Earth has mass ' $\mathrm{M}_{1}{ }^{\prime}$ and Radius ' $\mathrm{R}_{1}{ }^{\prime}$. Moon has mass ${…

Earth has mass ' $\mathrm{M}_{1}{ }^{\prime}$ and Radius ' $\mathrm{R}_{1}{ }^{\prime}$. Moon has mass ${ }^{\prime} \mathrm{M}_{2}{ }^{\prime}$ and radius ${ }^{\prime} \mathrm{R}_{2}{ }^{\prime}$. Distance between their centres is ' $\mathrm{r}^{\prime}$. A body of mass ' $\mathrm{M}^{\prime}$ is placed on the line joining them ata distance $\frac{\mathrm{r}}{3}$ from centre of the earth. To project the mass ' $\mathrm{M}^{\prime}$ to escape to infinity, the minimum speed required is
  1. $\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}-\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}$
  2. $\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}+\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}$
  3. $\left[\frac{3 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}+\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}$
  4. $\left[\frac{3 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}-\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}$

Solution

The binding energy of the body is given by $\text { B.E. } \begin{aligned} &=\frac{G M_{1} M}{\frac{r}{3}}+\frac{G M_{2} M}{\frac{2 r}{3}} \\ &=\frac{3 G M_{1} M}{r}+\frac{3 G M_{2} M}{2 r} \\ &=\frac{3 G M}{r}\left[M_{1}+\frac{M_{2}}{2}\right] \end{aligned}$ If $V$ is the velocity given to the body, then $\begin{aligned} & \frac{1}{2} m V^{2}=\frac{3 G M}{r}\left[M_{1}+\frac{M_{2}}{2}\right] \\ \therefore \quad & V=\left[\frac{6 G}{r}\left(M_{1}+\frac{M_{2}}{2}\right)\right]^{\frac{1}{2}} \end{aligned}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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