wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their center of


Open in App
Solution

Step 1: Given

  1. The equal mass of the particles is m
  2. The radius of the circle is R.

Step 2: Formula used

  1. Gravitational force =GMmR2
  2. Centrifugal force =mω2R

Step 3: Solution

Let the speed of rotation of each particle be v.

By equating gravitational force with centrifugal force

Gm22R2=mω2R

G= Gravitational constant

2R=The distance between the center of the two particles.

ω= Angular velocity of the particles

Now,

Gm24R2=mω2RGm24R2·mR=ω2Gm4R3=ω2ω=Gm4R3......1

From the relation between angular velocity and linear velocity

v=ωR

v=Gm4R3×RPuttingvalueofωfrom1v=GmR24R3v=Gm4R

Hence the required answer is v=Gm4R.


flag
Suggest Corrections
thumbs-up
123
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Newton's Universal law of Gravitation Tackle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon