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The masses and radii of the earth and moon are \left(\mathrm{M}_{1}, \mathrm{R}_{1}\right)$ and $\left(\mathrm{M}_{2}, \mathrm{R}_{2}\right) respectively. Their centres are at a distance ' r ' apart. Find the minimum escape velocity for a particle of mass ' \mathrm{m} ' to be projected from the middle of these two masses :
 
Option: 1 \mathrm{V}=\sqrt{\frac{4 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}}
Option: 2 \mathrm{V}=\frac{1}{2} \sqrt{\frac{2 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}}
Option: 3 \mathrm{V}=\frac{1}{2} \sqrt{\frac{4 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}}
Option: 4 \mathrm{V}=\frac{\sqrt{2 \mathrm{G}}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}

Answers (1)

best_answer


Point P is not midpoint between o_{1} & o_{2}
PE= \frac{-GM_{1}m}{\frac{r}{2}}+\left ( \frac{-GM_{2}m}{\frac{r}{2}} \right )
For minimum escape velocity,
TE= 0= PE+KE
0= \frac{-2GM_{1}m}{r}-\frac{2GM_{2}m}{r}+\frac{1}{2}mv^{2}_{e}
V_{e}^{2}= \frac{4GM_{1}}{r}+\frac{4GM_{2}}{r}
\rightarrow V_{e}=\sqrt{\frac{4G\left ( M_{1}+M_{2} \right )}{r}}
The correct option is (1)
 

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vishal kumar

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