Question

Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius 1.0% of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun's new rotation rate be? (Take the Sun's current period to be about 30 days.) What would be its final KE in terms of its initial KE of today?

Answer 1

Answer question

Answer 2

Now, the formula for angularmomentum is:
L= Iw
So L before = L after:
I₁w₁= I₂w₂

The trick to solving these isto figure out what the change in moment of inertia is, and thenapply the concept of conservation of angular momentum toit.
The sun and the white dwarf will both be spherical

A uniform sphere (p. 223) is²/₅mr²:
I₁= ²/₅mr²= ²/₅MR²

Where M and R are the current massand radius
And
w₁=1 rev/30days

After, the sun has half the mass,and 1/100 the radius:
I₂ = ²/₅(^M/₂)(^R/₁₀₀)²  = (²/₅MR²)/(20,000)

Plugging into conservation ofangular momentum:
I₁w₁= I₂w₂

(²/₅MR²)(1 rev/30 days) =(²/₅MR²)/(20,000)w₂
(1 rev/30 days) = w₂/_(20,000)
w₂ =(20,000)(1 rev/30 days)
w₂ =(20,000)(1 rev/30 days)(2p radians/revolution)(1day/24^.3600 sec) =0.048481368 rad/sec = .048 rad/sec



As for the ratio of kineticenergies,
Before:
E_{k rot}=¹/₂I_ow_o²= ¹/₂(²/₅MR²)(1 rev/30 days)²
After:
E_{k rot}=¹/₂I_ow_o²= ¹/₂(²/₅(^M/₂)(^R/₁₀₀)²)((20,000)(1rev/30 days))²
E_krot=¹/₂I_ow_o²= (20,000)²/(20,000){ ¹/₂(²/₅MR²)(1rev/30 days)²}

Which is just 20,000 times what it was before