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Two particles of masses \mathrm{m} and \mathrm{M} are initially at rest and infinitely separated from each other. Due to gravitational attraction the particles approach each other. Their relative velocity of approach at a separation \mathrm{r} between them is
 

Option: 1

\mathrm{\left(\frac{\mathrm{Gr}}{\mathrm{M}+\mathrm{m}}\right)^{\frac{1}{2}}}
 


Option: 2

\mathrm{\left[\frac{2 \mathrm{G}(\mathrm{M}+\mathrm{m})}{\mathrm{r}}\right]^{\frac{1}{2}}}


 


Option: 3

\mathrm{2 \mathrm{G}(\mathrm{M}+\mathrm{m})^{\frac{1}{2}}}
 


Option: 4

\mathrm{\frac{2 \mathrm{Gr}}{\mathrm{M}+\mathrm{m}}}


Answers (1)

best_answer

Suppose at an instant the masses \mathrm{m} and \mathrm{M} are at \mathrm{x} distance apart then gravitational force of attraction between them is

\mathrm{ \mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{x}^2} }

\mathrm{ \mathrm{a}_{\mathrm{m}}=\frac{\mathrm{GM}}{\mathrm{x}^2} ; \mathrm{a}_{\mathrm{m}}=\frac{\mathrm{GM}}{\mathrm{x}^2}}

Net acceleration of approach is

\mathrm{ \mathrm{a}=\mathrm{a}_{\mathrm{m}}+\mathrm{a}_{\mathrm{m}}=\frac{\mathrm{G}(\mathrm{M}+\mathrm{m})}{\mathrm{x}^2} }

As \mathrm{v} increases, when \mathrm{x} decreases, we have

\mathrm{\mathrm{a}=\frac{-\mathrm{v} \mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{G}(\mathrm{M}+\mathrm{m})}{\mathrm{x}^2} }

\mathrm{\mathrm{vdv}=\frac{-\mathrm{G}(\mathrm{M}+\mathrm{m})}{\mathrm{x}^2} \mathrm{dx}}

Integrating

\mathrm{ \int_0^v v d v=-G(M+m) \int_{\infty}^h x^{1 / 2} d x }

\mathrm{\frac{v^2}{v}=-G(M+m)\left[\frac{x^{-2+1}}{-2+1}\right]_{-\infty}^{\mathrm{r}}=-G(M+m)\left[\frac{-1}{r}\right]_{\infty}^{-r} }

\mathrm{ \therefore \quad v=\sqrt{\frac{2 G(M+m)}{r}}}

Hence option 2 is correct.




 

Posted by

Deependra Verma

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