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Satellite orbiting the earth

  • Thread starter mmh37
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  • #1
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Hi everyone!
I have some trouble with the following problem. Can anyone give me a hint?

This is the problem:

A sattelite is to be located always above the same point on Earth. What is the radius of its orbit and how is the plane of the orbit inclined wrt the North-South axis of Earth?


my attempt:

1) find angular velocity w of the earth (one rotation per day) and this must be the same for the sattelite

now,

[tex] T^2=(2*pi*r / v)^2 [/tex]

where v = w * R(earth)

HOWEVER: only m(earth) and not R(earth) is given in the question. Therefore I do not "officially" know R(earth)

2) I have no idea how to do this one. I have been thinking this through several times already, but I do not see to which known quantity the angle could be related

If anyone would be able to give me a hint (or two :wink: ), that would be amazing!
Thanks so much!
 

Answers and Replies

  • #2
Doc Al
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What conditions must be met for a circular orbit about the earth? Hint: Think Newton's 2nd law, gravity, and circular motion.
 
  • #3
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Thanks for this. I think it works like this:

[tex] F = \frac {G*m*M} {R} = m*v^2 /R [/tex]

solve for v

now [tex] a = v^2/R = w*v [/tex]

Therefore using Newton's 2nd law and zentripedal force:

[tex] m*a = m*w*v = mv^2/R [/tex]

So, now we just have to solve for R.

However, I still really don't see anything for part b). I don't see how the angle is related to any of the known quantities...if I try and visualize it I always come to the conclusion that it must be 90 degrees, as the satellite should stay above the same point on the earth.
 
Last edited:
  • #4
nrqed
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mmh37 said:
Thanks for this. I think it works like this:

[tex] F = \frac {G*m*M} {R} = m*v^2 /R [/tex]
Watch out...the force goes like 1/r^2

solve for v

now [tex] a = v^2/R = w*v [/tex]

Therefore using Newton's 2nd law and zentripedal force:

[tex] m*a = m*w*v = mv^2/R [/tex]

So, now we just have to solve for R.

However, I still really don't see anything for part b). I don't see how the angle is related to any of the known quantities...if I try and visualize it I always come to the conclusion that it must be 90 degrees, as the satellite should stay above the same point on the earth.
That' sright.. The satellite must be above the equator.


Patrick
 
  • #5
Doc Al
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mmh37 said:
Thanks for this. I think it works like this:

[tex] F = \frac {G*m*M} {R} = m*v^2 /R [/tex]

solve for v

now [tex] a = v^2/R = w*v [/tex]

Therefore using Newton's 2nd law and zentripedal force:

[tex] m*a = m*w*v = mv^2/R [/tex]

So, now we just have to solve for R.
As nrqed already explained, you missed a factor of R in your expression for gravitational force. Also, you might find a different expression for centripetal acceleration easier to work with:
[tex]a_c = v^2/R = \omega^2 R[/tex]
 
  • #6
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Right, you need to use the fact that the orbit time is a day, because it stays over the same point on the earth all the time (It must be over the equator for this to be possible i think).
then use the equation that has already been put on the thread, in words:

radius of the sattelites orbit to the power of three is equal to universal gravitational constant multiplied by the mass of the earth, divided by the satellites angular velocity squared.

r^3 = (GM)/w^2

G = 6.67x10^-11
mass of earth = 6x10^24 (approx)


I have worked it out below so if you want to do it for yourself then don't look!! hope this helped!




if it has to orbit over the same point above the earth all the time, it must have an orbit time period of 24 hours (approximately) which is 86400 seconds. so it's frequencey is 1/86400, which is (1.15 x10^-5).
angular velocity, w = 2pi x f = (7.27 x 10^-5)

if radius of orbit is r, r^3 = (GM)/w^2

r^3 = (6.67x10^-11 x 6x10^24)/(7.27x10^-5)^2
r^3 = 7.57x10^22
r = 4.23 x 10^7 to 3 significant figures
 
  • #7
Hootenanny
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maltesers said:
Right, you need to use the fact that the orbit time is a day, because it stays over the same point on the earth all the time (It must be over the equator for this to be possible i think).
then use the equation that has already been put on the thread, in words:

radius of the sattelites orbit to the power of three is equal to universal gravitational constant multiplied by the mass of the earth, divided by the satellites angular velocity squared.

r^3 = (GM)/w^2

G = 6.67x10^-11
mass of earth = 6x10^24 (approx)


I have worked it out below so if you want to do it for yourself then don't look!! hope this helped!




if it has to orbit over the same point above the earth all the time, it must have an orbit time period of 24 hours (approximately) which is 86400 seconds. so it's frequencey is 1/86400, which is (1.15 x10^-5).
angular velocity, w = 2pi x f = (7.27 x 10^-5)

if radius of orbit is r, r^3 = (GM)/w^2

r^3 = (6.67x10^-11 x 6x10^24)/(7.27x10^-5)^2
r^3 = 7.57x10^22
r = 4.23 x 10^7 to 3 significant figures
Complete solutions shouldn't generally be posted. It is more constructive for people to work through the problems themselves. It also deters the people who just want us to do their homework for them :frown:
 

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