Question 334: Gravitation

Question

A body of mass '$$\mathrm{m}$$' kg starts falling from a distance 3R above earth's surface. When it reaches a distance '$$R$$' above the surface of the earth of radius '$$R$$' and Mass '$$M$$', then its kinetic energy is

Accepted Answer

Initial height:
$$h=3 R+R=4 R$$
The potential energy of the body initially will be:
$$\mathrm{U}_1=-\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}$$
$$	herefore \quad$$ At the height $$\mathrm{R}$$,
$$\mathrm{h}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}$$
Potential energy:
$$\mathrm{U}_2=-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}$$
Gain in kinetic energy is equal to loss in potential energy.
$$\begin{aligned}
	herefore \quad \mathrm{KE} & =\mathrm{U}_1-\mathrm{U}_2 \
& =-\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}-\left(-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}ight)=\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}
\end{aligned}$$

Answer

$$\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}}$$

Answer

$$\frac{1}{3} \frac{\mathrm{GMm}}{\mathrm{R}}$$

Answer

$$\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}$$

Answer

$$\frac{1}{4} \frac{\mathrm{GMm}}{\mathrm{R}}$$

Gravitation

A body of mass '$\mathrm{m}' kg starts falling from a distance 3R above earth's surface. When it reaches a distance 'R' above the surface of the earth of radius 'R' and Mass 'M$', then its kinetic energy is

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