Q. The escape velocity of a body on the surface of the earth is $\text{11.2}kms^{- 1}$ . If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes
NTA AbhyasNTA Abhyas 2022
Solution:
$V_{e}=\sqrt{\frac{2 G M_{e}}{R_{e}}}=11.2 \, kms^{- 1}$
Now $M^{′}=2M_{e}$ and $R^{′}=\frac{R_{e}}{2}$
So $V_{e}^{′}=\sqrt{\frac{2 G \left(2 M_{e}\right)}{\left(\frac{R_{e}}{2}\right)}}=2\sqrt{\frac{2 G M_{e}}{R_{e}}}=2\times 11.2$
$=\text{22.4} \, kms^{- 1}$
Now $M^{′}=2M_{e}$ and $R^{′}=\frac{R_{e}}{2}$
So $V_{e}^{′}=\sqrt{\frac{2 G \left(2 M_{e}\right)}{\left(\frac{R_{e}}{2}\right)}}=2\sqrt{\frac{2 G M_{e}}{R_{e}}}=2\times 11.2$
$=\text{22.4} \, kms^{- 1}$