What is the angular momentum of the moon orbiting an Earth-like planet?

[NX805]

Hi, can anyone help me get started on this problem...I am totally lost and don't know what equation to use...thanks a lot

There is a moon orbiting an Earth-like planet. The mass of the moon is 6.23 x 10^22 kg, the center-to-center separation of the planet and the moon is 649000 km, the orbital period of the moon is 29.3 days, and the radius of the moon is 1040km. What is the angular momentum of the moon about the planet.

 

[Phymath]

they are talkin about point mass angular momentum

L = r x p (the linear v is 90 degrees to the radius therefore) = r*p while p = mv and u can figure out the linear veloctiy useing the orbital period.

29.3 days to go 2pi radians so... w = 2pi/(29.3 days) and v = wr so u get...

L = r^2 m (2pi)/(29.3 days)

 

[wais]

To solve this problem, we can use the equation for angular momentum, which is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

First, we need to calculate the moment of inertia of the moon. This can be done using the formula I = 2/5mr^2, where m is the mass of the moon and r is its radius. Plugging in the given values, we get I = 2/5(6.23 x 10^22 kg)(1040km)^2 = 1.37 x 10^33 kgm^2.

Next, we need to find the angular velocity of the moon. This can be calculated using the formula ω = 2π/T, where T is the orbital period. Plugging in the given value of 29.3 days (which needs to be converted to seconds), we get ω = 2π/(29.3 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute) = 2.66 x 10^-6 rad/s.

Now, we can plug in our calculated values into the equation for angular momentum: L = (1.37 x 10^33 kgm^2)(2.66 x 10^-6 rad/s) = 3.65 x 10^27 kgm^2/s.

Therefore, the angular momentum of the moon orbiting an Earth-like planet is 3.65 x 10^27 kgm^2/s.