JEE MAIN - Physics (2012 - No. 13)
The mass of a spaceship is $$1000$$ $$kg.$$ It is to be launched from the earth's surface out into free space. The value of $$g$$ and $$R$$ (radius of earth ) are $$10\,m/{s^2}$$ and $$6400$$ $$km$$ respectively. The required energy for this work will be:
$$6.4 \times {10^{11}}\,$$ Joules
$$6.4 \times {10^8}\,$$ Joules
$$6.4 \times {10^9}\,$$ Joules
$$6.4 \times {10^{10}}\,$$ Joules
Explanation
Potential energy at earth surface = $$ - {{GMm} \over R}$$
and at free space potential energy = 0
Work done for this = $$0 - \left( { - {{GMm} \over R}} \right)$$ = $${{{GMm} \over R}}$$
So the required energy for this work is
= $${{GMm} \over R}$$
=$${{{g{R^2}m} \over R}}$$ [ as $$g = {{GM} \over {{R^2}}}$$ ]
= $$mgR$$
$$ = 1000 \times 10 \times 6400 \times {10^3}$$
$$ = 6.4 \times {10^{10}}\,\,$$ Joules
and at free space potential energy = 0
Work done for this = $$0 - \left( { - {{GMm} \over R}} \right)$$ = $${{{GMm} \over R}}$$
So the required energy for this work is
= $${{GMm} \over R}$$
=$${{{g{R^2}m} \over R}}$$ [ as $$g = {{GM} \over {{R^2}}}$$ ]
= $$mgR$$
$$ = 1000 \times 10 \times 6400 \times {10^3}$$
$$ = 6.4 \times {10^{10}}\,\,$$ Joules
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