Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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]

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]

Grade:

1 Answers

vikas askiitian expert
509 Points
10 years ago

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 rotating...now 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/a2

                 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

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free