Question

Using data from the previous problem, find the increase in rotational kinetic energy, given the core’s mass is\({\rm{1}}{\rm{.3}}\)times that of our Sun. Where does this increase in kinetic energy come from?

Short answer

The required increase in rotation kinetic energy is \({\rm{1}}{\rm{.90}} \times {\rm{1}}{{\rm{0}}^{45}}\,{\rm{J}}\).

Step-by-step solution

Kinetic energy and moment of inertia

The expression for the rotational kinetic energy is given by,

\(KE = \frac{1}{2}I{\omega ^2}\) …… (i)

Here\(I\)is the moment of inertia,\(\omega \)is the angular frequency and\(KE\)is the kinetic energy.

The moment of a spherical part like star can be given by,

\(I = \frac{2}{5}M{r^2}\) …… (ii)

Here\(M\)is the mass of the star and\(r\)is the radius of the star.

The expression for the angular frequency is given by,

\(\omega = 2\pi f\)…… (iii)

Here\(f\)is the frequency.

Evaluating the increase in rotational kinetic energy

Substitute \(\frac{2}{5}M{r^2}\) for \(I\) and \(2\pi f\) for \(\omega \) into the equation (i)

\(KE = \frac{1}{2}\left( {\frac{2}{5}M{r^2}} \right){\left( {2\pi f} \right)^2}\)…… (iv)

Calculate initial kinetic energy,

Substitute\(5 \times {10^8}\,{\rm{m}}\)for\(r\),\(1.3 \times 1.99 \times {10^{30}}\,{\rm{kg}}\)for\(M\)and\(3.858 \times {10^{ - 7}}\,{\rm{s}}\)for\(f\)into the equation (iv)

\(\begin{array}{c}K{E_i} &= \frac{1}{2}\left( {1.3} \right)\left( {1.99 \times {{10}^{30}}\,{\rm{kg}}} \right){\left( {5.0 \times {{10}^8}\,{\rm{m}}} \right)^2}{\left( {2\pi } \right)^2}{\left( {3.858 \times {{10}^{ - 7}}\,{{\rm{s}}^{{\rm{ - 1}}}}} \right)^2}\\ &= 7.60 \times {10^{35}}\,{\rm{J}}\end{array}\)

Calculate the final kinetic energy,

Substitute\(10 \times {10^3}\,{\rm{m}}\)for\(r\),\(1.3 \times 1.99 \times {10^{30}}\,{\rm{kg}}\)for\(M\)and\(964.5\,{\rm{rev/s}}\)for\(f\)into the equation (iv)

\(\begin{array}{c}K{E_f} &= \frac{1}{2}\left( {1.3} \right)\left( {1.99 \times {{10}^{30}}\,{\rm{kg}}} \right){\left( {10.0 \times {{10}^3}\,{\rm{m}}} \right)^2}{\left( {2\pi } \right)^2}{\left( {964.5\,{{\rm{s}}^{{\rm{ - 1}}}}} \right)^2}\\ &= 1.90 \times {10^{45}}\,{\rm{J}}\end{array}\)

Calculate the increase in kinetic energy,

\(\begin{array}{c}\Delta KE &= K{E_f} - K{E_i}\\ &= 1.90 \times {10^{45}}\,{\rm{J}}\, - 7.60 \times {10^{35}}\,{\rm{J}}\\ &= {\rm{1}}{\rm{.90}} \times {\rm{1}}{{\rm{0}}^{45}}\,{\rm{J}}\end{array}\)

Therefore the required increase in rotation kinetic energy is \({\rm{1}}{\rm{.90}} \times {\rm{1}}{{\rm{0}}^{45}}\,{\rm{J}}\).