Question

A sky laboratory of mass $2\times {10}^3$kg has to be lifted from one circular orbit of radius $2$R into another circular orbit of radius $3$R. Calculate the minimum energy required if the radius of earth is $R=6.37\times {10}^6$m and $g=9.8m{s}^{-2}$.

• A. $1.04\times {10}^{10}$J.
• B. $4\times {10}^{10}$J.
• C. $3\times {10}^{10}$J.
• D. $2\times {10}^{10}$J.

Answer 1

The correct option is A $1.04\times {10}^{10}$J.

R.E.F image

Minimum energy

required $=$ Change in energy (potential and Kinetic)

for an Axiect orbitingat radius r, We get

$\frac{m{v}^2}{r}=\frac{GMm}{r^2}\Rightarrow \frac{1}{2}m{m}^2=\frac{1}{2}\frac{GMm}{r}=K.E$

Potential energy $=-\frac{GMm}{r}.$ So total energy

$=\frac{-1}{2}G\frac{Mm}{r}$

So, change in total energy $=\frac{GMm}{2}(-\frac{1}{3R}+\frac{1}{2R})=\frac{GMm}{12R}$

$=\frac{GM}{R^2}.\frac{MR}{12}=g\frac{MR}{12}=9.8\times \frac{2\times {10}^3\times 6.37\times {10}^6}{12}$

$=1.04\times {10}^{10}J$

Ans is (A)