Register to reply 
Speed of the moon at perigee 
Share this thread: 
#1
Apr614, 08:57 AM

P: 9

Hi
Im trying to calculate the speed of the moon at perigee but i dont know where im 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 im using are g= 6.67384e11 m = 5.97219e+24kg p = 363295km a = 384399km the result im 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 im working in. Also as a side note i pulled the data of wikipedia* 


#2
Apr614, 09:15 AM

P: 999

Show your units. There is at least one error in this regard that would be more easily visible if you did.



#3
Apr614, 09:17 AM

Mentor
P: 3,969

What units are you using?



#4
Apr614, 10:28 AM

Mentor
P: 15,204

Speed of the moon at perigee



#5
Apr614, 01:06 PM

P: 9

ok so
the new values im using G = 6.67384 × 10^11 m3 kg1 m = 5.97219*10^24 kg p = 363295000 m a = 384399000 m im getting a result of 1075.7984 m/s wrong? correct? 


#6
Apr614, 01:15 PM

Mentor
P: 15,204

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". 


#7
Apr614, 01:23 PM

P: 9

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?



#8
Apr614, 01:28 PM

Mentor
P: 15,204

Yes.



#9
Apr614, 01:29 PM

P: 9

What is the correct formula?



#10
Apr614, 01:44 PM

P: 9

is the correct answer 1076040.527 m/s ?



#11
Apr614, 02:09 PM

P: 999

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 backoftheenvelope 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..." 


Register to reply 
Related Discussions  
Calculation of the Distance to the Moons Apogee/Perigee ?  Astronomy & Astrophysics  2  
Speed of Satellite at Perigee and Apogee  Calculus & Beyond Homework  1  
Perigee Apogee Ratio  Introductory Physics Homework  12  
The Ratio of Velocities at Perigee and Apogee Proof  General Math  1  
Apogee and perigee  Astronomy & Astrophysics  2 