Gravitational potential and potential energy

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Question

A sky laboratory of mass 3 x 10³ kg has to be lifted from one circular orbit of radius 2R into another circular orbit of radius 3R. Calculate the minimum energy (in x 10¹⁰J ) required if the radius of earth is $R = 6 .4 \times 10^{6} m$ and $g = 10 {ms}^{- 2}$

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Solution

Total energy of sky laboratory, $E = - \frac{GMm}{2 r}$
$E_{1} = - \frac{GMm}{2 (2 R)} and E_{2} = - \frac{GMm}{2 (3 R)}$
Therefore, the energy required,
$\begin{array}{l} ΔE = E_{2} - E_{1} = GMm (\frac{1}{4 R} - \frac{1}{6 R}) = \frac{GMm}{12 R} \\ = \frac{(R^{2} g) m}{12 R} = \frac{mgR}{12} = 1 .60 \times 10^{10} J \end{array}$

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Gravitational potential and potential energy

Remember concepts with our Masterclasses.

A sky laboratory of mass 3 x 103 kg has to be lifted from one circular orbit of radius 2R into another circular orbit of radius 3R. Calculate the minimum energy (in x 1010J ) required if the radius of earth is R=6.4×106m and g=10ms−2

Total energy of sky laboratory, E=−GMm2rE1=−GMm2(2R) and E2=−GMm2(3R)Therefore, the energy required,ΔE=E2−E1=GMm14R−16R=GMm12R=R2gm12R=mgR12=1.60×1010J

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A sky laboratory of mass 3 x 10³ kg has to be lifted from one circular orbit of radius 2R into another circular orbit of radius 3R. Calculate the minimum energy (in x 10¹⁰J ) required if the radius of earth is $R = 6 .4 \times 10^{6} m$ and $g = 10 {ms}^{- 2}$