Orbital Period of Apollo Mission

Click For Summary
SUMMARY

The orbital period of the Apollo command module, placed in a low circular orbit approximately 60 km above the Moon's surface, is calculated using the formula t² = 4π²r³ / GM. With the Moon's radius at 7738 km and its mass at 7.34 x 10²² kg, the correct orbital period is approximately 17.2 hours or 6.18 x 10⁴ seconds. The confusion arose from incorrect unit conversions, specifically using kilometers instead of meters, which led to discrepancies in the calculated period. Ultimately, the correct approach confirms the orbital period aligns with the expected physical parameters of the Moon.

PREREQUISITES
  • Understanding of gravitational physics and orbital mechanics
  • Familiarity with the formula t² = 4π²r³ / GM
  • Knowledge of unit conversions, particularly between kilometers and meters
  • Basic proficiency in algebra and square root calculations
NEXT STEPS
  • Study gravitational physics and its applications in orbital mechanics
  • Learn about unit conversions in scientific calculations
  • Explore the implications of orbital periods in space missions
  • Investigate the historical context and details of the Apollo missions
USEFUL FOR

Students studying physics, aerospace engineers, and anyone interested in the calculations behind space missions and orbital mechanics.

Immy2000
Messages
8
Reaction score
0

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!
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K