Calculating the period of a low Earth satelite using Kepler

[barryj]

Homework Statement

Find the period of a low Earth orbiting satellite using Kepler Laws
earth radius = 6.38E6 meters
T^2/R^3 for Earth = 2.97E-19 (sec^2/m^3)

Homework Equations

2.97E-19 = T^2/(6.38E6)^3
T^2 = (2.97E-19)(6.38^6)^3 = 77.1
T = 8.72 sec

The Attempt at a Solution

This is not correct since I know the period of a low earther is about 90 minutes or about 5600 seconds.
?

 

[TSny]

T^2/R^3 for Earth = 2.97E-19 (sec^2/m^3)

Hello. The number on the right doesn't look correct. How did you get that value?

 

[barryj]

The number came from a chart that was supplied with a homework set for physics.
I realize that there are other variants of this number with different units such as
3.35 10E24km^3/yr^2) this shyulkd be the same value if the units are converted.
Maybe there is an error in one of he documents.

 

[TSny]

The value given in your table is incorrect. You can calculate the correct value yourself in terms of the mass of the earth.

See http://pvhslabphysics.weebly.com/keplers-3rd-law.html

Here $a$ is the radius of the orbit if the orbit is circular. If the orbit is elliptical, then $a$ is the length of the semi-major axis of the ellipse.

 

[TSny]

The value of the constant that you used appears to be for orbits around the sun rather than around the earth.

 

[Phylosopher]

T^2/R^3 for Earth = 2.97E-19 (sec^2/m^3)

Your constant is wrong

 

[barryj]

I get it now. Yes, I need a different constant for around earth. I was confused but better now. thanks