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`        	Two particles of maasses m1 and m2 seperated by a distnace (d) are coming towards each other by their mutual gravitational attraction. Find the direction and magnitude of their relative acceleration. Also find the ratio of their mutual attraction at any instant. `
one year ago

## Answers : (1)

Arun
23381 Points
```							Let v be the velocity of approach of two bodies at distance r apart . The reduced mass of the system of the two particles1/u = 1/ m1 + 1/m2 ,u = m1 m2 / m1 + m2According to law of conservation-Decrease in potential energy = Increase in kinetic energy-(Gm1 m2 /r ) = 1/2 uv2Gm1 m2 /r =1/2 (m1 m2 / m1+m2 ) v2v = (2G (m1 m2 )/r )-1/2
```
one year ago
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• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions