Question

Three stars each of mass M and radius R are initially at rest and the distance between centres of any

two stars is d and they form an equilateral triangle. They start moving towards the centroid due

to

mutual

force of attraction. What are the velocities of the stars just before the collision? Radius of each

star is R.

Answer 1

we know, all the three stars move around centre of mass. and here centre of mass of an equilateral triangle lies at centroid if all the three mass located in vertices of triangle are identical.

so, distance of each star from COM { centroid } = 2/3 × dcos30° = d/√3

now force on each star due to other two stars , F = (GM²/d² + GM²/d²)cos30° = √3GM²/d²

this force is acting along the radius where this acts as centripetal force.

F = √3GM²/2d²

Mv²/(d/√3) = √3GM²/d²

√3v² = √3GM/d

v² = GM/d

hence, v = √{GM/d}

Answer 2

Answer:

t=

v+vcosθ

d

=

3v

2d

v=

d

3Gmm

E=

2

1

mv

2

By energy conversation

d

3Gmm

=

2

1

mv

2

v=

d

6Gm

t=

3

6Gm

2

d

d

=

3

2d

6Gm

d

We get, t=

3

2d

6Gm

d

.