At a height $ h $ above the Earth's surface, the new distance from the center of the Earth becomes $ R + h $. The weight at this height, $ W_h $, is:
$  W_h = \frac{G M m}{(R + h)^2}  $
According to the problem, $ W_h = \frac{1}{16} W_0 $. Plugging in the expressions for $ W_0 $ and $ W_h $:
$ \frac{G M m}{(R + h)^2} = \frac{1}{16} \frac{G M m}{R^2}  $
Canceling $ G M m $ from both sides and solving for $ h $:
$  \frac{1}{(R + h)^2} = \frac{1}{16 R^2}  $
Taking the reciprocal gives:
$  (R + h)^2 = 16 R^2  $
Taking the square root of both sides:
$  R + h = 4R  $
Solving for $ h $:
$ h = 4R - R = 3R  $
Thus, the height at which the weight becomes $\frac{1}{16}$ of its weight on the surface is:
Option B: 3R

Answer

2 R

Answer

3 R

Answer

4 R

Answer

5 R

Gravitation

The height at which the weight of the body becomes \frac{1^{\text {th }}}{16} of its weight on the surface of the earth of radius ' R ' is

Choose the correct answer: