Finding the orbital radius of a GPS satellite with 8-hour period

[D H]

You're right too. That is a very big problem.

 

[Ryo124]

Alright, I changed it and got 2.56 x 10^7 m for the radius of the orbit. . . Can someone work the problem out and see if this is the correct answer?

Also, no one told me if my final equation was correct... [2(pi)^2]r^3 = GmeT^2
Is this equation correct?

 

[nrqed]

Alright, I changed it and got 2.56 x 10^7 m for the radius of the orbit. . . Can someone work the problem out and see if this is the correct answer?

Also, no one told me if my final equation was correct... [2(pi)^2]r^3 = GmeT^2
Is this equation correct?

I get $2.03 \times 10^7 m$ You forgot to square the 2. It's (2 pi)^2

 

[Ryo124]

OK. I did it and got 2.04 x10^7 m. I did you just round up in your calculation?

 

[nrqed]

OK. I did it and got 2.04 x10^7 m. I did you just round up in your calculation?

I keep getting 2.03 x 10^7 m. I did not round off.
(I mean I rounded off my final answer but not in the intermediate steps)

 

[Ryo124]

OK. Weird, I did it again and got 2.03 x 10^7. Cool.

Thanks for your help and patience everyone, I guess I made this problem harder than it needed to be.

 

[D H]

Since you only used two decimal places of accuracy for G and M_e, it would be better to say 2.0x10⁷ meters in this case.

As an aside, it is often better to use the planet gravitational constant (the product of universal gravitational constant and the planet mass rolled into one constant) rather than the product of universal gravitational constant and the planet mass as separate values. What's the difference? Accuracy. In the case of the Earth, we know the product G M_e to 9 places of accuracy: 398,600.4418±0.0008 km³/s². In comparison, we only know G to 4 decimal places.