91Ó°ÊÓ

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:

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