Question

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

Short answer

The fraction of the Solar System’s total angular momentum possessed by the planets is 0.965.

Step-by-step solution

Identification of the given data

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

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

Definition of angular momentum

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)

Calculation of the moment of inertia of each planet

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)

Calculation of the Sun’s spin angular momentum

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:

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

Calculation of the orbital angular momentum of Jupiter

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:

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

Calculation of the orbital angular momentum of Saturn

Similarly, the orbital angular momentum of Saturn is:

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

Calculation of orbital angular momentum of Uranus

The orbital angular momentum of Uranus is:

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

Calculation of the orbital angular momentum of Neptune

The orbital angular momentum of Neptune is:

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

Calculation of the fraction of the Solar System’s total angular momentum

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

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