Question

A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of radius $2$R(from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius $3$R. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are.

• A. $\frac{3}{4}mgR,\frac{mgR}{6}$
• B. $\frac{3}{4}mgR,\frac{mgR}{12}$
• C. $mgR,mgR$
• D. $2mgR,mgR$

Answer 1

The correct option is B $\frac{3}{4}mgR,\frac{mgR}{12}$

Energy required to place the lab in first orbit is the change in potential energy$+k_0{E}_0$

$△PE=\frac{-GMm}{2R}+\frac{GMm}{2R}=\frac{GMm}{2R}=△u$

Velocity given to lab is provided by gravitation force. So, centripetal force=gravitational force

$\frac{m{v}^2}{2R}=\frac{GMm}{{\left(2R\right)}^2}\Rightarrow v=\sqrt{\frac{GM}{2R}}$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}\frac{GM}{2R}=\frac{1}{4}\frac{GMm}{R}$

Total energy$=△u+{K.E}_0$

$=\frac{GMm}{2R}+\frac{1}{4}\frac{GMm}{R}$

$=\frac{3}{4}\frac{GMm}{R}$$g=\frac{GMm}{R^2}$

$T.E.=\frac{3}{4}mgR$

Similarly to shift to orbit of $3R$

$\frac{m{v}^2}{3R}=\frac{GMm}{{\left(3R\right)}^2}\Rightarrow v^2=\frac{GM}{3R}$

Total energy$=\frac{GMm}{R}-\frac{GMm}{3R}+\frac{1}{2}m{v}^2$

$=\frac{GMm}{R}-\frac{GMm}{3R}+\frac{1}{2}\frac{GMm}{3R}$

$=\frac{GMm}{R}(1-\frac{1}{3}+\frac{1}{6})$

$=\frac{5}{6}mgR$

Energy required$=\frac{5}{6}mgR-\frac{3}{4}mgR$

$=\frac{(20-18)mgR}{24}$

Minimum energy required$=\frac{mgR}{12}$