Calculating the period of a low Earth satelite using Kepler

  • Thread starter barryj
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  • #1
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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
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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
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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
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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
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The value of the constant that you used appears to be for orbits around the sun rather than around the earth.
 
  • #6
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T^2/R^3 for earth = 2.97E-19 (sec^2/m^3)
Your constant is wrong
 
  • #7
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I get it now. Yes, I need a different constant for around earth. I was confused but better now. thanks
 

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