Question

Suppose our Sun eventually collapses into a white dwarf, in the process losing about three-fourths of its mass and winding up with a radius 3.0 percent of its existing radius. What would its new rotation rate be? (Take the Sun's current period to be about 30 days.)
______ rad/s
What would be its final KE in terms of its initial KE of today?
(KE_final / KE_initial) = ______

Answer 1

Here, there are two conditions are given, sun original data and sun after collision converting into the white dwarf.

As per question we have to find Final K.E in terms of initial K.E. now, to achieve that first, we have to find angular velocity of final case. As, after the collision, mass is reduced by 0.75 of original and radius become 3% of the original.

Let's say original sun mass be M and radius R so after collision white dwarf mass is 0.75M and 0.03R.

Applying conservation of angular momentum to find the final angular velocity

I1W1=I2W2.............(1) WhereI1= Moment of inertia of sun and W1 is angular velocity of the sun; I2&W2 is for white dwarf respectively.

I1=2/5(M*R^2)

So, I2= 2/5(0.75M*0.03R^2)

W1= 2pi/30days= 2.42*10^-6 rad/sec

W2= ?

Inserting values in eq 1

W2= (2/5*M*R^2*2.42*10^-6)/(2/5*0.75M*0.03R^2)

W2=4*10^-3 rad/sec

Now as we know

K.E= 1/2*I*W^2

so, Initial K.E(i)= 1/2*((2/5)*M*R^2)*(2.42*10^-6)^2

Similarly, Final K.E(f)= 1/2*((2/5)*0.75M*0.03R^2)*(4*10^-3)

Now let's take the ratio of final K.E(f) to Initial K.E(i)

K.E(f)/K.E(i)= 614.7

or, K.E(f)=614.7*K.E(i)