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Two masses $m_1\, \& m_2$ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separation distance of $d$ is :
$\frac{{2Gd}}{{({m_1} + {m_2})}}$
$\frac{{({m_1} + {m_2})G}}{{2d}}$
${\left[ {({m_1} + {m_2})\frac{{2G}}{d}} \right]^{1/2}}$
${({m_1} + {m_2})^{1/2}}2Gd$
Solution
We use the energy balance
Initial potential energy equals final kinetic energy
$\frac{G m_{1} m_{2}}{d}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2}$
also from the conservation of momentum we have
$m_{1} v_{1}=m_{2} v_{2}$
or $v_{1}=\frac{m_{1} v_{1}}{m_{2}}$
Substituting this we get
$v_{1}=\sqrt{\frac{2 G m_{2}^{2}}{d\left(m_{1}+m_{2}\right)}}$
Similarly we have $v_{2}=\sqrt{\frac{2 G m_{1}^{2}}{d\left(m_{1}+m_{2}\right)}}$
Now as velocities are in opposite direction their relative velocity is
$v_{1}-\left(-v_{2}\right)=v_{1}+$$v_{2}$
or
$=\sqrt{\frac{2 G\left(m_{1}+m_{2}\right)}{d}}$
Similar Questions
Match list-$I$ with list-$II$:
| List-$i$ | List-$2$ |
| $(A)$Kinetic energy of plant | $(1)$ $-\frac{\mathrm{GMm}}{\mathrm{a}}$ |
| $(B)$ Gravaitatioin potentiyal energy of sun -plant system | $(2)$ $\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
| $(C)$Total mecaniacal energy of palnt | $(3)$ $\frac{\mathrm{Gm}}{\mathrm{r}}$ |
| $(D)$Escap energyat the surface of plant for unit mass object | $(4)$ $-\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
(Where $\mathrm{a}=$ radius of planet orbit, $\mathrm{r}=$ radius of planet, $M=$ mass of Sun, $m=$ mass of planet)
Choose the correct answer from the options given below:
