EBK AN INTRODUCTION TO MODERN ASTROPHYS
EBK AN INTRODUCTION TO MODERN ASTROPHYS
2nd Edition
ISBN: 9781108390248
Author: Carroll
Publisher: YUZU
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Question
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Chapter 2, Problem 2.6P

(a)

To determine

The total orbital momentum of the Sun-Jupiter System.

(a)

Expert Solution
Check Mark

Answer to Problem 2.6P

The momentum of the combined system is 1.9267×1043kg/m2s.

Explanation of Solution

Write the expression for the reduced mass of the system

    μ=MoMJMo+MJ        (1)

Here, μ is the reduced mass of the combined system, Mo is the mass of the Sun and MJ is the mass of the Jupiter.

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

    Lt=μG(Mo+MJ)a(1e2)

Here, lt 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 1.989×1030kg for Mo and 1.899×1027kg for MJ in equation (1).

    μ=(1.989×1030kg)(1.899×1027kg)(1.989×1030kg)+(1.899×1027kg)=1.897×1027kg

Substitute 1.897×1027kg for μ, 6.67×1011Nm2/kg2 for G, 1.989×1030kg for Mo, 1.899×1027kg for MJ, 5.20AU for a and 0.048 for e in equation (2).

    L=(1.897×1027kg)[(6.67×1011Nm2/kg)(1.989×1030kg+1.899×1027kg)×(5.20AU)(1.497×1011m1AU)(10.0482)]=1.9267×1043kgm2/s

Thus, the momentum of the combined system is 1.9267×1043kgm2/s.

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 2.6P

The contribution of the Sun in the angular momentum is 1.840×1040kgm2/s.

Explanation of Solution

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

    ao=μaMo        (3)

Here, ao is the distance of the Sun from the center.

Write the expression for the Sun’s orbital speed.

    vo=2πaoPJ        (4)

Here, vo is the orbital speed and PJ is the orbital period.

Write the expression for the momentum of orbital motion of Sun

    LSun=Moaovo        (5)

Here, LSun is the angular momentum of the Sun.

Conclusion:

Substitute 1.989×1030kg for Mo, 5.20AU for a and 1.897×1027kg for μ in equation (3).

    ao=(1.897×1027kg)(5.20AU)1.989×1030kg=(1.897×1027kg)(5.20AU)(1.497×1011m1AU)1.989×1030kg=7.426×108m

Substitute 7.426×108m for ao and 3.743×108s for PJ in equation (4).

    vo=2π(7.426×108m)3.743×108s=12.46ms1

Substitute 1.989×1030kg for Mo, 7.426×108m for ao and 12.46ms1 for vo in equation (5).

    LSun=(1.989×1030kg)(7.426×108m)(12.46ms1)=1.840×1040kgm2/s

Thus, the contribution of the Sun in the angular momentum is 1.840×1040kgm2/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).

(c)

Expert Solution
Check Mark

Answer to Problem 2.6P

The contribution of the Jupiter in the angular momentum is 1.929×1043kgm2/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

    ao=μaMo        (6)

Here, ao is the distance of the Jupiter from the center.

Write the expression for the Jupiter’s orbital speed.

    vo=2πaoPJ        (7)

Here, vo is the orbital speed and PJ is the orbital period.

Write the expression for the momentum of orbital motion of Jupiter

    LSun=Moaovo        (8)

Here, LSun is the angular momentum of the Sun.

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

    ΔL=LtLSun        (9)

Here, ΔL is the difference of momentum.

Conclusion:

Substitute 1.989×1030kg  for Mo, 5.20AU for a and 1.897×1027kg fir μ in equation (6).

    ao=(1.897×1027kg)(5.20AU)(1.497×1011m1AU)1.899×1027kg=7.778×1011m

Substitute 7.778×1011m for ao and 3.743×108s for PJ in equation (7).

    vo=2π(7.778×1011m)3.743×108s=1.306×104ms1

Substitute 1.899×1027kg for Mo, 7.778×1011m for ao and 1.306×104ms1 for vo in the equation (8).

    LSun=(1.899×1030kg)(7.778×1011m)(1.306×104ms1)=1.929×1043kgm2/s

Substitute 1.9267×1043kgm2/s for Lt and 1.840×1040kgm2/s for LSun in the equation (9).

    ΔL=1.9267×1043kgm2/s1.840×1040kgm2/s=1.925×1043kgm2/s

Thus, the contribution of the Jupiter in the angular momentum is 1.929×1043kgm2/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.

(d)

Expert Solution
Check Mark

Answer to Problem 2.6P

The angular momentum by using mass moment of inertia, for the Sun is 1.078×1042kgm2/s and for Jupiter 6.312×1038kgm2/s.

Explanation of Solution

Write the expression for the angular momentum of the Sun

    LSun=Ioωo        (10)

Here, Lo is the angular momentum of the Sun, Io is the mass moment of inertia of the Sun and ωo is the angular velocity.

Write the expression for the angular momentum of the Jupiter.

    LJ=IJωJ        (11)

Here, LJ is the angular momentum of the Jupiter, IJ is the mass moment of Inertia of the Jupiter and ωJ is the angular velocity of the Jupiter.

Conclusion:

Substitute 3.85×1047kgm2 for Io and 2.8×106rad/s in equation (10).

    LSun=(3.85×1047kgm2)(2.8×106rad/s)=1.078×1042kgm2/s

Substitute 3.62×1042kgm2 for IJ and 1.7436×104rad/s for ωJ in equation (11).

    LJ=(3.62×1042kgm2)(1.7436×104rad/s)=6.312×1038kgm2/s

Thus, the angular momentum by using mass moment of inertia, for the Sun is 1.078×1042kgm2/s and for Jupiter 6.312×1038kgm2/s.

(e)

To determine

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

(e)

Expert Solution
Check Mark

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 1.078×1042kgm2/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.

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