Calculating the moon's speed at perigee with the vis-viva equation

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ryanie
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Hi

Im trying to calculate the speed of the moon at perigee but i don't know where I am going wrong.

As i know the formula is v = sqrt[ GM( (2/p) - (1/a) ) ]

where
g = gravitation constant
m = mass of the earth
p = distance perigee from the earth
a = semi major axis

The values I am using are
g= 6.67384e-11
m = 5.97219e+24kg
p = 363295km
a = 384399km

the result I am getting is 75.5524m/s

Im not even sure if this is correct, could someone advice me in the correct method

*EDIT, showed the unit I am working in. Also as a side note i pulled the data of wikipedia*
 
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Show your units. There is at least one error in this regard that would be more easily visible if you did.
 
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ryanie said:
the result I am getting is 75.5524m/s
That's not even close.

*EDIT, showed the unit I am working in. Also as a side note i pulled the data of wikipedia*
You did not show all of the units you are working in. G has units, too, and this is the source of your problem. You are using inconsistent units.
 
ok so

the new values I am using

G = 6.67384 × 10^-11 m3 kg-1
m = 5.97219*10^24 kg
p = 363295000 m
a = 384399000 m

im getting a result of 1075.7984 m/s

wrong? correct?
 
That's much closer, but it's still wrong. That result would be correct if the Moon's mass was much, much less than the mass of the Earth. It isn't. The mass of the Moon is about 0.0123 times the mass of the Earth. That's small but it is not negligible.

BTW, please stop using text speech. It's "I', not "i", "I'm", not "im".
 
The formula I'm using does not take into account the mass of the moon, only the mass of the earth. Am I using the wrong formula?
 
You have proposed three answers now, varying from 75 meters per second to over one million meters per second. A tiny bit of sanity checking would show you that at least two of those answers are unreasonable.

You know that the radius of the moon's orbit is roughly 360,000 km.
You should know that the period of the moon's orbit is roughly 28 days.

That is enough for a back-of-the-envelope estimate of 360,000,000*2*pi meters in 28*24*60*60 seconds. That's roughly 1000 meters per second.

DH has given you a big hint by saying of your second answer, the one close to 1000 meters per second: "that answer would be correct..."