A probe to sun - 4

Amount of work that must be done on a 100 100 Kg object to elevate it to a height 1000 km above Earth surface is x × 1 0 8 J x\times 10^8 J Find x x


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The answer is 8.5.

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1 solution

Anandhu Raj
Feb 19, 2015

Work done, dW= F.dr

Integrating gives , Δ w = F . d r \Rightarrow \Delta w=\int { F.dr }

Δ w = r 1 r 2 G M m r 2 w h e r e , M m a s s o f e a r t h m m a s s o f o b j e c t r 1 i n i t i a l s e p a r a t i o n f r o m c e n t e r o f e a r t h r 2 f i n a l s e p a r a t i o n f r o m c e n t e r o f e a r t h \Rightarrow \Delta w=\int _{ { r }_{ 1 } }^{ { r }_{ 2 } }{ \frac { -GMm }{ { r }^{ 2 } } } \quad \quad where,\\ M\longrightarrow mass\quad of\quad earth\\ m\longrightarrow mass\quad of\quad object\\ { r }_{ 1 }\longrightarrow initial\quad separation\quad from\quad center\quad of\quad earth\\ { r }_{ 2 }\longrightarrow final\quad separation\quad from\quad center\quad of\quad earth

Δ w = G M m r 1 r 2 r 2 \Rightarrow \Delta w=-GMm\int _{ { r }_{ 1 } }^{ { r }_{ 2 } }{ { r }^{ -2 } }

Δ w = G M m [ r 1 1 ] r 1 r 2 \Rightarrow \Delta w=-GMm{ { \left[ \frac { { r }^{ -1 } }{ -1 } \right] } }_{ { r }_{ 1 } }^{ { r }_{ 2 } }\quad

Δ w = G M m [ 1 r 2 1 r 1 ] \Delta w=GMm{ { \left[ \frac { 1 }{ { r }_{ 2 } } -\frac { 1 }{ { r }_{ 1 } } \right] } }

On substituting values we get Δ w = 8.48 × 10 8 J \Delta w=\boxed{8.48\times { 10 }^{ 8 }J}

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