Calculating the period of a low Earth satelite using Kepler

  • Thread starter barryj
  • Start date
  • #1
796
44

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.
????
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
13,482
3,681
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?
 
  • #3
796
44
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.
 
  • #4
TSny
Homework Helper
Gold Member
13,482
3,681
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.
 
  • #5
TSny
Homework Helper
Gold Member
13,482
3,681
The value of the constant that you used appears to be for orbits around the sun rather than around the earth.
 
  • #7
796
44
I get it now. Yes, I need a different constant for around earth. I was confused but better now. thanks
 

Related Threads on Calculating the period of a low Earth satelite using Kepler

Replies
5
Views
682
Replies
2
Views
3K
Replies
2
Views
3K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
5
Views
7K
Top