Question 323: Gravitation

Question

A body of mass '$$\mathrm{m}$$' is raised through a height above the earth's surface so that the increase in potential energy is $$\frac{\mathrm{mgR}}{5}$$. The height to which the body is raised is ( $$\mathrm{R}=$$ radius of earth, $$\mathrm{g}=$$ acceleration due to gravity)

Accepted Answer

When a particle of mass $$m$$ is taken from the Earth's surface to a height $$h=n R$$, then the change in P.E. can be calculated as,
$$\begin{array}{ll} 
& \Delta \mathrm{U}=\mathrm{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}ight) \
	herefore & \frac{\mathrm{mgR}}{5}=\operatorname{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}ight) \
	herefore & \mathrm{n}+1=5 \mathrm{n} \
	herefore & \mathrm{n}=\frac{1}{4} \
	herefore & \mathrm{h}=\frac{\mathrm{R}}{4}
\end{array}$$

Answer

$$\mathrm{R}$$

Answer

$$\frac{\mathrm{R}}{2}$$

Answer

$$\frac{\mathrm{R}}{4}$$

Answer

$$\frac{\mathrm{R}}{8}$$

Gravitation

A body of mass '$\mathrm{m}' is raised through a height above the earth's surface so that the increase in potential energy is \frac{\mathrm{mgR}}{5}. The height to which the body is raised is ( \mathrm{R}= radius of earth, \mathrm{g}=$ acceleration due to gravity)

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