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

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Homework Help Overview

The discussion revolves around determining the radius of a satellite's orbit required for it to complete five revolutions around the Earth in one day, while neglecting the moon's influence. The participants are exploring gravitational equations and period calculations related to orbital mechanics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the orbital radius using the gravitational equation and period derived from the number of seconds in a day divided by five. Others question the validity of the period calculation and compare it to a teacher's different approach.

Discussion Status

Participants are actively discussing discrepancies in period calculations and interpretations of the problem. Some suggest that the teacher's method may be incorrect, while others are seeking clarity on the reasoning behind the different calculations. There is no explicit consensus yet, but the conversation is focused on understanding the assumptions and calculations involved.

Contextual Notes

There is a noted difference in how the period is calculated, with one participant using a method that results in a period of 0.2 days for a complete revolution, while another participant references a calculation that suggests a longer period. The discussion is constrained by the need to clarify these differing interpretations.

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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