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Two particles of equal mass $(m)$ move in a circle of a radius $(r)$ under the action of their mutual gravitational attraction. Which of the following option will be correct for the speed of each particle?

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\sqrt{\frac{GM}{2 r}}

\sqrt{\frac{GM}{4 r}}

\sqrt{\frac{Mr}{2 G}}

\sqrt{\frac{4 r}{GM}}

answer is B.

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Detailed Solution

The speed of each particle is

\sqrt{\frac{GM}{4 r}}

.
When two particles remain directly opposed, they will move in a circular path, with gravitational pull on one particle applied along the radius. Question Image

It provides the required gravitational force.
Given by,
⇨

F = \frac{m v^{2}}{r}

-----------------(1)
Here,

m = mass, v = velocity and r = radius and F = gravitational force

And,
⇨

F =

\frac{Gmm}{{(r + r)}^{2}}

--------------------(2)
Here,

m = mass

G = gravitational constant

and

F = gravitational force

On equation (1) and (2), we have,
⇨

\frac{m v^{2}}{r}

\frac{Gmm}{{(r + r)}^{2}}

On solving we have,
⇨

v = \sqrt{Gm / 4 r}

Hence, the correct option is 2.

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