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10 years ago

509 Points
```							let particle are at A,B,C...
force on A due to B is F & force on A due to C is also F...
F = Gm2/a2
net force  = (F2+F2+2F2cos@)1/2                                 (by vector method)
angle bw them is @ = 60
so net force = Fn = 31/2F       ...........1
due to this force ,each particle circulates aroung their center of mass & center of mass lies at the center of equilateral triangle ......
Fc = Fn = 31/2Gm2/a2
mv2/r = 31/2Gm2/a2
v = {31/2 Gm/a2r}1/2
now r = a/31/2
now v becomes
v = {3Gm/a3}1/2                                             ans1
time period = 2pir/v
=2pi(a/31/2){3Gm/a3)1/2
= 2pi(gm/a)1/2                             ans2
```
10 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions