Radius of Orbits

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Homework Statement



A GPS satellite is put in a high circular orbit around the earth. The period of revolution is 8 hours. Calculate the radius r of the orbit, compared to the radius R of the earth.

Homework Equations



Rearth= 6.4 x 10^6 m
mearth= 6 x 10^24 kg

The Attempt at a Solution



I have no idea how to attempt this problem. Can someone please help?
 
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Answers and Replies

  • #2
Dick
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Write down the acceleration of an object moving in circular orbit of radius r with period T. Equate that to the acceleration caused by gravity at radius r.
 
  • #3
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It is not clear to me what you mean. Don't you mean velocity? How does the mass of the earth come into play? So, step-by-step, what do I do? Whats first? Please Help.
 
  • #4
Dick
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No, I mean acceleration. An object moving in a circular path experiences a centripetal acceleration. There's a formula for it in terms of v and r. Now equate that to the gravitational acceleration. There's a formula for that as well. Find those formulas and set them equal.
 
  • #5
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Ohhh... I see now that it is ac. mv^2/r right? I'll try to work it out and respond if i get stuck, thanks.
 
  • #6
nrqed
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Ohhh... I see now that it is ac. mv^2/r right? I'll try to work it out and respond if i get stuck, thanks.
Right (m is the mass of the satellite). Now set this equal to the force of gravity of the Earth on the satellite (using the universal law of gravitation). You will also need to use that for circular motion, [tex] v = \frac{2 \pi r }{ T} [/tex]
 
  • #7
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[tex] v = \frac{2 \pi r }{ T} [/tex]
Do I find the velocity for earth's orbit AND the satellite with this equation? Or just the satellite? Isn't the mass of the satellite negligible, so set it to 1?
 
  • #8
Dick
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I think finding the radius of the satellite is the question. You can set the mass of the satellite equal to one if you wish, it will cancel from both sides of the equation you have yet to write down. And sure, you'll want to assume it's negligible compared with the earth so you can assume the earth is stationary while the satellite orbits. WRITE DOWN THE EQUATION, ok?
 
  • #9
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mv^2/r = Gm1m2/R^2

Is this the equation? And all values are known except r, correct? And you find the v from [tex] v = \frac{2 \pi r }{ T} [/tex] correct?

So this would mean that the centripetal force from the satellite equals the gravitational force between the earth and the satellite??
 
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  • #10
Dick
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Finally. Yes. One of the m1 and m2 on the right is equal to m on the left. Cancel it. The remaining m is mass of the earth, right?
 
  • #11
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How are you supposed to find the v of the satellite using [tex] v = \frac{2 \pi r }{ T} [/tex] if you don't know what r is???
 
  • #12
D H
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From post #9, you have [itex]v^2/r=GM_e/r^2[/itex]. You don't know [itex]v[/itex] but you do have an expression for it: [itex]v=2\pi r/T[/itex]. So substitute this expression in the equation from post #9 and solve for [itex]r[/itex].
 
  • #13
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I keep getting 8.08 x 10^14 m for r, but this is incorrect. Is my math just wrong? You set [2(pi)r/(T)]^2 = GME/RE^2 right???
 
  • #14
Dick
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No. i) the quantity on the left should be v^2/r. I think you are missing the r. And the R_E on the right should be r of the satellite, right?
 
  • #15
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Sorry. When I did the calculation I had the r on the left side and the right side, but I still keep getting the answer incorrect.

The equation is [itex]v^2/r=GM_e/r^2[/itex] with [tex] v = \frac{2 \pi r }{ T} [/tex]

I know this and keep getting the wrong answer of 8.08 x 10^14. What are you getting??
 
  • #16
Dick
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I get something round about 10^7m. Can you post your final equation for r and maybe some intermediate results?
 
  • #17
Dick
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You should be getting that r^3 is proportional to T^2. That's one of Kepler's laws.
 
  • #18
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I am getting [2(pi)r]^2 x [1/T]^2 = GME/r^2

From here I can't simplify and I am not getting 10^7. I'm getting around 10^14 instead.

When I tried to simplify, I got [2(pi)r]/(8.29x10^8) = [4.002x10^36]/(r^2). Then I crossed multiplied and got the wrong answer.

I am not simplifying correctly, can someone help me with this, I feel dumb.
 
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  • #19
Dick
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The 1/r in v^2/r has disappeared again. You should be able to simplify it to r^3=(something)*T^2. What's the (something)?
 
  • #20
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Is it GME/ 2(pi) ???
 
  • #21
Dick
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Would you believe GM_E/(2pi)^2??
 
  • #22
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I plug that in and get 5.5 x 10^14 m, which is wrong.
 
  • #23
D H
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Show us exactly what you did. Believe it or not, Dick's expression is quite correct.
 
  • #24
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OK. Centripetal force from the satellite equals the gravitational force between the earth and the satellite:

mv^2/r = Gm1m2/r^2

The mass of the satellite is negligible:

v^2/r = Gme/r^2

The velocity in terms of radius and period is 2(pi)r/T so:

[2(pi)r/T]^2/r = Gme/r^2

An r cancels on the left side and you have:

[2(pi)^2(r)]/T^2 = Gme/r^2

Cross multiply and get:

[2(pi)^2]r^3 = GmeT^2

I plug in the numbers and get 5.5 x 10^14 m, which is wrong.
 
  • #25
Dick
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Tell us what numbers you are plugging in where. Otherwise, are we supposed to figure out why you are getting a wrong number?
 

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