SATHEE: Gravitation Question 282

Gravitation Question 282

Question: Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995; RPMT 2003]

Options:

A) $[v = \frac{1}{2 R} \sqrt{\frac{1}{G m}}]$

B) $[v = \sqrt{\frac{G m}{2 R}}]$

C) $[v = \frac{1}{2} \sqrt{\frac{G m}{R}}]$

D) $[v = \sqrt{\frac{4 G m}{R}}]$

Show Answer

Answer:

Correct Answer: C

Solution:

Centripetal force provided by the gravitational force of attraction between two particles i.e. $[\frac{m v^{2}}{R} = \frac{G m \times m}{{(2 R)}^{2}}]$ $[\Rightarrow v = \frac{1}{2} \sqrt{\frac{G m}{R}}]$

Gravitation Question 282

Question: Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995; RPMT 2003]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Gravitation Question 282

Question: Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995; RPMT 2003]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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