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

Driving past your opponent in a space race

  1. Apr 21, 2006 #1
    The situation is this: Two space vehicles A and B are doing a race in space. A is in front of B and they are both in orbit around the earth. For simplicity let this orbit be a circle (i.e. neglectable eccentricity). Now, B wants to get past A, that is, B wants to cross the "line" connecting the center of the earth and A in his orbit. That means, he'll catch up a higher angular velocity [tex]\omega[/tex]. B has two choices: he can break or he can accelerate (both in tangential dierction). What should B do?
    For simplicity ignore all details: B won't lose any mass, and he'll stay in an (approximately) circular orbit.

    One approach would be this: B could brake instantaneously and thereby lower his angular momentum [tex]L=mrv=mr^{2} \omega[/tex], which would get him behind and then make him "fall" to a lower orbit, picking up speed. Looking at [tex]\omega[/tex] as a function of [tex]L[/tex], one gets: [tex]\omega(L)=C\cdot L^{-3}[/tex], [tex]C[/tex] being some constant involving the masses etc. but not the radii. This would lead to argue: If [tex]L[/tex] goes down (braking) [tex]\omega[/tex] must go up and B will pass by A.

    Now, I'm not so sure about the validity of this argument, as there are several critical points. First of all: If B brakes instantaneously, i.e. his speed [tex]v[/tex] decreases, so does his [tex]L[/tex] (that's ok so far). But this also means, that [tex]\omega[/tex] would decrease. How could I tell that the increase of [tex]\omega[/tex] caused by falling towards the earth is bigger than this decrease?
    Second: Are the simplifications made here ignoring important aspects or even not valid at all? This whole low eccentricity thing might spoil our argument.

    Thanks for any comments. Best regards...Cliowa
     
  2. jcsd
  3. Apr 22, 2006 #2
    To the admins: I'm sorry I wasn't careful enough, but: Maybe this thread should be moved to the "Classical Physics" section (?).
     
  4. Apr 24, 2006 #3
    instead of breaking, he should try rocketing towards the center, he would then accelerate in the direction of the race
     
  5. Apr 24, 2006 #4
    I'm sorry this is not an option, I'm talking about braking/accelerating exclusively in the direction tangential to the orbit.

    Thanks anyway.
    Best regards...Cliowa
     
  6. Apr 24, 2006 #5

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If B brakes instantanously it will begin to fall in toward the Earth. As it does so, it will speed up, exchanging potential energy for kinetic. After it has traveled 180° in its orbit, it will be at its lowest point and moving its fastest. It will then start to pull away from the Earth and 180° later end up right back where it started with the same velocity it had after it braked. IOW, it will enter a elliptical orbit with the staring point as the apogee.

    Since the period of an orbit decreases with a decrease in its average radius, and the new orbit's average raduis is smaller than A's, B will catch and pass A.
     
  7. Apr 25, 2006 #6
    Thanks alot, Janus.
    How did you get those exact numbers ("After it has traveled 180° in its orbit..." and so forth)? When I said B brakes I meant that be doesn't "stop", but simply lowers his speed.

    What do you think would happen, if B were only allowed to brake a little tiny bit (so that his orbit would stay more or less circular)? Same argumentation, right?
     
  8. Apr 25, 2006 #7

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    After braking, a new orbit is made. One of a different eccentricity, but will return to the same point. This makes the point of Braking the Apogee of the New orbit. Apogee and Perigee are always 180° apart.
    As did I. When he lowers his speed he no longer has enough velocity to maintain a circular orbit at that altitude, hence he begins to drift in towards the Earth.
    He would still change the eccentricity and average radius of his orbit slighty, and the difference in period will undergo a like small change. The difference would be that it would just take longer for B to overtake A.
     
  9. Apr 25, 2006 #8
    Ok, thanks alot.
    Best regards...Cliowa
     
  10. Apr 25, 2006 #9
    janus you said that if he braked he would begin to fall twards earth increasing his speed, I think that would only happen if he was going below his terminal velocity? am I wrong?
     
  11. Apr 26, 2006 #10

    DaveC426913

    User Avatar
    Gold Member

    You mean terminal velocity due to atmospheric friction?? Does not apply here. This is orbital mechanics, not ballistics.

    He will fall towards Earth as his orbit takes him close, just as ALL bodies in elliptical orbits abide by Kepler's Laws.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Driving past your opponent in a space race
  1. Gyroscopic space-drive (Replies: 14)

Loading...