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Velocity at Perihelion

  1. Dec 22, 2009 #1
    A comet is observed to have a perihelion distance of 0.6Au and a period of 341 years. Calculate it's Aphelion distance. What is the velocity of the comet at Perihelion?


    The first thing I did was remind myself of some important terminology:

    Perihelion - closest distance to the sun/star
    Aphelion - furthest distance from the sun/star.

    What I'm confused is whether Aphelion is the same as the Major Semi-axis or different.. Because I've done allsorts of calculations, and I keep getting a relativistic speed which is just dumb. So my values for Aphelion are all wrong.

    How would I calculate Aphelion?

    I can directly determine the semi-major axis by using [tex]T^2 = a^3[/tex] and that gives me a value of 48.8AU as the semi major axis. Do I add this to the Perihelion and say that the sum is the furthest distance away from the sun? Or am I missing a crucial piece here?

    I even drew a diagram, I have no way of finding the eccentricity of the orbit without Aphelion :-(.

    Help please. And lastly for working out the velocity at perihelion, what is 'r' in the equation:

    Assuming that 'a' is the semi major axis.
  2. jcsd
  3. Dec 22, 2009 #2
    K I did it again and this time used a diagram, I have Aphelion currently as 97AU..
  4. Dec 22, 2009 #3
    K I got a final velocity of 54km..

    I used:


    u = GM

    But for some reason you have to put 1.5E11m under the the GM (I don't know why but I've done it before and it resolves things). I think I put in wrong values for a and r though. I know a is semi major axis, but what would r be?
  5. Dec 22, 2009 #4
    Btw, how would I use 'the orbit sweeps equal areas in equal times' to solve this question if possible?
  6. Dec 22, 2009 #5


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    It's because GM is measured in the MKS system and the distances are given in AU. 1.5e11m is 1 AU, So it gets everything into the same units.
    r is your perihelion distance.
  7. Dec 23, 2009 #6
    Yeah I thought it would be something along those lines since 1.5E11m is 1AU.

    Thank you for your help, I have 1 last question.

    A ship from earth is sent via Hohmann transfer orbit to rendevous with the comet in it's perihelion position. How long would the spacecraft take to reach the comet and at what speed would it be travelling at when it arrives? Assume the comet and Earth move in the same direction and on the same plane.



    I drew a diagram and subsequently became confused. It looks as if the craft would have to slingshot around the Sun to even attempt to get into the orbit of the comet. The question doesn't specify at which point in the orbit of the earth/comet is the craft shot...

    The diagram I drew is merely an assumption (a pretty good one I think), but now I don't know what to do. How would I calculate the time taken? And the speed at that point?

    Ok I've just worked out how I would find the time period and the speed (I would simply use vis viva again), but the diagram is confusing me. The comet has a Perihelion of less than 1AU, so how would I draw this diagram to make sure I'm doing the correct calculations?
  8. Dec 23, 2009 #7
    K I drew a shoddy diagram. If the Hohmann transfer orbit is eliptical (obviously), then it will have a perihelion of 0.6AU (which is the comets perihelion) and then an Aphelion of 1AU (since it was shot from Earth and that is the origin of the orbit).

    It's major axis would therefore be 1.6, half of that is a semi-major axis of 0.8AU.

    Now that this is an elliptical orbit, we can use Keplers relationship T^2 = a^3 to find the time for 1 cycle in years, half of which should be when the craft rendevous with the comet which it what I want. I get a time period of 0.357 years... WTF?

    I then used vis viva and got a velocity of 40kms-1

    Something has gone wrong here, and I blame it on my diagram...

    Also, if this is correct, then my craft is going slower than the comet at rendezvous and I need to give it some energy. How would I calculate the energy needed if the craft is 500kg?

    Thanks for any help, I will get a good mark in this exam.
    Last edited: Dec 23, 2009
  9. Dec 23, 2009 #8


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    Looks good to me.
    Both numbers are right in there as far as being correct.
    Straight forward. What is the difference in velocity between comet and craft when they meet? How much kinetic energy does that equate to for a 500 kg mass?
  10. Dec 23, 2009 #9

    You are a star ^^

    I just use the kinematic energy equation with the relative velocites and solve for k.e.

    Many Thanks! ^^
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