(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 - QAmmunity.org

(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

asked Mar 23, 2021 117k views

1 Answer

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.

Step-by-step explanation:

The angular momentum
of a rigid body is

$L = I\omega$ ,

where
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
the mass,
the radius, and
$\omega_1$ the angular velocity of the sun before it becomes a white dwarf, then its linear momentum is

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

$I_2\omega_2 = (2)/(5) (0.8M)(0.0115R)^2 \omega_2$

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

$(2)/(5)MR^2 \omega_1 = (2)/(5) (0.8M)(0.0115R)^2 \omega_2$

which we solve for
$\omega_2$ :

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

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

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

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

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

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

Now the rotation period
is

$T_2 = (2\pi)/(\omega_2)$

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

since
$(2\pi)/( \omega_1) =T_1$

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

Similarly, the rotation kinetic energy will be

$K_2 = (1)/(2)I_2\omega_2^2$

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

K_2 =0.8*0.0115^2*9452^2 [(1)/(2)*(2)/(5) mR^2w_1^2]

$\boxed{K_2 =9452 K_1}$

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

Raj answered Mar 30, 2021

by Raj

4.7k points

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