First slide

Gravitational potential and potential energy

Question

A small mass 2kg moved slowly from the surface of earth to a height of 6.4 x 10⁶ m above the earth. The work done [in Mega Joule] is __________.

Moderate

Solution

$\begin{array}{l} W_{ext} = ΔU = U_{f} - U_{i} \\ = (\frac{- GM}{(R + h)}) - (\frac{- GM}{R}) \\ ΔU = \frac{mgh}{1 + \frac{h}{R}} = \frac{mgR}{1 + \frac{R}{h}} = \frac{mgR}{2} \\ ΔU = \frac{2 \times 10 \times 6 .4 \times 10^{6}}{2} \Rightarrow ΔU = 6 .4 \times 10^{7} joule \\ \Rightarrow W = ΔU = 64 Mega Joule \end{array}$

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