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  • A body of mass is taken from earth surface to the height h equal to twice the radius of earth (), the increase in potential

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

Option: 1

3 mgR_{e}


Option: 2

\frac{1}{3}mgR_{e}


Option: 3

\frac{2}{3}mgR_{e}


Option: 4

\frac{1}{2}mgRe


Answers (1)

best_answer

h = 2 Re
                        
\begin{aligned} & \Delta \mathrm{U}=\mathrm{U}_{\mathrm{B}}-\mathrm{U}_{\mathrm{A}} \\ & =\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}-\left(\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}\right) \\ & =\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}+2 \mathrm{R}_{\mathrm{e}}}+\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}=\frac{2}{3} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}} \\ & =\frac{2}{3} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^2} \mathrm{R}_{\mathrm{e}} \\ & \Delta \mathrm{U}=\frac{2}{3} \mathrm{mg} \mathrm{R}_{\mathrm{c}} \end{aligned}

Posted by

Divya Prakash Singh

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