1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Centripetal Acceleration of a satellite above a planet of the Earth's radius.

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data
    A satellite is in orbit just above the surface of a spherical planet which has the same radius as Earth and the same acceleration of free fall at it's surface. Calculate:
    i) Speed
    ii) Time for 1 complete orbit.

    Radius of Earth = 6400km or 6400000m and accel. of free fall = 9.81ms-2
    2. Relevant equations
    Speed (v) = (2pi)r/T

    Cent. Accel. (a) = v2/r

    Cent. Force. F = mv2/r = mw2r

    3. The attempt at a solution

    First time on here so finding it difficult to locate symbols so sorry for loss in translation.

    My attempt of a solution is: v = (2xpi)x 6400000/3.1e7 (no of s in 1 year (365.25days))

    v = 1.29ms-1
    then plugged this in to v2/r but this is heading down a blind alley.
    Please Help.
  2. jcsd
  3. Jul 6, 2011 #2


    User Avatar
    Science Advisor

    If you can't use LaTeX, "^" is standard for powers. Or [ sup]2[/sup] (without the first space).

    Why would you believe "v = (2xpi)x 6400000/3.1e7" Are you assuming the satellite takes one year to orbit the planet???

    The simplest thing to do is to calculate the centripetal force on the satellite- that is just the force holding in orbit which is just the gravitational force on the satellite:
    [tex]F= -\frac{GmM}{r^2}[/tex]
    "G" and "M" are those values for the earth and m is the mass of the satellite, r is 6400 km.

    And then, of course, F= ma so
    [tex]a= -\frac{GM}{r^2}[/tex]

    Now, on the surface of the earth, the acceleration due to gravity is -9.81 and the radius is, to two significant figures, 6400 km! Adding another 6400 for the altitude of the satellite, we have r= 2(6400) so [itex]r^2= 4(6400)^2[/itex].
  4. Jul 6, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    hi mogsplanet! welcome to pf! :smile:

    (try using the X2 icon just above the Reply box :wink:)
    what does the year have to do with it? :confused:

    as usual, use good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" F = ma …

    what is the acceleration? what is the force? :wink:
    Last edited by a moderator: Apr 26, 2017
  5. Jul 6, 2011 #4
    The question has no mention of mass so this should not be used. All the information I have is in the original question. It is A satellite in low orbit around a planet with same radius as the Earth.
  6. Jul 6, 2011 #5
    Thanks for help. It has come to me now.

    9.81 = v^2/6400000

    So v = 7.9e3 ms-1

    and so time for 1 orbit is (2pi)r/v
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook