Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be [2023]

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Solution

Q.
Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be [2023]

Solution

Q.
Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be [2023]

1 $\sqrt{\frac{G M}{2 r}}$

2 $\sqrt{\frac{4 G M}{r}}$

3 $\sqrt{\frac{G M}{r}}$

4 $\sqrt{\frac{G M}{4 r}}$

Ans.
(4)

$\frac{G m^{2}}{4 r^{2}} = \frac{m v^{2}}{r} \Rightarrow v = \sqrt{\frac{G m}{4 r}}$

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Solution

Q. Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be [2023]

1 GM2r

2 4GMr

3 GMr

4 GM4r

Ans. (4) Gm24r2=mv2r⇒v=Gm4r

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Q.
Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be [2023]

1 $\sqrt{\frac{G M}{2 r}}$

2 $\sqrt{\frac{4 G M}{r}}$

3 $\sqrt{\frac{G M}{r}}$

4 $\sqrt{\frac{G M}{4 r}}$

Ans.
(4)

$\frac{G m^{2}}{4 r^{2}} = \frac{m v^{2}}{r} \Rightarrow v = \sqrt{\frac{G m}{4 r}}$