Question

Question: (II) 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, (a) what would the Sun’s new rotation rate be? Take the Sun’s current period to be about 30 days. (b) What would be its final kinetic energy in terms of its initial kinetic energy of today?

Short answer

(a) The Sun’s new rotation rate would be\(4.8 \times {10^{ - 2}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

(b) The final kinetic energy in terms of its initial kinetic energy today would be \(\left( {2 \times {{10}^4}} \right)K{E_1}\).

Step-by-step solution

Determination of angular momentum

The angular momentum of a rotating body can be calculated by multiplying the values of the body’s mass, velocity, and the radius of the rounded path.

Given information

Given data:

The final time of rotation is \({T_1} = 30\;{\rm{days}}\).

Let the initial mass of the Sun be \({M_1} = M\) and its radius be \({R_1} = R\).

So, the final mass will be \({M_2} = \frac{M}{2}\) , the radius will be \({R_2} = 0.01R\) , and the final time of rotation will be \({T_2}\).

Calculate the Sun’s new rotation rate

(a)

Apply the conservation of angular momentum to calculate the new rotation rate of the Sun.

\(\begin{aligned}{c}{I_1}{\omega _1} &= {I_2}{\omega _2}\\\left( {\frac{2}{5}{M_1}R_1^2} \right) \times \left( {\frac{{2\pi }}{{{T_1}}}} \right) &= \left( {\frac{2}{5}{M_2}R_2^2} \right) \times \left( {\frac{{2\pi }}{{{T_2}}}} \right)\\\left( {\frac{2}{5}{M_1}R_1^2} \right) \times \left( {\frac{{2\pi }}{{{T_1}}}} \right) &= \left( {\frac{2}{5}\frac{M}{2}{{\left( {0.01R} \right)}^2}} \right) \times \left( {\frac{{2\pi }}{{{T_2}}}} \right)\\{T_2} &= \frac{1}{2} \times {10^{ - 4}} \times {T_1}\\{T_2} &= \frac{1}{2} \times {10^{ - 4}} \times \left( {30\;{\rm{days}}} \right)\end{aligned}\)

The final angular velocity of the Sun can be calculated as:

\(\begin{aligned}{l}{\omega _2} &= \frac{{2\pi }}{{{T_2}}}\\{\omega _2} &= \frac{{2\pi }}{{\left( {\frac{1}{2} \times {{10}^{ - 4}} \times \left( {30\;{\rm{days}}} \right)} \right)\left( {\frac{{86400\;{\rm{s}}}}{{1\;{\rm{day}}}}} \right)}}\\{\omega _2} &= 4.8 \times {10^{ - 2}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the final angular velocity of the Sun is \(4.8 \times {10^{ - 2}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

Calculate the final kinetic energy in terms of its initial kinetic energy today

(b)

The expression for the initial rotational kinetic energy of the system can be written as:

\(\begin{aligned}{c}K{E_1} &= \frac{1}{2}{I_1}\omega _1^2\\K{E_1} &= \frac{1}{2}\left( {\frac{2}{5}{M_1}R_1^2} \right){\left( {\frac{{2\pi }}{{{T_1}}}} \right)^2}\\K{E_1} &= \frac{1}{2} \times \frac{2}{5}M{R^2} \times {\left( {2\pi } \right)^2} \times {\left( {\frac{1}{{30\;{\rm{days}}}}} \right)^2}\end{aligned}\) … (i)

The expression for the final rotational kinetic energy of the system can be written as:

\(\begin{aligned}{c}K{E_2} &= \frac{1}{2}{I_2}\omega _2^2\\K{E_2} &= \frac{1}{2}\left( {\frac{2}{5}{M_2}R_2^2} \right){\left( {\frac{{2\pi }}{{{T_2}}}} \right)^2}\\K{E_2} &= \frac{1}{2}\left( {\frac{2}{5}\frac{M}{2}{{\left( {0.01R} \right)}^2}} \right){\left( {2\pi } \right)^2}{\left( {\frac{1}{{\frac{1}{2} \times {{10}^{ - 4}} \times \left( {30\;{\rm{days}}} \right)}}} \right)^2}\end{aligned}\) … (ii)

Now, divide equation (ii) by equation (i).

\(\begin{aligned}{c}\frac{{K{E_2}}}{{K{E_1}}} &= \frac{{\frac{1}{2}\left( {\frac{2}{5}\frac{M}{2}{{\left( {0.01R} \right)}^2}} \right){{\left( {2\pi } \right)}^2}{{\left( {\frac{1}{{\frac{1}{2} \times {{10}^{ - 4}} \times \left( {30\;{\rm{days}}} \right)}}} \right)}^2}}}{{\left( {\frac{1}{2} \times \frac{2}{5}M{R^2} \times {{\left( {2\pi } \right)}^2} \times {{\left( {\frac{1}{{30\;{\rm{days}}}}} \right)}^2}} \right)}}\\\frac{{K{E_2}}}{{K{E_1}}} &= 2 \times {10^4}\\K{E_2} &= \left( {2 \times {{10}^4}} \right)K{E_1}\end{aligned}\)

Thus, the final kinetic energy in terms of its initial kinetic energy today would be \(\left( {2 \times {{10}^4}} \right)K{E_1}\).