Two identical particles each of mass go round a circle of radius under the action of their mutual gravitational attraction. The angular speed of each particle will be [2023]

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Solution

Q.
Two identical particles each of mass $' m'$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be [2023]

Solution

Q.
Two identical particles each of mass $' m'$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be [2023]

1 $\sqrt{\frac{G m}{2 a^{3}}}$

2 $\sqrt{\frac{G m}{8 a^{3}}}$

3 $\sqrt{\frac{G m}{4 a^{3}}}$

4 $\sqrt{\frac{G m}{a^{3}}}$

Ans.
(3)

$F = m ω^{2} r$

$\Rightarrow \frac{G m m}{{(2 a)}^{2}} = m ω^{2} a$

$\Rightarrow ω = \sqrt{\frac{G m}{4 a^{3}}}$

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Solution

Q. Two identical particles each of mass 'm' go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be [2023]

1 Gm2a3

2 Gm8a3

3 Gm4a3

4 Gma3

Ans. (3) F=mω2r ⇒Gmm(2a)2=mω2a ⇒ω=Gm4a3

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Q.
Two identical particles each of mass $' m'$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be [2023]

1 $\sqrt{\frac{G m}{2 a^{3}}}$

2 $\sqrt{\frac{G m}{8 a^{3}}}$

3 $\sqrt{\frac{G m}{4 a^{3}}}$

4 $\sqrt{\frac{G m}{a^{3}}}$

Ans.
(3)

$F = m ω^{2} r$

$\Rightarrow \frac{G m m}{{(2 a)}^{2}} = m ω^{2} a$

$\Rightarrow ω = \sqrt{\frac{G m}{4 a^{3}}}$