Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.
kanchana Rao , 14 Years ago
Grade
2 Answers
vikas askiitian expert
Last Activity: 14 Years ago
net force on any of particle due to remaining three = GM2 (81/2+1)/4R2
centripital force is provided by mutual gravitational force
GM2 (81/2+1)/4R2 = MV2/R
V = [GM(81/2+1)/4R]
sai raghava rajeev penagamuri
Last Activity: 14 Years ago
SQUARE ROOT OF GM/R(2(1.414)+1)/4
Provide a better Answer & Earn Cool Goodies
Enter text here...
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free