Scatch space mod prob. is need a formula and answer check

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SUMMARY

The discussion centers on calculating the orbital speed and period of a command module around the Moon. The speed was determined to be 1634 m/s using the formula Vcm² = GMm/Ro, where the mass of the Moon is 7.35E22 kg and the radius is 1.74E6 m. The period was initially calculated using Kepler's 3rd law, resulting in a time of 7063 seconds (or 118 minutes), which was later confirmed to be accurate. Participants emphasized the importance of verifying calculations and considering simpler methods for determining orbital period.

PREREQUISITES
  • Understanding of gravitational constants, specifically G = 6.67 x 10^-11 m^3 kg^-1 s^-2
  • Familiarity with Kepler's 3rd law of planetary motion
  • Basic knowledge of orbital mechanics and circular motion
  • Ability to perform calculations involving mass, radius, and speed
NEXT STEPS
  • Study the derivation and applications of Kepler's 3rd law in orbital mechanics
  • Learn about gravitational force calculations and their implications in space missions
  • Explore alternative methods for calculating orbital periods, such as using circumference and speed
  • Research the impact of altitude on orbital speed and period calculations
USEFUL FOR

Astronomy students, aerospace engineers, and anyone involved in orbital mechanics or space mission planning will benefit from this discussion.

cowgiljl
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i think i got it right it wants to know the speed of the command module and the time it takes to do 1 revolution
mass moon = 7.35E22 kg
radius moon = 1.74E6 m
altitude 60 mi

fromula used was Vcm^2 = GMm/Ro

the speed of the module was 1634 m/s

The time formula thans to enigma was

P = 2\pi \sqrt{\frac{a^3}{\mu}}}
T = 6511 seconds/60
T = 108 min

is this right?
 
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Your speed looks correct, using your value for mass of the moon.

But I don't follow what you're doing to get the period.

I'm using Kepler's 3rd law:
T^2 = (\frac{4\pi^2}{GM_m})r^3
and here r = (1.74 x 106 + 1609 x 60) so using G=6.67x10-11 and Mm=7.35x1022 I'm getting
T = 7063 s or 118 min.

(Incidentally, my book lists the mass of the moon as 7.36x1022 kg, but what's a mere 100,000,000,000,000,000,000 kg among friends?)
 
you re right gnome

I wrote down and typed the wrong answer but my calculations were correct. the mind was quicker than the pen

thanks again
 
Sometimes its good to step back & think about what we're doing.

While I was walking my dog, I was wondering why we're using that crazy formula to get the time of 1 revolution. (Heresy -- calling Kepler's 3rd law a crazy formula! )

We already know the speed. Just divide the circumference of the orbit by the speed & get the same answer.]
 

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