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Satellite Problem

  • #1

Homework Statement



A satellite traveling in circular obit about the equator of the earth in the opposite direction of the rotation of the earth. It passes over a position of the earth every 12 hours. What is the velocity and altitude of the satellite?

Homework Equations



G (universal gravitational constant) = [tex]6.6742 \times 10^{-11} \frac{Nm^2}{kg}[/tex]

[tex]M_E[/tex] (mass of the earth) = [tex]5.97 \times 10^{24} kg[/tex]

[tex]R_E[/tex] (radius of the earth) = [tex]6.38 \times 10^6 m[/tex]

The Attempt at a Solution



My attempt is attached below. Can you check this to make sure I am doing it right? No need to whip out the calculator and check my calculations, rather just check to see if my steps are right. If anything looks wrong please tell me. Thanks a lot.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
ideasrule
Homework Helper
2,266
0
That's right, but I think the question asked for the speed of the satellite relative to the ground. You calculated its speed relative to an inertial reference frame.
 
  • #3
It just said what is the speed and altitude of the satellite so I just assumed the speed was its distance divided by the time it took. Does velocity by default mean relative to the ground? How would I go about solving that? Thanks!
 
  • #4
ideasrule
Homework Helper
2,266
0
It just said what is the speed and altitude of the satellite so I just assumed the speed was its distance divided by the time it took. Does velocity by default mean relative to the ground? How would I go about solving that? Thanks!
Actually, the question is badly worded; velocity usually means relative to an inertial reference frame. But if that's the case, whether the satellite orbits opposite to Earth's rotation would be irrelevant, so I assumed the question was talking about speed relative to ground.

To find it, you just add Earth's rotational speed to the orbital speed, since the two are in opposite directions.
 
  • #5
But wouldn't how high it is off the ground play into its relative speed or is that irrelevant? I was thinking maybe it was talking about velocity with respect to a certain point on the trajectory of the circle and they just told us the rotational information so we can find the period. I don't know do you think that could be a possibility? So I have 3071 m/s as my velocity so relative to ground it is 3071m/s + [2pi(R of earth + h)/(24 hrs=86400 sec)? Well I used the velocity, 3071m/s, to calculate my altitude so is my altitude wrong? What if the satellite was in synchronous orbit? Then would the velocity be just the earth's speed or would it be zero? A lot of questions, sorry.
 

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