# Calculating the Radius of the Moon's orbital path .. with a twist

Hi,

one of the exercises in my Physics Book, is to calculate the radius of the moon, given the following:

Earth's mass = 5.974e+24 kg
Moon's mass = 7.3483e+22 kg

Moon's average speed = 1023 m/s

Using what you know about Centripetal and Gravitational Force..

Now, I figured that the radius of the moon's orbit, could be calculated with:

r = (moon's mass) * speed^2 / Fc

where:
Fc = Centripetal Force

However, I don't know what the centripetal force of the moon is (in relation to the earth) .. I'm guessing that this is where my (supposed) knowledge of gravity is supposed to come into play.. i.e. Replace the Fc with the Gravitational force between the earth and moon... However, the book doesn't give that value... and surely it can't be calculated without knowing the distance between the earth and the moon (which is the radius of orbit -- which is what we're trying to calculate here!).

Is it possible to calculate this with the info provided? Thanks.
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Commented:
Fc = (6.6726e-11m^3/kg/s^2) * (Earth's mass)*(Moon's mass)/radius^2

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Commented:
Or,  Fc = (Moon's mass)*(acceleration of gravity at the Earth's surface)*(radius of Earth/radius of the moon's orbit)^2
Author Commented:
Hey Ozo,

The problem is though, that those equations use the radius of the orbit (don't they?) .. which is what I'm trying to work out..
Commented:
Yes.
Can you manipulate the equations to isolate the radius on one side of the equation?
Author Commented:
Ah, of course! :-) hehe. Thanks, Ozo!
Author Commented:
..Just for myself (in the future) and for any other users that may be interested.. It can be rearranged to create this equation:

r = (6.6726e-11m^3/kg/s^2) * (Earth's Mass) / Velocity^2

Which comes to about:

380,898,306.7 m

:-)  Cheers, Ozo.
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Math / Science

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