Question 329: Gravitation

Question

The height at which the weight of the body becomes $$\left(\frac{1}{9}ight)^{	ext {th }}$$ its weight on the surface of earth is $$(\mathrm{R}=$$ radius of earth)

Accepted Answer

$$\begin{array}{ll} 
& \mathrm{W}_{\mathrm{h}}=\frac{\mathrm{W}}{9} \
& \mathrm{mg}_{\mathrm{h}}=\frac{\mathrm{mg}}{9} \
& \mathrm{~g}_{\mathrm{h}}=\frac{\mathrm{g}}{9} \
	herefore \quad & \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{GM}}{9 \mathrm{R}^2} \
	herefore \quad & \mathrm{R}+\mathrm{h}=3 \mathrm{R} \
	herefore \quad & \mathrm{h}=2 \mathrm{R}
\end{array}$$

Answer

8R

Answer

4R

Answer

3R

Answer

2R

Gravitation

The height at which the weight of the body becomes $\left(\frac{1}{9}\right)^{\text {th }} its weight on the surface of earth is (\mathrm{R}=$ radius of earth)

Choose the correct answer: