Question

For a planet in our solar system, assume that the axis of orbit is at the sun and is circular. Then the angular momentum about that axis due to the planet’s orbital motion is L= MvR. (a) Derive an expression for Lin terms of the planet’s mass M, orbital radius R, and period Tof the orbit. (b) Using Appendix F, calculate the magnitude of the orbital angular momentum for each of the eight major planets. (Assume a circular orbit.) Add these values to obtain the total angular momentum of the major planets due to their orbital motion. (All the major planets orbit in the same direction in close to the same plane, so adding the magnitudes to get the total is a reasonable approximation.) (c) The rotational period of the sun is 24.6 days. Using Appendix F, calculate the angular momentum the sun has due to the rotation about its axis. (Assume that the sun is a uniform sphere.) (d) How does the rotational angular momentum of the sun compare with the total orbital angular momentum of the planets? How does the mass of the sun compare with the total mass of the planets? The fact that the sun has most of the mass of the solar system but only a small fraction of its total angular momentum must be accounted for in models of how the solar system formed. (e) The sun has a density that decreases with distance from its centre. Does this mean that your calculation in part (c) overestimates or underestimates the rotational angular momentum of the sun? Or doesn’t the nonuniform density have any effect?

Short answer

1. Expression for L in terms of M is ,$L=\frac{2\mathrm{\pi }m{R}^2}{T}$.
2. The magnitude of the orbital angular momentum for each planets are listed below as,
3. The orbital angular momentum of Mercury is $(9.14x{10}^8\mathrm{kg}.m^2/s)$ .
4. The orbital angular momentum of Venus is, $L=1.84\times {10}^{40}kg{m}^2/s$.
5. The orbital angular momentum of Earth is, $L=2.67\times {10}^{40}kg{m}^2/s$.
6. The orbital angular momentum of Mars is, $L=3.53\times {10}^{39}kg{m}^2/s$.
7. The orbital angular momentum of Jupiter is, role="math" localid="1668070553462" $L=1.93\times {10}^{43}kg{m}^2/s$.
8. The orbital angular momentum of Saturn is, $L=7.86\times {10}^{42}kg{m}^2/s$.
9. The orbital angular momentum of Uranus is, $L=1.73\times {10}^{42}kg{m}^2/s$.
10. The orbital angular momentum of Neptune is, $L=2.50\times {10}^{42}kg{m}^2/s$.

(c) The rotational angular momentum of sun due to rotation about axis is, $L=1.14\times {10}^{42}kg{m}^2/s$.

(d) The fraction of angular momentum of sun, $\frac{L_s}{L_m}=0.0363$.

Step-by-step solution

Identification of given data

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’s orbital motion is, L = MvR.

Concept of angular momentum

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 … (i)

Determine relation between angular momentum and average speed

The average speed is given by,

$V=\frac{2\mathrm{\pi }R}{T}$ … (ii)

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

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

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

Determine the orbital angular of planets

The orbital angular momentum of Mercury can be evaluated by,

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

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,

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

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,

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

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

The orbital angular momentum of Mars can be evaluated by,

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

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,

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

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,

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

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,

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

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,

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

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 $L_{total}=313x{10}^{43}kg{m}^2/s$.

(c) Determine the rotational angular momentum of sun due to rotation about axis

Assuming a spherical representation of the sun,

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

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.

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

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,

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

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.