JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Gravitation

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 :

(A) $\frac{GM}{2 r}$
(B) $\frac{4 GM}{r}$
(C) $\frac{GM}{r}$
(D) $\frac{GM}{4 r}$ ✓ Correct

Answer: (D)

Solution

Both masses circle the common centre at radius

r

, so their separation is

2 r

. The mutual gravitational pull supplies the centripetal force on each:

\frac{G m ^{2}}{( 2 r ) ^{2}} = \frac{m v ^{2}}{r}

. This gives

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

, so

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

. The speed of each particle is

\frac{G m}{4 r}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option

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