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Time it takes for a satellite to orbit the Earth

  • Thread starter balletgirl
  • Start date
  • #1
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Homework Statement



Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. The definition of "near-Earth" orbit is one which is at a height above the surface of the Earth which is small compared to the radius of the Earth, so that you may take the acceleration due to gravity as essentially the same as that on the surface. Does your result depend on the mass of the satellite?

M(earth)= 5.98x10^24 kg
R(earth)=6.38x10^6 m
G= 6.67x10^-11

Homework Equations



g= GM/r^2

FG= Gm1m2/r^2

The Attempt at a Solution



I am not sure where to start since I don't know what equation to use.
 

Answers and Replies

  • #2
44
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One more equation might help you. The satellite in orbit will have an acceleration equal to [tex]a = \frac{v^2}{r}[/tex] which will equal earth's gravity, [tex]g = \frac{GM}{r^2}[/tex]

Setting those equal to each other, can you determine the velocity of the satellite and use that to find the period?
 
  • #3
rock.freak667
Homework Helper
6,230
31
The gravitational force between the Earth and the satellite provides the centripetal force of the satellite. Are you able to make an equation?
 
  • #4
36
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Okay, I'm not sure if this is right, but here's my attempt:

I did v^2/r = GM/r^2
& plugged in:
V^2/(6.38x10^6) = (6.67x10^-11)(5.98x10^24)/(6.38x10^6)

and ended up with:
V= 1.99x10^7 m/s

when i plugged it in to a=v^2/r i got a= 6.25x10^7

this doesn't seem right, isn't is suppose to be near 9.8 m/s?
 
  • #5
44
0
You didn't square the radius on the right side.
 
  • #6
36
0
Wow, dumb mistake...I redid it by squaring it and got 9.796 m/s. Thank you for your help!
 

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