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Eccentricity of orbits and gravitational energy

  1. Apr 5, 2012 #1

    sk9

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    hello. i recently came across this question in which a satellite is put into orbit at a point around a planet with velocity 1.2V, where V is the speed for a circular orbit at that point and they ask for ratio of max to min distance. [Ans. 2.57]
    what i want to know is how do we relate eccentricity to the velocity in these type of cases
     
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  3. Apr 6, 2012 #2

    haruspex

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    Let the initial radius and velocity be r1, v1 (assume tangential only).
    Let the max radius and corresponding velocity be r2, v2.
    The gravitational potential at radius r is -g/r, some constant g.
    You can write the down the total energy at min and max radius in terms of the above, since r' = 0 at r1 and r2, and by conservation of energy equate them.
    By conservation of angular momentum, r1v1 = r2v2.
    If v is the speed for circular orbit at radius r1 then the centripetal force v^2/r1 = g/(r1^2).
    We are told v1 = K.v where K = 1.2.

    From the above equations you should obtain
    r2/r1 = K^2/(1-K^2) = 2.57....
     
  4. Apr 6, 2012 #3

    sk9

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    two things
    1) what is r' = 0
    2)i understand tha r1 is the initial radius, but in last step you have used it for min distance which it cannot be becauseat min distance all its velocity must be perpendicular.

    Thanks btw!!
     
  5. Apr 6, 2012 #4

    D H

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    If you don't make that assumption the question is not answerable.

    That obviously isn't correct. With K=1.2, K^2/(1-K^2) = -3.27.

    The correct result is [itex]r_2/r_1 = K^2/(2-K^2)[/itex] .
     
  6. Apr 6, 2012 #5

    haruspex

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    Yes, sorry, typo.
     
  7. Apr 6, 2012 #6

    haruspex

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    I wrote r' for dr/dt because I couldn't be bothered to switch to equation mode.
    You didn't say which direction the initial speed is in, but you have to assume a direction or there's not enough info. I hazarded that the original question specified tangential but that you had omitted that in stating it.
     
  8. Apr 6, 2012 #7

    sk9

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    Okay i found the question
    7974172.jpg

    it does not say anything about the direction of velocity to be tangential or otherwise, so WHY are we taking it as tangential i have tried taking it at theta but it just didn't solve.

    and sorry it has been a while since i used r' notations.
     
  9. Apr 6, 2012 #8

    D H

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    Welcome to the world of poorly written questions. This is the rest of your life. If only the real world was as clear as the ambiguity in this question.

    The question as written does not have a unique answer. The question does provide enough information to determine the semi major axis of the orbit, but that is not what was asked for. You need to make some reasonable assumptions to be able to determine the apogee, and the obvious assumption is that the satellite is at perigee at the point in question.
     
  10. Apr 7, 2012 #9

    sk9

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    I am not going to start an argument but
    1) i never asked to solve the question, my interest was in the approach ,which i thankfully got and was also the reason that i didn't bother about (2-K^2) part. It was just an eg. to indicate what was my topic.

    2) i was writing it from my memory and so i agree it was not properly put up by me.

    3) though as dumb as i may seem to you , please explain me that what gives you this idea about semi major axis. let me point out that it has SPEED 1.2V where V is also a SPEED, it does not give any indication to me about its direction.

    Thanks for your rudeness and knowledge.
     
  11. Apr 7, 2012 #10

    D H

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    You misunderstood my last post. The question as written in the book does not provide enough information to give a definitive answer. It only provide enough information to say that the ratio is 2.57 or greater. The ratio is 2.57 if the satellite is at the periapsis point but is greater than 2.57 otherwise.

    You don't need to know the direction of the velocity vector to determine the semi major axis. The vis viva equation, [itex]v^2 = GM(2/r - 1/a)[/itex] , can be used to solve for the semi major axis a. This equation is essentially conservation of energy for an elliptical orbit.
     
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