Question 131: Gravitation

Question

A body of mass is taken from earth surface to the height h equal to twice the radius of earth (R$$_e$$), the increase in potential energy will be :
(g  = acceleration due to gravity on the surface of Earth)

Accepted Answer

$$
\begin{aligned}
& V_{	ext {surface }}=-\left(\frac{G M m}{R_{e}}ight) \\
& V_{p}=-\frac{G M m}{3 R_{e}} \\
& \Delta V=\frac{G M m}{R_{e}}\left(1-\frac{1}{3}ight) \\
& =\frac{2}{3} \frac{G M m}{\left(R_{e}^{2}ight)} 	imes R_{e} \\
& \Delta V=\frac{2}{3} m g R_{e}
\end{aligned}
$$

Answer

$$\frac{1}{2}mgR_e$$

Answer

$$3~mgR_e$$

Answer

$$\frac{1}{3}mgR_e$$

Answer

$$\frac{2}{3}mgR_e$$

Gravitation

A body of mass is taken from earth surface to the height h equal to twice the radius of earth (R$_e$), the increase in potential energy will be : (g = acceleration due to gravity on the surface of Earth)

Choose the correct answer: