Question

(28 points) In a little over 5 billion years, our star will slough off ~20% of its mass and collapse to a white dwarf star of radius 8,000 km. We will model it as a sphere. What will its angular momentum be in terms of its current angular momentum? What will its rotation period be in terms of its current period? b) (12 points) What will its rotational kinetic energy be in terms of its current rotational kinetic energy?

Answer 1

Answer:

The angular momentum is same as it was before.

The rotation period is $1.058*10^{-4}$ times the original period.

The rotational kinetic energy is 9452 times greater.

Explanation:

The angular momentum $L$ of a rigid body is

$L = I\omega$,

where $I$ is the moment of inertia and $\omega$ is the angular velocity.

Now, the law of conservation of momentum demands that

$I_1\omega_1 = I_2\omega_2$,

in words this means the angular momentum before must equal the angular momentum after.

Let us call $M$ the mass, $R$ the radius, and $\omega_1$ the angular velocity of the sun before it becomes a white dwarf, then its linear momentum is

$I_1\omega_1 = \dfrac{2}{5}MR^2 \omega_1$     (Remember for a solid sphere $I = \dfrac{2}{5} MR^2$)

After it has become a white dwarf, the suns mass is 80% of what it had before (went off by 20%), and its radius has become 0.0115% its initial value (8000 km is 0.0115% of the radius of the sun ); therefore, the angular momentum is

$I_2\omega_2 = \dfrac{2}{5} (0.8M)(0.0115R)^2 \omega_2$

which must be equal to the angular momentum it had before; therefore

$\dfrac{2}{5}MR^2 \omega_1 = \dfrac{2}{5} (0.8M)(0.0115R)^2 \omega_2$

which we solve for $\omega_2$:

$MR^2 \omega_1 = (0.8M)(0.0115R)^2 \omega_2$

$MR^2 \omega_1 = (0.8)(0.0115)^2 MR^2\omega_2$

$\omega_1 = (0.8)(0.0115)^2 \omega_2$

$\omega_2= \dfrac{\omega_1}{(0.8)(0.0115)^2 }$

$\boxed{ \omega_2 = 9452\omega_1.}$

which is about whopping 9500 times larger than initial angular velocity!!

Now the rotation period $T$ is

$T_2 = \dfrac{2\pi}{\omega_2}$

$T = \dfrac{2\pi}{ 9452\omega_1}= 1.058*10^{-4} (\dfrac{2\pi}{ \omega_1})$

since $\dfrac{2\pi}{ \omega_1} =T_1$

$\boxed{T_2 = 1.058*10^{-4} T_1}$

Similarly, the rotation kinetic energy will be

$K_2 = \dfrac{1}{2}I_2\omega_2^2$

$K_2 = \dfrac{1}{2}*\dfrac{2}{5} (0.8M)(0.0115R)^2 ( 9452\omega_1})^2$

$K_2 =0.8*0.0115^2*9452^2 [\dfrac{1}{2}*\dfrac{2}{5} mR^2w_1^2]$

$\boxed{K_2 =9452 K_1}$

which is about 9500 times larger than initial rotational kinetic energy!