Question

Two stars each of one solar mass (= 2× 10³⁰ kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Answer 1

Mass of each star, M = 2 × 10³⁰ kg

Radius of each star, R = 10⁴ km = 10⁷ m

Distance between the stars, r = 10⁹ km = 10¹²m

For negligible speeds, v = 0 total energy of two stars separated at distance r

`=(-GMM)/r = 1/2 mv^2`

=`(-(GMM))/r + 0` ...(i)

Now, consider the case when the stars are about to collide:

Velocity of the stars = v

Distance between the centers of the stars = 2R

Total kinetic energy of both stars      ` =1/2 mv^2+ 1/2Mv^2 = Mv^2`

Total potential energy of both stars `= (-GMM)/(2R)`

Total energy of the two stars = `Mv^2 -  (GMM)/(2R)` ...(ii)

Using the law of conversion of energy we can write

`Mv^2 - (GMM)/(2r) = (-GMM)/r`

`v^2 = (-GM)/r + (GM)/(2R) = GM(-1/r+1/(2R))`

`=6.67xx10^(-11) xx 2xx 10^30[-1/(10)^12 +1/(2xx10^7)]`

`=13.34 xx 10^(19)[-10^(-12)+5xx10^(-8)]`

`~13.34xx10^19xx5xx10^(-8)`

`~6.67 xx 10^(12)`

`v = sqrt(6.67 xx 10^(12)) = 2.58 xx 10^6` m/s

Answer 2

Here, mass of each star, M = 2 x  10³⁰ kg

Initial potential between two stars, r =  10⁹ km =  10¹² m.

Initial potential energy of the system = -GMm/r

Total K.E. of the stars = 1/2Mv² + 1/2Mv²

where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r’ = 2 R.

:. Final potential energy of two stars = -GMm/2R

Since gain in K.E. is at the cost of loss in P.E

`:.Mv^2 = -(GMM)/r -(-(GMM)/(2R)) = -(GMM)/r+(GMM)/(2R)`

or `2 xx 10^30 v^2 = - (6.67xx10^(-11) xx (2xx10^(30))^2)/10^(12) + (6.67xx10^(-11)xx(2xx10^(30))^2)/(2xx10^7)`

`= -2.668xx10^38 + 1.334 xx 10^(43)`

`=1.334 xx 10^(43)` J

v= `sqrt(1.334 xx 10^(43))/(2xx10^30) = 2.583xx 10^6 ms^(-1)`