Most of our Solar System’s mass is contained in the Sun, and the planets possess almost all of the Solar System’s angular momentum. This observation plays a key role in theories attempting to explain the formation of our Solar System. Estimate the fraction of the Solar System’s total angular momentum that is possessed by planets using a simplified model which includes only the large outer planets with the most angular momentum. The central Sun (mass\(1.99 \times {10^{30}}\;{\rm{kg}}\), radius\(6.96 \times {10^8}\;{\rm{m}}\)) spins about its axis once every 25 days and the planets Jupiter, Saturn, Uranus, and Neptune move in nearly circular orbits around the Sun with orbital data given in the Table below. Ignore each planet’s spin about its own axis.

Planet	Mean Distance from Sun\(\left( { \times {{10}^6}\;{\rm{km}}} \right)\)	Orbital Period (Earth Years)	Mass \(\left( { \times {{10}^{25}}\;{\rm{kg}}} \right)\)
Jupiter	778	11.9	190
Saturn	1427	29.5	56.8
Uranus	2870	84.0	8.68
Neptune	4500	165	10.2

The mass of the Sun is\({M_{\rm{s}}} = 1.99 \times {10^{30}}\;{\rm{kg}}\).

The radius of the Sun is \({R_{\rm{s}}} = 6.96 \times {10^8}\;{\rm{m}}\)

Angular momentum is the rotational equivalent to the linear momentum of a body.It is expressed as the product of the moment of inertia and the angular velocity.

\(L = I\omega \) … (i)

The four given planets and the Sun are solid spheres. The moment of inertia of a solid sphere of mass M and radius R,rotating on its axis is:

\(I = \frac{2}{5}M{R^2}\) … (ii)

The angular speed of a rotating rigid body is given as:

\(\omega = \frac{{2\pi }}{T}\)

Therefore, the angular speed of the Sun will be:

\(\begin{aligned}{c}{\omega _{{\rm{sun}}}} = \frac{{2\pi }}{{{T_{{\rm{sun}}}}}}\\ = \frac{{2\pi }}{{25\;{\rm{days}}}} \times \left( {\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}} \right)\end{aligned}\)

From equations (1) and (2), the spin angular momentum of the Sun is determined as:

\(\begin{aligned}{c}{L_{{\rm{sun}}}} = {I_{{\rm{sun}}}}{\omega _{{\rm{sun}}}}\\ = \left( {\frac{2}{5}{M_{{\rm{sun}}}}R_{{\rm{sun}}}^2} \right){\omega _{{\rm{sun}}}}\\ = \frac{2}{5}\left( {1.99 \times {{10}^{30}}\;{\rm{kg}}} \right) \times {\left( {6.96 \times {{10}^8}\;{\rm{m}}} \right)^2} \times \frac{{2\pi }}{{25\;{\rm{days}}}} \times \left( {\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}} \right)\\ = 1.122 \times {10^{42}}\;{\rm{kg}}\;{\rm{m/s}}\end{aligned}\)

The orbit of Jupiter and other planets moving around the Sun can be considered as a hoop whose moment of inertia is\({I_{{\rm{orbit}}}} = M{R^2}\), where R is the radius of the orbit.

From equation (1), the orbital angular momentum of Jupiter is:

\(\begin{aligned}{c}{L_{\rm{J}}} = {I_{\rm{J}}}{\omega _{\rm{J}}}\\ = {M_{\rm{J}}}R_{\rm{J}}^2\left( {\frac{{2\pi }}{{{T_{\rm{J}}}}}} \right)\\ = \left( {190 \times {{10}^{25}}\;{\rm{kg}}} \right) \times {\left( {778 \times {{10}^9}\;{\rm{m}}} \right)^2} \times \left( {\frac{{2\pi }}{{11.9\;{\rm{yr}}}} \times \frac{{1\;{\rm{yr}}}}{{3.156 \times {{10}^7}\;{\rm{s}}}}} \right)\\ = 1.9240 \times {10^{43}}\;{\rm{kg}}\;{\rm{m/s}}\end{aligned}\)

Similarly, the orbital angular momentum of Saturn is:

\(\begin{aligned}{c}{L_{\rm{S}}} = {I_{\rm{S}}}{\omega _{\rm{S}}}\\ = {M_{\rm{S}}}R_{\rm{S}}^2\left( {\frac{{2\pi }}{{{T_{\rm{S}}}}}} \right)\\ = \left( {56.8 \times {{10}^{25}}\;{\rm{kg}}} \right) \times {\left( {1427 \times {{10}^9}\;{\rm{m}}} \right)^2} \times \left( {\frac{{2\pi }}{{29.5\;{\rm{yr}}}} \times \frac{{1\;{\rm{yr}}}}{{3.156 \times {{10}^7}\;{\rm{s}}}}} \right)\\ = 7.806 \times {10^{42}}\;{\rm{kg}}\;{\rm{m/s}}\end{aligned}\)

The orbital angular momentum of Uranus is:

\(\begin{aligned}{c}{L_{\rm{U}}} = {I_{\rm{U}}}{\omega _{\rm{U}}}\\ = {M_{\rm{U}}}R_{\rm{U}}^2\left( {\frac{{2\pi }}{{{T_{\rm{U}}}}}} \right)\\ = \left( {8.68 \times {{10}^{25}}\;{\rm{kg}}} \right) \times {\left( {2870 \times {{10}^9}\;{\rm{m}}} \right)^2} \times \left( {\frac{{2\pi }}{{84\;{\rm{yr}}}} \times \frac{{1\;{\rm{yr}}}}{{3.156 \times {{10}^7}\;{\rm{s}}}}} \right)\\ = 1.695 \times {10^{42}}\;{\rm{kg}}\;{\rm{m/s}}\end{aligned}\)

The orbital angular momentum of Neptune is:

\(\begin{aligned}{c}{L_{\rm{N}}} = {I_{\rm{N}}}{\omega _{\rm{N}}}\\ = {M_{\rm{N}}}R_{\rm{N}}^2\left( {\frac{{2\pi }}{{{T_{\rm{N}}}}}} \right)\\ = \left( {10.2 \times {{10}^{25}}\;{\rm{kg}}} \right) \times {\left( {4500 \times {{10}^9}\;{\rm{m}}} \right)^2} \times \left( {\frac{{2\pi }}{{165\;{\rm{yr}}}} \times \frac{{1\;{\rm{yr}}}}{{3.156 \times {{10}^7}\;{\rm{s}}}}} \right)\\ = 2.492 \times {10^{42}}\;{\rm{kg}}\;{\rm{m/s}}\end{aligned}\)

The fraction of the Solar System’s total angular momentum possessed by the planets is given by:

\(\begin{aligned}{c}f = \frac{{{L_{{\rm{planets}}}}}}{{{L_{{\rm{planets}}}} + {L_{{\rm{sun}}}}}}\\ = \frac{{{L_{\rm{J}}} + {L_{\rm{S}}} + {L_{\rm{U}}} + {L_{\rm{N}}}}}{{\left( {{L_{\rm{J}}} + {L_{\rm{S}}} + {L_{\rm{U}}} + {L_{\rm{N}}}} \right) + {L_{{\rm{sun}}}}}}\\ = \frac{{\left( {19.240 + 7.806 + 1.695 + 2.492} \right) \times {{10}^{42}}\;{\rm{kg}}\;{\rm{m/s}}}}{{\left( {19.240 + 7.806 + 1.695 + 2.492 + 1.122} \right) \times {{10}^{42}}\;{\rm{kg}}\;{\rm{m/s}}}}\\ = 0.965\end{aligned}\)