1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Total Energy of a satellite

  1. Jan 23, 2007 #1
    Calculate the total energy of a geosynchronous satellite (one that orbits over a fixed spot) with a mass of 1500kg, orbiting Earth at a height of 325km with an orbital speed of 5000m/s

    This question is starting to drive me a little mad. First of all, the satellite can't be in a geosynchronous orbit AND traveling at 5000m/s 325km above the earth. The two just don't jive.

    Secondly, I am completely at a loss as to what formulas I should be using.

    Should I simply be using 1/2mv^2 + mgh (where g = equals the reduced gravity of 8.8756069 at that height) for the question or is it much more involved than that?

    I've read over articles involving critical velocities, gravitational attraction, circular motion, and centrifugal force until my head is spinning.

    Can some please please give me a nudge in the right direction?

  2. jcsd
  3. Jan 24, 2007 #2

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    Oh my god.

    Whoever framed this question won't have a clue as what the answer ought to be. You are totally correct in that a geosychrynous satellite won't be at that altitude or speed. Not only that, a satellite at the altitude given, won't have the speed that's given either.

    And you are also correct in that the "mgh" formula is too simple. Since g will not be constant. I bet, however, this is what the sorry excuse for a teacher wants. I'd say give him/her the 1/2mv^2 + mgh answer, then explain how wrong the question is.


    A geosynchronous satellite must be at an orbital radius of about 40,000,000 meters, and it's speed is about 3000 m/s. This is by using the period (T)of one earth day (in seconds), and satisfying three formulas:
    F= GMm/r^2
    and v=(2(pi)r)/T

    for any given T, there is only one v and one r that works.

    The Total gravitational potential energy of an object is actually zero at a large distance from the earth, and is considered negative the closer you get to earth, using the equation U=-GMm/r (not r^2)
    Last edited: Jan 24, 2007
  4. Jan 24, 2007 #3
    Thanks Chi, much appreciated
  5. Jan 24, 2007 #4
    If you're really ambitious, you could assume that he's talking about some other planet coincidentally named "earth," and satisfy the three requirements that Chi mentioned.
  6. Jan 24, 2007 #5
    Yeah, I had considered that, I would rather work with the numbers they gave me seeing has it clearly is not based on the Earth I know.

    Update: In order to make the equation work, I have changed the mass of my bizarro Earth to: 2.51311844078*10^24 kg and a full day only lasts 2.34 hours.
    Last edited: Jan 24, 2007
  7. Jan 25, 2007 #6
    Ok, after making the changes to the mass of earth as stated above I have:

    G = 6.67*10^-11 (N*m^2)/kg^2
    M = 2.51311844078*10^24 kg ***revised mass to fit the question***
    m = 1500 kg
    v = 5000 m/s


    F = (GMm)/r^2

    F = 5592.84116331 N

    U = -(GMm)/r ***not r^2***

    U = -3.75*10^2 J

    Ek = (1/2)(GMm/r)

    Ek = 1.875*10^10 J

    Et = Ek +U

    Et = -1.875*10^10 J

    The Total energy of the satellite is -1.875*10^10 J.

    Does that look right to you guys? Thanks in advance
    Last edited: Jan 26, 2007
  8. Nov 6, 2008 #7
    I used 2 different formulas to find Ek, and got a different number from yours...

    I dont know what number you used for the radius, but I used 6378.1 km, which I found off google...

    Ek = mv2
    Ek = (1500kg)(5000m/s)2
    Ek = 3.75 x 1010 J

    Ek = -GMm/r
    Ek = -[(6.67x10-11Nm2/kg2)(2.513x1024kg)(1500kg)]/6 378 100m
    Ek = 2.514x1017Nm/6 379 100
    Ek = 3.94x1010Nm

    The answers are slightly different, possibly due to rounding... I dont understand why your equation is halved...

    When I use the earth's real mass, I get Ek = 9.37 Nm, which does not match my first equation at all...

    Also, is there any use for finding F?
  9. Nov 6, 2008 #8

    oopss, nevermind this response, it makes sense
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Total Energy of a satellite
  1. Energy of a satellite (Replies: 11)