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 $$__________$$.

Wext $$=$$Δ$$U=Uf$$−$$Ui=$$−$$GM(R+h)$$−−GMRΔ$$U=mgh1+hR=mgR1+Rh=mgR2$$Δ$$U=2$$×$$10$$×$$6.4$$×$$1062$$⇒Δ$$U=6.4$$×$$107$$ joule ⇒$$W=$$Δ$$U=64$$ Mega Joule

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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 __________.

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

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Gravitational potential and potential energy

Remember concepts with our Masterclasses.

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 __________.

Wext =ΔU=Uf−Ui=−GM(R+h)−−GMRΔU=mgh1+hR=mgR1+Rh=mgR2ΔU=2×10×6.4×1062⇒ΔU=6.4×107 joule ⇒W=ΔU=64 Mega Joule

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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 __________.