Question 383: Gravitation

Question

A body weighs 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:

Accepted Answer

To find the gravitational force on a body at a height equal to one-third the Earth's radius from its surface, we start with the weight of the body on the EarthÃ¢ÂÂs surface, which is 48 N.
The gravitational force at the surface, $ W $, is given by:
$ W = mg $
Where $ g = \frac{GM}{R^2} $.
At a height $ h $ above the Earth's surface, the gravitational force $ g_h $ is:
$ g_h = \frac{GM}{(R+h)^2} $
To find the weight $ W_h $ at this height, the ratio of gravitational forces at height $ h $ and at the surface is:
$ \frac{W_h}{W} = \frac{g_h}{g} = \frac{R^2}{(R+h)^2} $
Given $ h = \frac{R}{3} $, substitute $ h $ into the equation:
$ \frac{W_h}{W} = \frac{R^2}{\left(R+\frac{R}{3}ight)^2} = \frac{R^2}{\left(\frac{4R}{3}ight)^2} = \frac{9}{16} $
Thus, the new weight $ W_h $ is:
$ W_h = \frac{9}{16} 	imes W = \frac{9}{16} 	imes 48 \, 	ext{N} $
$ W_h = 27 \, 	ext{N} $
Therefore, at this height, the gravitational force experienced by the body is 27 N.

Answer

32 N

Answer

36 N

Answer

16 N

Answer

27 N

Gravitation

A body weighs 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:

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