Question 95: Gravitation

Question

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as $$R$$.)

Accepted Answer

$$\begin{aligned}
& g_p=\frac{g R^2}{(R+h)^2} \
& g_q=g\left(1-\frac{h}{R}ight) \
& g_p=g_q \
& \frac{g}{\left(1+\frac{h}{R}ight)^2}=g\left(1-\frac{h}{R}ight) \
& \left(1-\frac{h^2}{R^2}ight)\left(1+\frac{h}{R}ight)=1
\end{aligned}$$
Take $$\frac{\mathrm{h}}{\mathrm{R}}=\mathrm{x}$$
So
$$\begin{aligned}
& \mathrm{x}^3-\mathrm{x}+\mathrm{x}^2=0 \
& \mathrm{x}=\frac{\sqrt{5}-1}{2} \
& \mathrm{~h}=\frac{\mathrm{R}}{2}(\sqrt{5}-1)
\end{aligned}$$

Answer

$$\frac{\sqrt{3} R-R}{2}$$

Answer

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

Answer

$$\frac{\sqrt{5} R-R}{2}$$

Answer

$$\sqrt{5} R-R$$

Gravitation

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as $R$.)

Choose the correct answer: