What is the Angular Momentum of a Planet Rotating and Orbiting a Star?

Click For Summary
SUMMARY

The discussion focuses on calculating the angular momentum of a planet that rotates once daily and orbits a star annually. For part (a), the moment of inertia (I) is calculated using the formula I = 2/5MR², yielding I = 1.01 x 1038 kg·m2. The angular momentum (L) about the planet's rotation axis is determined to be L = 7.34 x 1033 kg·m2/s. For part (b), the angular momentum in the planet's orbit is calculated using L = m*r²*ω, where ω is derived from the orbital period.

PREREQUISITES
  • Understanding of angular momentum and moment of inertia
  • Familiarity with rotational dynamics and uniform spheres
  • Knowledge of orbital mechanics and planetary motion
  • Basic proficiency in physics equations and unit conversions
NEXT STEPS
  • Study the derivation of angular momentum formulas for rotating bodies
  • Learn about the conservation of angular momentum in celestial mechanics
  • Explore the effects of mass distribution on moment of inertia
  • Investigate the relationship between orbital radius and angular velocity
USEFUL FOR

Students in physics, astrophysics enthusiasts, and anyone interested in understanding the dynamics of planetary motion and angular momentum calculations.

Bones
Messages
108
Reaction score
0

Homework Statement


Consider a planet that rotates one revolution in one day and orbits a star in one year. The planet has mass = 5.6 1024 kg, radius = 6.7 106 m, and is 1.7 1011 km from the star.
(a) Determine the angular momentum for the rotating planet about its rotation axis (assume a uniform sphere).
kg · m2/s

(b) Determine the angular momentum for the planet in its orbit around its star (treat the planet as a particle orbiting the star).
kg · m2/s






Homework Equations





The Attempt at a Solution



I started this problem, but I am not sure how to do part b.
a)
I=2/5MR^2
I=2/5(5.6*10^24)(6.7*10^6)^2
I=1.01*10^38
w=1rev/day*2pi/1rev*1day/24hrs*1hr/60min*1min/60sec=7.27*10^-5
L=Iw
L=(1.01*10^38)(7.27*10^-5)
L=7.34*10^33
 
Physics news on Phys.org
The moment of inertia that you calculated in a) is for the rotation about the axis of the planet.

In b) they are asking you to calculate the angular momentum about it's star. They also suggest to treat it as a point mass.

I = m*r²

L = m*r²*ω = 2π*m*r²/t
 
Thanks!
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K