Test Index

Gravitation Part 1

Section: Physics

© examsnet.com

Question : 6 of 100

Marks: +1, -0

The masses and radii of the Earth and Moon are

(m_{1}, R_{1})

(m_{2}, R_{2})

, respectively. Their centres are at a distance r apart. Find the minimum escape velocity for a particle of mass m to be projected from the middle of these two masses.

[31 Aug 2021 Shift 1]

$v = \frac{1}{2} \sqrt{\frac{4 G (m_{1}+m_{2})}{r}}$
$v = \sqrt{\frac{4 G (m_{1}+m_{2})}{r}}$
$v = \frac{1}{2} \sqrt{\frac{2 G (m_{1}+m_{2})}{r}}$
$v = \frac{\sqrt{2 G} (m_{1}+m_{2})}{r}$

Solution:

Given, masses and radii of Earth and Moon are

(m_{1}, R_{1}),(m_{2}, R_{2})

and separation between their centre of mass = r

If v be the minimum escape velocity of particle of mass m, then by using law of conservation of energy,

Initial energy = Final energy = 0

\frac{1}{2} m v^{2} - \frac{G m_{1} m}{\frac{r}{2}} - \frac{G m_{2} m}{\frac{r}{2}} = 0

\Rightarrow \frac{1}{2} m v^{2} = \frac{2 G m}{r} (m_{1}+m_{2})

\Rightarrow v = \sqrt{\frac{4 G}{r}(m_{1}+m_{2})}

© examsnet.com

Go to Question:

Prev Question

Next Question