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

[Nightrider55]

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?

 

[Dick]

That seems pretty reasonable to me. Why do you think there is something wrong with it?

 

[Nightrider55]

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.

 

[Dick]

(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'.

 

[Nightrider55]

Yea I don't know why he did it either. Can any else find a reason for it?

 

[mysqlpress]

I think you are right
5 times a day means 0.2 day for a complete revolution...

 

[tony873004]

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