Geostationary Satellite and Orbiting Satellite problem

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Homework Statement


A satellite GeoSAT is in a circular geostationary orbit of radius RG above a point P on the equator. Another satellite ComSAT is in a lower circular orbit of radius 0.81RG. At 7 P.M. on January 1, ComSAT is sighted directly above P. On which day among the following can ComSAT be sighted directly above P between 7 P.M. and 8 P.M. ??
(a) Jan 3
(b) Jan 9
(c) Jan 15
(d) Jan 21

2. The attempt at a solution
I am sorry but I am unable to think of any possible way to solve this problem. Will Keplar's equations help?? :confused: I am totally confused!! Please help!!
For your assistance, the answer given is January 9.
 
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Answers and Replies

  • #2
phyzguy
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What is the period of a geosynchronous orbit with radius RG? Using Kepler's laws, what is the period of an orbit with radius 0.81 RG?
 
  • #3
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Umm...would that help?? :/
 
  • #4
phyzguy
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Umm...would that help?? :/

Yes, it will help - that's why I asked the questions. Can you answer them?
 
  • #5
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Well, the formula to be applied can be.....
T2/R3 = (4 * π2) / (G * M)

where M= 5.98x1024 kg i.e. the mass of the Earth.
Would it help now??
 
  • #6
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The above formula is for geosynchronous as I found out.... but the question tells about geostationary orbit.....
 
  • #7
phyzguy
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The above formula is for geosynchronous as I found out.... but the question tells about geostationary orbit.....
Geosynchronous and geostationary really mean the same thing. Do you know what a geostationary orbit is? If you do, you should be able to tell me the period of a geostationary orbit without doing any calculations.
 
  • #8
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As Wikipedia states, the time period can be
T = 2π √(r3/μ)

What say??
 
  • #10
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Geosynchronous and geostationary really mean the same thing. Do you know what a geostationary orbit is? If you do, you should be able to tell me the period of a geostationary orbit without doing any calculations.

Yeah it's 24 hours! :tongue:
 
  • #11
phyzguy
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Yeah it's 24 hours! :tongue:

OK, good. It's actually 23 hours and 56 minutes, but 24 hours is probably close enough. So if the period of the satellite at radius R = RG is T = 24 hours, and if we know from what you wrote earlier that
[tex]\rm \frac{T^2}{R^3} = constant[/tex]
then what is T when R = 0.81RG?
 
  • #12
D H
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You don't need to know M or G. What you need to do is to answer phyzguy's questions.

What's the period of a geostationary satellite?
What do Kepler's laws say about the period of a satellite? Hint: Only one law says anything about the period.

From that, what is the period of that satellite whose orbital radius is 0.81 RG?
 
  • #13
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oh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?
 
  • #14
phyzguy
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oh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?

Right.
 
  • #15
D H
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oh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?
There's more to it than that. The satellite will be directly overhead after an integral number of orbits.

Here's what you need to solve for: For what integers N is the time needed to make N orbits between M days and M days plus one hour, where M is another integer?
 
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