Satellite Orbit Radius for 5 Revs/Day: Get the Answer

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SUMMARY

The discussion centers on calculating the orbital radius of a satellite that completes 5 revolutions per day around the Earth. The correct period for this scenario is 0.2 days, leading to an orbital radius of approximately 1.44 × 107 meters, as derived from the gravitational equation T2 = (4π2/(GM))r3. A misunderstanding arose from differing interpretations of the period calculation, where one participant incorrectly calculated the period as 5 x 12 x 3600 seconds instead of the correct (24 x 60 x 60)/5 seconds. The accurate radius calculation confirms the participant's approach, while the teacher's method was flawed.

PREREQUISITES
  • Understanding of gravitational equations, specifically T2 = (4π2/(GM))r3
  • Basic knowledge of orbital mechanics and satellite motion
  • Familiarity with unit conversions between days, seconds, and meters
  • Ability to perform calculations involving periods and frequencies of motion
NEXT STEPS
  • Research the derivation of Kepler's Third Law of planetary motion
  • Learn about the implications of orbital radius on satellite stability
  • Explore online calculators for orbital mechanics, such as the one mentioned in the discussion
  • Study the effects of gravitational forces on satellite trajectories
USEFUL FOR

Aerospace engineers, physics students, and anyone involved in satellite design and orbital mechanics will benefit from this discussion.

Nightrider55
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QUESTION:
Suppose we want a satellite to revolve around the Earth 5 times a day. What should the radius of its orbit be? (Neglect the presence of the moon.)

WORK:

Found the number of second in 24 hours and divided it by 5 to get the period. Then rearranged the gravitational equation with T^2=((4pi^2)/(GM))r^3 and solved for r. I came up with 1.44 × 10^7 m but its wrong. Any insight?
 
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That seems pretty reasonable to me. Why do you think there is something wrong with it?
 
My teacher calculated 7.78 x 10^7 m. The difference is our calculations of the period. He did 5 x 12 x 3600 for the period while I did (24 x 60 x 60)/5. I don't understand where he came up with 12. Could someone help me see where my teacher or I made a mistake.
 
(24*60*60)/5 is 5 times a day. 5*12*3600 is once every 60 hours. If you are reading the same problem I don't see how he could interpret that as '5 times a day'.
 
Yea I don't know why he did it either. Can any else find a reason for it?
 
I think you are right
5 times a day means 0.2 day for a complete revolution...
 
Here's an online calculator I made. It does the unit conversions for you. Choose units of days, Earth masses, and seconds. Plug in 0.2 for period (1/5 of a day), 1 for Mass, and it will compute seconds for you: 1.44E7. Your teacher is wrong. You are right.

http://orbitsimulator.com/gravity/articles/formula55.html
 

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