Question

Two bodies of masses $m_1$ and $m_2$ initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance $r$ is (G = Universal gravitational constant)

• A. $\left[ \frac{2G\left(m_{1}-m_{2}\right)}{r}\right]^{\frac{1}{2}} $
• B. $\left[ \frac{2G\left(m_{1} + m_{2}\right)}{r}\right]^{\frac{1}{2}} $
• C. $\left[ \frac{r}{2G\left(m_{1} m_{2}\right)}\right]^{\frac{1}{2}} $
• D. $\left[\frac{r}{2 G} m_{1} m_{2}\right]^{1 / 2}$

Answer 1

The Correct Option is B

Initially when the two masses are at an infinite distance from each other, their gravitational potential energy is zero. When they are at a distance $r$ from each, the gravitational $PE$ is
$PE =\frac{-G m_{1} m_{2}}{r^{2}}$
The minus sign-indicates that there is a decrease in PE. This gives rise to an increase in kinetic energy.
If $v_{1}$ and $v_{2}$ are their respective velocities when they are a distance $r$ apart then, from the law of conservation of energy, we have
$\frac{1}{2} m_{1} v_{1}^{2}=\frac{G m_{1} m_{2}}{r}$ or $v_{1}=\sqrt{\frac{2 G m_{2}}{r}}$
and $\frac{1}{2} m_{2} v_{2}^{2}=\frac{G m_{1} m_{2}}{r}$ or $v_{2}=\sqrt{\frac{2 G m_{1}}{r}}$
Therefore, their relative velocity of approach is
$ v_{1}+v_{2} =\sqrt{\frac{2 G m_{2}}{r}}+\sqrt{\frac{2 G m_{1}}{r}}=\sqrt{\frac{2 G}{r}\left(m_{1}+m_{2}\right)}$
$=\left(\frac{2 G\left(m_{1} \times m_{2}\right)}{r}\right)^{\frac{1}{2}}$