What is the ratio of the orbital angular momentum of Jupiter to its spin angular momentum? Jupiter's orbit has a semimajor axis of 5 AU and period of 12 years,and Jupiter has a rotation period of 0.4 day and a radius of 70,000 km.
A) 10
B) 26 thousand
C) 36 million
D) 50 billion

Question

Quizplus · Accepted Answer

The orbital angular momentum of Jupiter is given by:

L_orbit = mJupiter * v * r

where mJupiter is the mass of Jupiter, v is the orbital velocity of Jupiter, and r is the distance from the center of Jupiter to its orbit. We can find v using Kepler's third law:

P^2 = (4*pi^2 / G*M) * a^3

where P is the period of Jupiter's orbit, a is the semimajor axis, G is the gravitational constant, and M is the mass of the sun. Solving for v, we get:

v = (2*pi*a) / P

Plugging in the values for P and a, we get:

v = 13.1 km/s

Plugging in the values for mJupiter and r, we get:

L_orbit = 1.898 * 10^27 kg * 13.1 km/s * 7.785 * 10^8 m

L_orbit = 1.31 * 10^40 kg m^2/s

The spin angular momentum of Jupiter is given by:

L_spin = I * w

where I is the moment of inertia of Jupiter and w is its angular velocity. For a sphere, the moment of inertia is:

I = (2/5) * m * r^2

Plugging in the values for mJupiter and r, we get:

I = (2/5) * 1.898 * 10^27 kg * (7.785 * 10^7 m)^2

I = 4.15 * 10^47 kg m^2

Plugging in the value for w, we get:

L_spin = 4.15 * 10^47 kg m^2 * (2*pi / 0.4 day)

L_spin = 4.12 * 10^40 kg m^2/s

Therefore, the ratio of the orbital angular momentum to the spin angular momentum is:

L_orbit / L_spin = (1.31 * 10^40 kg m^2/s) / (4.12 * 10^40 kg m^2/s)

L_orbit / L_spin = 0.318

L_orbit is only about a third the size of L_spin, so the best choice is B) 26 thousand.

What Is the Ratio of the Orbital Angular Momentum of Jupiter