PhysicsA small 2 kg mass moved slowly from the surface of earth to a height of 6.4 × 106m above theearth. Find the work done [in mega joule].

A small 2 kg mass moved slowly from the surface of earth to a height of $6.4 \times 10^{6} m$ above the
earth. Find the work done [in mega joule].

$W_{ext} = ΔU = U_{f} - U_{1} = (\frac{- GM}{(R + h)}) - (\frac{- GM}{R})$

$ΔU = \frac{mgh}{1 + \frac{h}{R}} = \frac{mgR}{1 + \frac{R}{h}} = \frac{mgR}{2} [h = R = 6.4 \times 10^{6} k m]$

$ΔU = \frac{2 \times 10 \times 6.4 \times 10^{6}}{2}$

$\Rightarrow ΔU = 6.4 \times 10^{7} Joule \Rightarrow W = ΔU = 64 MJ$

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