Question 254: Gravitation

Question

A body weighs 45 N on the surface of the earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be

Accepted Answer

The gravitational force on a body is given by this formula:
$$ \mathrm{F} = \frac{\mathrm{GMm}}{\mathrm{r}^2} $$
On the Earth's surface, the distance from the center of Earth is $R$. So, the weight (force) is:
$$ \mathrm{W} = \frac{\mathrm{GMm}}{\mathrm{R}^2} = 45~\mathrm{N} $$
If the body is lifted to a height that equals half the Earth's radius ($h = \frac{R}{2}$), then the total distance from the center becomes:
Distance from center = $R + \frac{R}{2} = \frac{3R}{2}$
The new force of gravity at this distance is:
$$ \mathrm{F}' = \frac{\mathrm{GMm}}{(\frac{3R}{2})^2} $$
This simplifies to:
$$ \mathrm{F}' = \frac{\mathrm{GMm}}{\frac{9R^2}{4}} = \frac{4}{9} \cdot \frac{\mathrm{GMm}}{R^2} $$
We already know that $\frac{\mathrm{GMm}}{R^2} = 45~\mathrm{N}$, so substitute this value:
$$ \mathrm{F}' = \frac{4}{9} 	imes 45 = 20~\mathrm{N} $$

Answer

20 N

Answer

22.5 N

Answer

30 N

Answer

36 N

Gravitation

A body weighs 45 N on the surface of the earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be

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