# 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
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.
?

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?

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.

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.

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

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