How Long Does It Take to Orbit the Moon at an Altitude of 100km?

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SUMMARY

The discussion centers on calculating the orbital period of a spacecraft orbiting the Moon at an altitude of 100 km. Using the gravitational constant (G = 6.67 x 10^-11), the mass of the Moon (Mm = 7.35 x 10^22 kg), and the Moon's radius (Rm = 1.74 x 10^6 m), participants derived the orbital velocity (v) and period (T). The correct calculation yielded an orbital period of approximately 7,112 seconds, correcting earlier miscalculations of 2.9 x 10^9 seconds and 1.89 x 10^10 seconds.

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Homework Statement



During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of about 100km. How long did it take to go around the Moon once?

Rm= 1.74x10^6 m
Mm= 7.35x10^22 kg
G= 6.67x10^-11
r= Rm + 100,000 m

Homework Equations



v= sqrt(Gm/r)

V=2pi*r/t

The Attempt at a Solution



after using the first equation, i got v= 1,679 m/s

i plugged this into the second equation to solve for T, and got
T= 2.9x10^9 s

This seems way too high to be the answer...
 
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It would appear you forgot to consider "r= Rm + 100,000 m"
 
You're right I didn't, but this time got 1.89x10^10.

Here is my work:
v=sqrt(6.67x10^-11)(7.35x10^22)/[(1.74x10^6)+100,000]
v= sqrt(2.66x10^6)
v= 1,631 m/s

1,631= 2(3.14)(1.74x10^6 + 100,000)/T
1631= 1.16x10^7/T

T=1.89x10^10 s
 
Continuing your work:

1631 = \frac{1.16 \times 10^7}{T}

T = \frac{1.16 \times 10^7}{1631} \neq 1.89 \times 10^{10} s

You have a simple arithmetic error in there somewhere.
 
Okay I believe I figured it out, and got 7,112 s instead, which seems much more reasonable. Thank you!
 

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