Earth has mass ' $M_1$ ' radius ' $R_1$ ' and for moon mass ' $\mathrm{M}_2$ ' and radius ' $\mathrm{R}_2$ '…

Earth has mass ' $M_1$ ' radius ' $R_1$ ' and for moon mass ' $\mathrm{M}_2$ ' and radius ' $\mathrm{R}_2$ '. Distance between their centres is ' $r$ '. A body of mass ' $M$ ' is placed on the line joining them at a distance $\frac{r}{3}$ from the centre of the earth. To project a mass ' M ' to escape to infinity, the minimum speed required is
  1. $\left[\frac{2 G}{r}\left(M_2+\frac{M_1}{2}\right)\right]^{1 / 2}$
  2. $\left[\frac{4 G}{r}\left(M_1+\frac{M_2}{2}\right)\right]^{1 / 2}$
  3. $\left[\frac{3 G}{r}\left(M_1+M_2\right)\right]^{1 / 2}$
  4. $\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_1+\frac{\mathrm{M}_2}{2}\right)\right]^{1 / 2}$

Solution

The binding energy of the body is given by $\begin{aligned} \text { B.E. } & =\frac{\mathrm{GM}_1 \mathrm{M}}{\frac{\mathrm{r}}{3}}+\frac{\mathrm{GM}_2 \mathrm{M}}{\frac{2 \mathrm{r}}{3}}=\frac{3 \mathrm{GM}_1 \mathrm{M}}{\mathrm{r}}+\frac{3 \mathrm{GM}_2 \mathrm{M}}{2 \mathrm{r}} \\ & =\frac{3 \mathrm{GM}}{\mathrm{r}}\left[\mathrm{M}_1+\frac{\mathrm{M}_2}{2}\right] \end{aligned}$
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 2024 (03 May Shift 1)

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