# Orbital Period of Apollo Mission

## Homework Statement

On each of the apollo missions the command module was placed in a very low aprox circular orbit above the moon. Assum the avrg hieght was 60km above surface of moon and moons radius is 7738km. (Mass of moon=7.34x10^22kg)

What was the command modules orbital period?

M=7.34x10^22kg
r=7738+60
G=6.67x10^-11

## Homework Equations

t^2=4pi^2r^3/GM(source)

## The Attempt at a Solution

T^2=4pi^2(7738+60)^3 / (6.67x10^-11)(7.34x10^22)

T^2= (1.87...x10^13) / (4.89...x10^12)

T^2= 3.8237...

T=Square Root(ANS)

T= (1.9554.. Days x 24 Hours) = (46.93.. Hours)

But the back of my worksheet says the answer is 17.2 hours or 6.18x10^4 seconds

GAH I am so confused! >_< Your help is much appreciated!

LowlyPion
Homework Helper
Isn't the radius of the moon 1,738 km?

Yeah it is but I am assuming I am supposed to use what the question gives me.. let me redo the calculation with 1738km's though...

Alright so I get to the part where you have to square root T^2 and I get 0.21649... days (I think) and if I multiply that by 24, I should get hours, but when I do so I get 5.19...which is off from 17.2 hours..

JEEZ I think I made a really silly mistake. Radius is in "m" right? I plugged in Km! Oh my!

Yeppers, it was the conversion. Wow. Please Lock or Delete this thread. :)

Last edited:
Why do you think the result of the calculation will be in days?
If you use the quantities in SI units, the period will be in seconds.
Then you can convert in hours, of course.
The answer is around 1.8 hours.

Yeah I later realized it was in seconds but as far as the answer goes, its 17.2 hours. Thanks for the help guys! :)

If you take the radius 7738 then you get something close to 17 hours. But I thought this value is just a typo. This will mean the Moon is bigger than Earth....

I know its quite silly however the answer was calculated with that number. Therefore I assume that it was typoed and then answered, then printed and handed out to the students.