
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!

Circular Orbits and Period

  1. Sep 28, 2008 #1

    The Hubble Space Telescope orbits Earth 614 km above Earth's surface. What is the period of the telescope's orbit?

    2. Relevant equations

    3. The attempt at a solution

    No attempt, Using that equation, I'd have to know the velocity, and that is where I am stuck.. Any help is appreciated!
  2. jcsd
  3. Sep 28, 2008 #2
    The centripetal acceleration is a = v^2/r...all you need to know is Earth's mass to solve for the acceleration. Then you could find the velocity.
  4. Sep 28, 2008 #3
    Ok, I have the mass of the earth = 5.9742 × 10^24 kilograms, but where do I plug that in to?
  5. Sep 28, 2008 #4
    F = G*Me*Ms/R^2 = Ms*a, Me is mass of earth and Ms is mass of satellite
    a = G*Me/R^2...so for the acceleration you need the mass of the earth and the radial distance, which is R = Re + h, Re being the radius of the earth and h being the altitude above earth's surface (614 km in this case).
    Once you have a and R, using a = v^2/R, you could find the velocity
  6. Sep 28, 2008 #5
    Ok, but the mass of the satellite is not given...
  7. Sep 28, 2008 #6
    No need for the mass of the satellite...it cancels so that a = G*Me/R^2.
  8. Sep 28, 2008 #7
    what is the capital G, is it gravity coefficient (9.81 m/s^2)?
  9. Sep 28, 2008 #8
    No, its the universal gravitational constant...G = 6.673 x 10^-11 N*m^2/kg^2. If you haven't learned this yet then I guess we'd have to solve this another way.
  10. Sep 28, 2008 #9
    No, I haven't learned that yet. You see my professor is a jackass. He's atleast 65 years old, and he is so far behind in lectures as compared to our homework, and we mention this to him and he don't care, all he cares about is his paycheck, so I am stuck trying to learn stuff on my own. None of this stuff is in my book either, which is a E-Book.
  11. Sep 28, 2008 #10
    Using the information you gave me I am still getting the wrong answer. I need the final answer in hours, I am so confused.
  12. Sep 28, 2008 #11
    oh...well...that makes things a bit complicated. I'll go ahead and scratch a bit of the surface of what we know about gravity for you. Gravity is a field force that occurs between masses. To calculate the force of gravity between masses, the general classical expression for is F = G*m1*m2/r^2, m1 and m2 being the 2 masses and r being the distance between their centers. The value of this force is both the force on m1 and m2 and it is an attractive force. In this case, the planet earth is exerting a force on the satellite. At the same time, the satellite has a tangential velocity, which is what you're looking for, that is keeping it from heading straight towards the earth (instead, it orbits the planet). So in this case, the force is point toward the center of the satellite's orbit, making the force a centripetal force....but I'm not sure if you're supposed to know this or not.

    what answer are you getting (so that I can check it with what I have)?
  13. Sep 28, 2008 #12
    Thanks for that information. I found something related to a similar problem using Kepler's third law, are you aware of what this i and how I could apply it to this problem?
  14. Sep 28, 2008 #13
    Kepler came up with his laws empirically before Newton came up with his work. Decades after Kepler's work, Newton did something pretty amazing, which was to prove Kepler's laws with his own work...this further proved the accuracy of his findings and Kepler's. Altogether, using Kepler's laws would get you the same answer as with Newton's. If you were to use the equation for Kepler's third law, all you'd need would be earth's mass and the distance between the satellite and earth, which is Re + h. With that you could find the period.
    Last edited: Sep 28, 2008
  15. Sep 28, 2008 #14
    Ok, I am getting nowhere, I know the answer is supposed to be around 1.6 hrs, can you give me a few extra tips? Thanks!
  16. Sep 28, 2008 #15
    Alright...that seems to be the right answer.
    The force between the satellite and Earth is F = G*Me*Ms/(R^2)...it is the force on the satellite and on the earth. Me is the mass of the earth, Ms is the mass of the satellite, and R is the distance between the two (which is the addition of the radius of the earth and the altitude of the satellite above the earth's surface...so you'll need to know the radius of the earth...the problem gives you the altitude, just be sure to convert it to meters). Force is also equal to M*a...so in the case of the satellite, F = G*Me*Ms/(R^2) = Ms*a. The Ms cancels so that a = G*Me/(R^2), so you do not need the mass of the satellite. Once you have a (and R)... you could solve for v in a = (v^2)/R. The period is T = 2*pi*R/v...the answer you'll get will be in seconds...you'll need to convert to hours.
  17. Sep 28, 2008 #16
    OOOk, I've finally figured it out using Kepler's third law, Thank you for your help Gear300, I really do appreciate it.
  18. Sep 28, 2008 #17
    alright then...looks like you did it differently...as long as you understand the concept of things, you'll be alright.

Have something to add?

Similar Discussions: Circular Orbits and Period
  1. Orbital Periods (Replies: 1)

  2. Orbital Period (Replies: 13)

  3. Orbital Period (Replies: 6)