Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Arjit Raj Grade:

3 particles of equal mass M eacha re moving on a circular path with a radius R under their mutual gravitational attraction. what is the speed of each particle. ans given : root[(GM)/1.732R]

7 years ago

Answers : (1)

vikas askiitian expert
510 Points

dear arjit

           place these masses at the corners of equilateral triangle of side a...   then draw circumcircle of this triangle....

     this is the circle in which these particles are calculate length of side in terms of radius by using simple geometry...

    a = sqrt3.R      (after solving)

 now ,let force bw any of the pair is  F then 


                 F=GM/3R2 ..........1        

now calculate net force due to two particles on other particle ,let it be F1 ..

 F1 = (F2 +F2 +2F2 cos60)1/2

   F1=sqrt3.F        or

  F1 =sqrt3.GM/3R2 ...................2

now this resultant force will provide necessary centripital force for each particle......

so F1 = Mv2 /R (Fc)

    sqrt3.GM/3R2 = Mv2 /R                        

                  v = (GM/sqrt3R)1/2           sqrt3 =1.732

 so         v= (GM/1.732R)1/2       ans


7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details