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The given data can be listed below as,

• The mass of planet is M.
• The orbital radius of planet is R.
• The time period taken by planet is T.
• The angular momentum due to planet鈥檚 orbital motion is, L = MvR.

The object's distance from the axis of rotation, multiplied by its linear momentum, yields the angular momentum.

The angular momentum is given by,

L = MvR 鈥� (i)

The average speed is given by,

$V=\frac{2\mathrm{蟺}R}{T}$ 鈥� (ii)

Substitute the equation (ii) in theequation (i).

$\begin{array}{r}L=mvR\\ L=m\left(\frac{2蟺R}{T}\right)R\\ L=\frac{2蟺m{R}^2}{T}\end{array}$

Thus, the expression is $L=\frac{2\mathrm{蟺}m{R}^2}{T}$.

The orbital angular momentum of Mercury can be evaluated by,

$\begin{array}{r}{L}_M=\frac{2蟺m_M{R}_M^2}{T_M}\\ L=\frac{2蟺\left(3.30脳{10}^{23}\mathrm{kg}\right){\left(5.79脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=9.14脳{10}^{38}\mathrm{kg}鈰�{\mathrm{m}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Mercury is $9.14x{10}^{38}kg{m}^2/s$.

The orbital angular momentum of Venus can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(4.867脳{10}^{24}\mathrm{kg}\right){\left(6.0518脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=1.84脳{10}^{40}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Venus is $1.84x{10}^{40}kg{m}^{2}ls$.

The orbital angular momentum of Earth can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(5.972脳{10}^{24}\mathrm{kg}\right){\left(6.371脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=2.67脳{10}^{40}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Earth is role="math" localid="1668070989291" $2.67脳{10}^{40}kg{m}^{2}ls$.

The orbital angular momentum of Mars can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(6.39脳{10}^{23}\mathrm{kg}\right){\left(3.3895脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=3.53脳{10}^{39}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Mars is $3.53x{10}^{39}kg{m}^2/s$.

The orbital angular momentum of Jupiter can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(1.89脳{10}^{27}\mathrm{kg}\right){\left(6.991脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=1.93脳{10}^{43}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Jupiter is, role="math" localid="1668071153620" $1.93x{10}^{43}kg{m}^2/s$.

The orbital angular momentum of Saturn can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(5.683脳{10}^{28}\mathrm{kg}\right){\left(5.823脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=7.86脳{10}^{42}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Saturn is, $7.86x{10}^{2}kg{m}^2/s$.

The orbital angular momentum of Uranus can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(8.681脳{10}^{25}\mathrm{kg}\right){\left(2.5362脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=1.73脳{10}^{42}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Uranus is, $1.73x{10}^{42}kg{m}^2/s$.

The orbital angular momentum of Neptune can be evaluated by,

$\begin{array}{r}L=\frac{2蟺m{R}^2}{T}\\ L=\frac{2蟺\left(1.024脳{10}^{28}\mathrm{kg}\right){\left(2.462脳{10}^{10}\mathrm{m}\right)}^2}{\left[\right(88\mathrm{d}\left)\right(24\mathrm{h}/\mathrm{d}\left)\right(3600\mathrm{s}/\mathrm{h}\left)\right]}\\ L=2.50脳{10}^{42}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the orbital angular momentum of Neptune is $2.50x{10}^{42}kg{m}^2/s$.

Since the planets all revolve around the sun in the same direction, the total angular momentum equals the sum of them $L_{total}=313x{10}^{43}kg{m}^2/s$.

Assuming a spherical representation of the sun,

$\begin{array}{r}L=I蝇\\ L=\left(\frac{2}{5}M{R}^2\right)\left(\frac{2蟺}{T}\right)\\ L=\frac{4蟺M{R}^2}{5T}\\ L=\frac{4蟺\left(1.99脳{10}^{30}\mathrm{kg}\right){\left(6.96脳{10}^8\mathrm{m}\right)}^2}{5\left(24.6\mathrm{d}\right)(24\mathrm{h}/\mathrm{d})(3600\mathrm{s}/\mathrm{h})}\\ L=1.14脳{10}^{42}{\mathrm{kgm}}^2/\mathrm{s}\end{array}$

Thus, the rotational angular momentum of sun due to rotation about axis is role="math" localid="1668071417375" $1.14x{10}^{42}kg{m}^2/s$.

Step 4: Determinethe fraction of angular momentum of sun.

$\begin{array}{r}\frac{L_S}{L_P}=\left(\frac{1.14脳{10}^{42}{\mathrm{kgm}}^2/\mathrm{s}}{2.669脳{10}^{43}{\mathrm{kgm}}^2/\mathrm{s}}\right)\\ \frac{L_S}{L_P}=0.0363\end{array}$

Thus,only 3.63 percent of the planets' angular momentum is accounted for by the sun.

Total mass of planet,$m_p=2.669x{10}^{27}\mathrm{kg}$

So,

$\begin{array}{r}\frac{m_s}{m_p}=\frac{1.99脳{10}^{30}\mathrm{kg}}{2.669脳{10}^{27}\mathrm{kg}}\\ \frac{m_s}{m_p}=746\end{array}$

This means that the sun is 746 times more massive than all of the planets put together.

Thus, the sun is only responsible for 3.6 percent of the solar system's rotational momentum. The angular momentum of the planets does not seem to change as we go from the inner planets to the outer ones.