MHT CET · Physics · Gravitation
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
- A \(\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}-\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}\)
- B \(\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}+\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}\)
- C \(\left[\frac{3 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}+\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}\)
- D \(\left[\frac{3 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}-\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{6 \mathrm{G}}{\mathrm{r}}\left(\mathrm{M}_{1}+\frac{\mathrm{M}_{2}}{2}\right)\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
(C)
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
\(\begin{aligned}
& \frac{1}{2} m V^{2}=\frac{3 G M}{r}\left[M_{1}+\frac{M_{2}}{2}\right] \\
\therefore \quad & V=\left[\frac{6 G}{r}\left(M_{1}+\frac{M_{2}}{2}\right)\right]^{\frac{1}{2}}
\end{aligned}\)
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
\(\begin{aligned}
& \frac{1}{2} m V^{2}=\frac{3 G M}{r}\left[M_{1}+\frac{M_{2}}{2}\right] \\
\therefore \quad & V=\left[\frac{6 G}{r}\left(M_{1}+\frac{M_{2}}{2}\right)\right]^{\frac{1}{2}}
\end{aligned}\)
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