What is the angular momentum of the Moon in its orbit around Earth in kg m2/s?
I do not have any information about the moon, so I looked it up, but I got the question wrong so I don't know what values I should be using...I got 5.7 * 10^35.
Thank you!
you need the radius, and the period
Momentum=1/2 m r^2 2PI/periodinSeconds
you can look up mass moon, radius of orbit, and calculat the period in seconds (you know its orbit period is 28days).
angular momentum of the moon is
... mass * velocity * orbital radius
orbital radius is earth-moon distance
... 385000 km ... (time averaged)
velocity (average) ... 1022 m/s
mass ... 7.342E22 kg
values courtesy of wikipedia
To calculate the angular momentum of the Moon in its orbit around Earth, you need to know the mass of the Moon, its orbital radius, and its orbital velocity.
The mass of the Moon is approximately 7.35 x 10^22 kg.
The mean orbital radius of the Moon around the Earth is about 3.84 x 10^8 meters.
The orbital velocity of the Moon is approximately 1.022 km/s or 1,022 meters per second.
Now we have all the required values to calculate the angular momentum using the formula:
Angular Momentum = (Mass of the Moon) x (Orbital radius of the Moon) x (Orbital velocity of the Moon)
Plugging the values into the formula:
Angular Momentum = (7.35 x 10^22 kg) x (3.84 x 10^8 meters) x (1,022 meters per second)
By performing the calculation, you will find the correct answer for the angular momentum of the Moon in its orbit around Earth.