Acceleration given distance from earth (r)

  • Thread starter Pajarito
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  • #1
Pajarito
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Question: Satellites are placed in a circular orbit that is 3.59 X 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

So far I've been using g=(M/r2) and get .31m/s2. There must be something wrong with either using this equation or that g isn't the acceleration. Can you explain how to reach acceleration from this question?
 

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  • #2
Nessdude14
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In the equation you posted, you're missing the gravitational constant "G." However, I suspect you used it in your calculation because I got the same answer when I included G. You seem to be assuming that your answer of .31 m/s2 is wrong, but it seems fine to me.

I wouldn't use small g to denote the acceleration in orbit though. Small g usually refers to the acceleration due to gravity on Earth's surface. Instead, you could call it "a" for acceleration.
 
  • #3
Pajarito
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Normally I'd use the traditional gravitational formula, but since I don't have a mass for the second object (satellite) I'm not sure if I can use it. F=(GM1M2)/r2. Since F=ma, I wonder if once I solve the equation I should divide it by the mass of Earth again to isolate a, that would lead to a very small number though and I don't think that would be accurate, any suggestions about doing it that way?
 
  • #4
Nessdude14
172
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Normally I'd use the traditional gravitational formula, but since I don't have a mass for the second object (satellite) I'm not sure if I can use it. F=(GM1M2)/r2. Since F=ma, I wonder if once I solve the equation I should divide it by the mass of Earth again to isolate a, that would lead to a very small number though and I don't think that would be accurate, any suggestions about doing it that way?

Sure, you can use Newton's law of gravitation no problem. That's what the equation you used is derived from. Here's how:

[itex]\displaystyle F=G\frac{mM_E}{r^2}[/itex]

[itex]\displaystyle a=F/m[/itex]

Substitute F:

[itex]\displaystyle a=G\frac{mM_E}{r^2}/m[/itex]

and m cancels to give:

[itex]\displaystyle a=G\frac{M_E}{r^2}[/itex]

Then, plugging in the radius given by the problem, and putting in the numerical values for G and ME (mass of earth):

[itex]\displaystyle a=(6.673×10^{-11})\frac{5.972×10^{24} kg}{(3.59×10^7 m)^2} = 0.31\frac{m}{s^2}[/itex]

Are you convinced that you got the right answer yet? :)
 
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  • #5
Pajarito
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I'm convinced but it's this online system that won't mark the answer as correct, I'll check with my professor. Thanks for you help.
 

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