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MHT CET · Physics · Gravitation

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. A \(\left[\frac{2 G}{r}\left(M_2+\frac{M_1}{2}\right)\right]^{1 / 2}\)
  2. B \(\left[\frac{4 G}{r}\left(M_1+\frac{M_2}{2}\right)\right]^{1 / 2}\)
  3. C \(\left[\frac{3 G}{r}\left(M_1+M_2\right)\right]^{1 / 2}\)
  4. D \(\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_1+\frac{\mathrm{M}_2}{2}\right)\right]^{1 / 2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_1+\frac{\mathrm{M}_2}{2}\right)\right]^{1 / 2}\)

Step-by-step Solution

Detailed explanation

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
\(\begin{aligned}
& \frac{1}{2} M v^2=\frac{3 G M}{r}\left[M_1+\frac{M_2}{2}\right] \\
\therefore \quad & v=\left[\frac{6 G}{r}\left(M_1+\frac{M_2}{2}\right)\right]^{\frac{1}{2}}
\end{aligned}\)