Question 369: Gravitation

Question

Earth has mass $M_1$ and radius $R_1$. Moon has mass $M_2$ and radius $R_2$. Distance between their centre is $r$. A body of mass $M$ is placed on the line joining them at a distance $\frac{r}{3}$ from centre of the earth. To project the mass $M$ to escape to infinity, the minimum speed required is

Accepted Answer

The given situation can be drawn as
The gravitational potential at $P$ is
$$\begin{aligned}
V_p & =-\left(\frac{G M_1}{\frac{r}{3}}+\frac{G M_2}{\frac{2 r}{3}}ight) \
& =\frac{-3 G\left(2 M_1+M_2ight)}{2 r}
\end{aligned}$$
The work done to escape the mass $M$ to infinity is
$$W=M\left(V_{\infty}-V_pight)=\frac{3 G M\left(2 M_1+M_2ight)}{2 r}$$
As, work done is equal to kinetic energy of mass $M$.
$$\begin{array}{rlrl}
\Rightarrow& \frac{1}{2} M v_e^2 =\frac{3 G M\left(2 M_1+M_2ight)}{2 r} \
& v_e =\left[\frac{3 G}{r}\left(2 M_1+M_2ight)ight]^{1 / 2} \
	ext { or } & v_e =\left[\frac{6 G}{r}\left(M_1+\frac{M_2}{2}ight)ight]^{1 / 2}
\end{array}$$

Answer

$\left[\frac{3 G}{r}\left(M_1+\frac{M_2}{2}\right)\right]^{\frac{1}{2}}$

Answer

$\left[\frac{6 G}{r}\left(M_1+\frac{M_2}{2}\right)\right]^{\frac{1}{2}}$

Answer

$\left[\frac{6 G}{r}\left(M_1-\frac{M_2}{2}\right)\right]^{\frac{1}{2}}$

Answer

$\left[\frac{3 G}{r}\left(M_1-\frac{M_2}{2}\right)\right]^{\frac{1}{2}}$

Gravitation

Earth has mass M_1 and radius R_1. Moon has mass M_2 and radius R_2. Distance between their centre is r. A body of mass M is placed on the line joining them at a distance \frac{r}{3} from centre of the earth. To project the mass M to escape to infinity, the minimum speed required is

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