First slide

Gravitational potential and potential energy

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 $g = 10 {ms}^{- 2}$

Moderate

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,
$\begin{array}{l} ΔE = E_{2} - E_{1} = GMm (\frac{1}{4 R} - \frac{1}{6 R}) = \frac{GMm}{12 R} \\ = \frac{(R^{2} g) m}{12 R} = \frac{mgR}{12} = 1 .60 \times 10^{10} J \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App