Chapter 2, Problem 2.6P

(a)

To determine

The total orbital momentum of the Sun-Jupiter System.

Answer to Problem 2.6P

The momentum of the combined system is $1.9267\times {10}^{43}\text{kg}/{\text{m}}^2\text{s}$.

Explanation of Solution

Write the expression for the reduced mass of the system

    $\mu =\frac{M_o{M}_J}{M_o+M_J}$        (1)

Here, $\mu$ is the reduced mass of the combined system, $M_o$ is the mass of the Sun and $M_J$ is the mass of the Jupiter.

Write the expression for the total angular momentum of the Sun and Jupiter system.

    $L_t=\mu \sqrt{G\left(M_o+M_J\right)a\left(1-e^2\right)}$

Here, $l_t$ is the total angular momentum, $G$ is the gravitational constant, $a$ is the length of the semi-major axis and $e$ is the eccentricity.

Conclusion:

Substitute $1.989\times {10}^{30}\text{kg}$ for $M_o$ and $1.899\times {10}^{27\text{kg}}$ for $M_J$ in equation (1).

    $\begin{array}{c}\mu =\frac{\left(1.989\times {10}^{30}\text{kg}\right)\left(1.899\times {10}^{27}\text{kg}\right)}{\left(1.989\times {10}^{30}\text{kg}\right)+\left(1.899\times {10}^{27}\text{kg}\right)}\\ =1.897\times {10}^{27}\text{kg}\end{array}$

Substitute $1.897\times {10}^{27\text{kg}}$ for $\mu$, $6.67\times {10}^{-11}\text{N}{\text{m}}^2/{\text{kg}}^2$ for $G$, $1.989\times {10}^{30}\text{kg}$ for $M_o$, $1.899\times {10}^{27\text{kg}}$ for $M_J$, $5.20\text{AU}$ for $a$ and $0.048$ for $e$ in equation (2).

    $\begin{array}{c}L=\left(1.897\times {10}^{27}\text{kg}\right)\sqrt{\left[\begin{array}{l}\left(6.67\times {10}^{-11}\text{N}{\text{m}}^2/\text{kg}\right)\left(1.989\times {10}^{30}\text{kg}+1.899\times {10}^{27}\text{kg}\right)\\ \times \left(5.20\text{AU}\right)\left(\frac{1.497\times {10}^{11}\text{m}}{1\text{AU}}\right)\left(1-{0.048}^2\right)\end{array}\right]}\\ =1.9267\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Thus, the momentum of the combined system is $1.9267\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}$.

(b)

To determine

The contribution of the Sun in the total orbital angular momentum.

Answer to Problem 2.6P

The contribution of the Sun in the angular momentum is $1.840\times {10}^{40}\text{kg}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

Write the expression for the distance of Sun from the centre of mass

    $a_o=\frac{\mu a}{M_o}$        (3)

Here, $a_o$ is the distance of the Sun from the center.

Write the expression for the Sun’s orbital speed.

    $v_o=\frac{2\pi a_o}{P_J}$        (4)

Here, $v_o$ is the orbital speed and $P_J$ is the orbital period.

Write the expression for the momentum of orbital motion of Sun

    $L_{\text{Sun}}=M_o{a}_o{v}_o$        (5)

Here, $L_{\text{Sun}}$ is the angular momentum of the Sun.

Conclusion:

Substitute $1.989\times {10}^{30}\text{kg}$ for $M_o$, $5.20\text{AU}$ for $a$ and $1.897\times {10}^{27}\text{kg}$ for $\mu$ in equation (3).

    $\begin{array}{c}{a}_o=\frac{\left(1.897\times {10}^{27}\text{kg}\right)\left(5.20\text{AU}\right)}{1.989\times {10}^{30}\text{kg}}\\ =\frac{\left(1.897\times {10}^{27}\text{kg}\right)\left(5.20\text{AU}\right)\left(\frac{1.497\times {10}^{11}\text{m}}{1\text{AU}}\right)}{1.989\times {10}^{30}\text{kg}}\\ =7.426\times {10}^8\text{m}\end{array}$

Substitute $7.426\times {10}^8\text{m}$ for $a_o$ and $3.743\times {10}^8\text{s}$ for $P_J$ in equation (4).

    $\begin{array}{c}{v}_o=\frac{2\pi \left(7.426\times {10}^8\text{m}\right)}{3.743\times {10}^8\text{s}}\\ =12.46{\text{ms}}^{-1}\end{array}$

Substitute $1.989\times {10}^{30}\text{kg}$ for $M_o$, $7.426\times {10}^8\text{m}$ for $a_o$ and $12.46{\text{ms}}^{-1}$ for $v_o$ in equation (5).

    $\begin{array}{c}{L}_{\text{Sun}}=\left(1.989\times {10}^{30}\text{kg}\right)\left(7.426\times {10}^8\text{m}\right)\left(12.46\text{m}\cdot {\text{s}}^{-1}\right)\\ =1.840\times {10}^{40}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Thus, the contribution of the Sun in the angular momentum is $1.840\times {10}^{40}\text{kg}\cdot {\text{m}}^2/\text{s}$.

(c)

To determine

The contribution of the Jupiter in the orbital angular momentum and the difference with calculation of part (a) and part(b).

Answer to Problem 2.6P

The contribution of the Jupiter in the angular momentum is $1.929\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}$ and difference of part (a) and part (b) are good match.

Explanation of Solution

Write the expression for the distance of Jupiter from the centre of mass

    $a_o=\frac{\mu a}{M_o}$        (6)

Here, $a_o$ is the distance of the Jupiter from the center.

Write the expression for the Jupiter’s orbital speed.

    $v_o=\frac{2\pi a_o}{P_J}$        (7)

Here, $v_o$ is the orbital speed and $P_J$ is the orbital period.

Write the expression for the momentum of orbital motion of Jupiter

    $L_{\text{Sun}}=M_o{a}_o{v}_o$        (8)

Here, $L_{\text{Sun}}$ is the angular momentum of the Sun.

Write the expression for the difference in momentum of part (a) and part (b).

    $\Delta L=L_t-L_{\text{Sun}}$        (9)

Here, $\Delta L$ is the difference of momentum.

Conclusion:

Substitute $1.989\times {10}^{30}\text{kg}$  for $M_o$, $5.20\text{AU}$ for $a$ and $1.897\times {10}^{27}\text{kg}$ fir $\mu$ in equation (6).

    $\begin{array}{c}{a}_o=\frac{\left(1.897\times {10}^{27}\text{kg}\right)\left(5.20\text{AU}\right)\left(\frac{1.497\times {10}^{11}\text{m}}{1\text{AU}}\right)}{1.899\times {10}^{27}\text{kg}}\\ =7.778\times {10}^{11}\text{m}\end{array}$

Substitute $7.778\times {10}^{11}\text{m}$ for $a_o$ and $3.743\times {10}^8\text{s}$ for $P_J$ in equation (7).

    $\begin{array}{c}{v}_o=\frac{2\pi \left(7.778\times {10}^{11}\text{m}\right)}{3.743\times {10}^8\text{s}}\\ =1.306\times {10}^4\text{m}\cdot {\text{s}}^{-1}\end{array}$

Substitute $1.899\times {10}^{27}\text{kg}$ for $M_o$, $7.778\times {10}^{11}\text{m}$ for $a_o$ and $1.306\times {10}^4\text{m}\cdot {\text{s}}^{-1}$ for $v_o$ in the equation (8).

    $\begin{array}{c}{L}_{\text{Sun}}=\left(1.899\times {10}^{30}\text{kg}\right)\left(7.778\times {10}^{11}\text{m}\right)\left(1.306\times {10}^4\text{m}\cdot {\text{s}}^{-1}\right)\\ =1.929\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Substitute $1.9267\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L_t$ and $1.840\times {10}^{40}\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L_{\text{Sun}}$ in the equation (9).

    $\begin{array}{c}\Delta L=1.9267\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}-1.840\times {10}^{40}\text{kg}\cdot {\text{m}}^2/\text{s}\\ =1.925\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Thus, the contribution of the Jupiter in the angular momentum is $1.929\times {10}^{43}\text{kg}\cdot {\text{m}}^2/\text{s}$ and difference of part (a) and part (b) are good match.

(d)

To determine

The angular momentum of Sun and Jupiter by using moment of inertia.

Answer to Problem 2.6P

The angular momentum by using mass moment of inertia, for the Sun is $1.078\times {10}^{42}\text{kg}\cdot {\text{m}}^2/\text{s}$ and for Jupiter $6.312\times {10}^{38}\text{kg}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

Write the expression for the angular momentum of the Sun

    $L_{\text{Sun}}=I_o{\omega }_o$        (10)

Here, $L_o$ is the angular momentum of the Sun, $I_o$ is the mass moment of inertia of the Sun and ${\omega }_o$ is the angular velocity.

Write the expression for the angular momentum of the Jupiter.

    $L_J=I_J{\omega }_J$        (11)

Here, $L_J$ is the angular momentum of the Jupiter, $I_J$ is the mass moment of Inertia of the Jupiter and ${\omega }_J$ is the angular velocity of the Jupiter.

Conclusion:

Substitute $3.85\times {10}^{47}\text{kg}\cdot {\text{m}}^2$ for $I_o$ and $2.8\times {10}^{-6}\text{rad}/\text{s}$ in equation (10).

    $\begin{array}{c}{L}_{\text{Sun}}=\left(3.85\times {10}^{47}\text{kg}\cdot {\text{m}}^2\right)\left(2.8\times {10}^{-6}\text{rad}/\text{s}\right)\\ =1.078\times {10}^{42}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Substitute $3.62\times {10}^{42}\text{kg}\cdot {\text{m}}^2$ for $I_J$ and $1.7436\times {10}^{-4}\text{rad}/\text{s}$ for ${\omega }_J$ in equation (11).

    $\begin{array}{c}{L}_J=\left(3.62\times {10}^{42}\text{kg}\cdot {\text{m}}^2\right)\left(1.7436\times {10}^{-4}\text{rad}/\text{s}\right)\\ =6.312\times {10}^{38}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Thus, the angular momentum by using mass moment of inertia, for the Sun is $1.078\times {10}^{42}\text{kg}\cdot {\text{m}}^2/\text{s}$ and for Jupiter $6.312\times {10}^{38}\text{kg}\cdot {\text{m}}^2/\text{s}$.

(e)

To determine

The part which is making largest contribution to the angular momentum of the Sun-Jupiter combines system.

Answer to Problem 2.6P

The Sun is making largest contribution to the angular momentum of the Sun-Jupiter combines system.

Explanation of Solution

From part (d) and part (c) it is clear that Sun has the highest contribution in the angular momentum of the Sun-Jupiter combined system. Since the value of the angular momentum is $1.078\times {10}^{42}\text{kg}\cdot {\text{m}}^2/\text{s}$ which is much greater than the value of angular momentum of Jupiter.

Conclusion:

Thus, the Sun is making largest contribution to the angular momentum of the Sun-Jupiter combines system.