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Help with Kepler's laws and satellite motion.

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    mass of the earth = 5.97 * 10^24 kg
    Polar Radius of Earth = 6.36 * 10^6 m
    Satellite = 1.08 * 10^3 Kg
    Altitude = 2.02 * 10^7 m

    3) for any object orbting around a primary body R^3 ∝ T^2
    where R is the radius of the orbit and T is the time period for the orbit.

    show that this is true and in doing so:
    - state the conditions required for a stable orbit
    - show that the conditions do not depend on the mass of the orbiting object.

    4) discuss the particular requirements for an orbit that will keep the a satellite vertically above a certain point on earths surface.


    2. Relevant equations

    1) calculate the net force on a 1.08 * 10^3 Kg Satellite when it is in a polar orbit 2.02 * 10^7 m above the earths orbit....

    So I think I get this one - as F=GMm/r^2 which gives the net force of (F= 610 N) 3 sig.fig

    2)show that the only stable orbit for the satellite orbiting at an altitude of 2.02 * 10^7 m has a period of appoximatly 12 hours.

    This next one I think it's correctly done so I've said working out velocity of the satellite I've come up with F(centripetal)=F(gravitational) so mv^2/r=FMm/r^2 so mv^2/r= the non-rounded answer in question 1 which is 609.6320 so 609.6230/1.08 * 10^3= velocity^2 so velocity = 3872.006122. Now with that substituting it into the formula of v = d/t you can find that the T - time period = 12 hours or 11.97 hours.
    3. The attempt at a solution

    Now these two questions 3) and 4) I really don't understand. I just don't know where to even start. I know however that it has something to do with Kepler's Laws of motion.? What I do think for question 4) is the velocity must be greater than the vertical acceleration... but I am still unsure..
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    ehild

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    Do the same you have done symbolically, without plugging in the numbers. You have got the equation

    mv^2/r=GMm/r^2


    Can you simplify the equation, by dividing both sides with the common factor?

    How is the speed of the satellite related to the radius of the orbit and the time period ? Plug in for v, simplify and arrange the equation with R on one side and T on the other.


    As for 4) what is the time period when the satellite is vertically above a fixed point of the equator?

    ehild
     
  4. Mar 11, 2012 #3
    thanks for that so here's what is got from plugging in v as 2πr/T into mv^2/r. so i got 4π^2r^3=GMT^2 so is that right and can i just say that - pi(π), G and M are all constants?? so therefore r^3 is proportional to T^2. Oh and what are the conditions exactly still don't get it?
     
  5. Mar 11, 2012 #4

    ehild

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    Yes, pi, M and G are constants, m cancels out, so the r3/T2 is the same for all planets and satellites orbiting around the same star.

    Question 4) is about geostationary orbits. See http://en.wikipedia.org/wiki/Geostationary_orbit

    Geostationaryjava3D.gif

    ehild
     
  6. Mar 11, 2012 #5
    thanks heaps ehild just reading the website you gave me about orbital stability...

    Although you said it was about geostationary orbits however the time period in the question is 12 hours thats half a sidereal day. So wouldn't that mean the satellite in this context would not actually be geosynchronous.?
     
    Last edited: Mar 11, 2012
  7. Mar 11, 2012 #6

    ehild

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    The satellite in the first part is not geostationary. You answered questions 1-2-3 correctly.

    I meant the last question
     
  8. Mar 13, 2012 #7
    Oh right I understand now. Cool understood.
     
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