Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orbit Dynamics

  1. Aug 10, 2011 #1
    Hi guys,

    i cant seem to find the answer .

    A satellite in earth orbit has a perigee velocity of 8 km=s and period of 2 hours. From this information,
    determine all the orbit parameters that you can. From those parameters, determine its altitude at perigee

    my vp=8km/sec
    Vp=sqrt(GM/r)---is this right and is r =rp?

    how do i calculate my semi major axis? is it from kepler's 3rd law? im stuck at finding rp and eccenctricity and ive been going at this for hours without proper examples in books/online.

    Someone please help!
  2. jcsd
  3. Aug 10, 2011 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That equation is valid only for circular orbits. The vis viva equation provides a more general answer:

    [tex]\frac{v^2}{GM} = \frac 2 r - \frac 1 a[/tex]
    Correct. So show some work so we can help you out a bit.
  4. Aug 10, 2011 #3
    but if its vis viva will my eccentricity value be more than 0.2..thats the value im getting, this is how im doing my calculations

    1) equation 1 : mag(h)=mag(rp)*mag(v)
    equation 2: mag(rp)= [mag(h)^2/GM]/(1+ecos(v) - trajectory eqn
    equation 3: T= 2phi/sqrt(GM) * (mag(h))^3/2

    i am to solve 3 unknowns mag(h),rp and e by simultaneous eqns by substituting all the equations into one another.

    ive done it a couple of times, either im getting a negative value for e(-0.1) which shldnt be the case.

    But my qn here is. if v=8000m/sec what is mag(v)? and in this case will a=mag(h) ?
  5. Aug 10, 2011 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Then you are doing it wrong. The eccentricity is less than 0.2. Show your work.

    Your equation 1 is only valid for circular orbits, and for elliptical orbits at perifocus and apofocus. (Since the given data point is perigee, this equation is okay here.) Your equation 3 however is only valid for circular orbits.

    Try finding a formula that relates semi-major axis (rather than specific angular momentum) to the period.

    The magnitude of the velocity vector is of course 8000m/s. As far as a=mag(h), no. Look at the units. Specific angular momentum(h) has units of length2/time. Semi-major axis (a) has units of length.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook