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Radial Impulse in Orbit

  1. Feb 11, 2016 #1
    I understand that if there is a body in a circular orbit around the Earth (or any other thing), and there is an impulse directed radially out away from the Earth against the orbiting body, the body will end up in an elliptical orbit, which will pass through the original circular orbit. This got me thinking about the ISS again. I'm not sure why I have been thinking about it so much recently. I guess learning leads to questions, which leads to more learning, which leads to questions...

    If an astronaut is standing on the outward facing side of the ISS, and jumps up (away from the Earth), they will now be in an elliptical orbit. When the astronaut crosses paths with the ISS again it will be at the same point at the same time? Meaning, will the astronaut come back down to the same spot they jumped up from?

    Also, it seems like the two orbits (the original circular orbit, and the new elliptical orbit), will always cross at two points on opposite sides of the Earth. So if the astronaut jumps up a little they will slowly drift up away from the ISS, and then slowly back down after one half of an orbit. And if they spring up with a really violent impulse they will go really far out, but still come back down, and intersect at the same point they would have intersected had the jumped very gently. Is that correct?
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  3. Feb 12, 2016 #2


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    The two new orbits will intersect at the same point at which the two bodies initially separated. But I expect the orbits may now have different periods, in which case the bodies won't pass that point at the same time, so the two bodies won't come together again.

    I don't think the orbits will also cross at an antipodal point. When I first drew the diagram it looked like that would happen. But that was based on the mistaken (I think) assumption that the two new ellipses are centred on the centre of the Earth. But they won't be. Rather, the centre of the Earth will be one of the two focii of each of the two orbits. So neither orbit will have any rotational symmetries around the centre of the Earth, and hence the reason for expecting an exactly antipodal intersection of the two orbits disappears.
  4. Feb 12, 2016 #3
    I think things are simpler than they really are. From Kepler's law the orbital period is:


    Since the astronaut's new orbit (after they took the leap) would be elliptical (assuming the ISS has a circular orbit), it would take the astronaut longer to get around the Earth.

    So let's say the astronauts just jumps using nothing but their own legs. In this case I think we can assume the ISS's orbit isn't really affected, since the astronaut doesn't weight much, and the force isn't very great. In this case I don't think the astronaut's orbit would be affected too much. The ISS is really close to the Earth, so the strength of gravity pulling the astronaut down is more or less the same as it is for us. Doesn't this mean that the astronaut wouldn't be able to launch themselves any higher off the surface of the ISS than if they were on the surface of the Earth? Because if that's true, it seems like the astronaut would come right back down at nearly the same place, since the jump wouldn't add much to the length of the semi-major axis compared to how big it would be from the start.

    I guess the questions I really have are, how high could one jump off the surface of the ISS? And how long would to take to cross the ISS's orbit again?
  5. Feb 12, 2016 #4


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    andrewkirk is correct that the period of the two orbits will be different. For example, if you were to jump radially from the ISS at 10 m/s your orbital period would increase by a bit over 0.014 sec. Thus you would return to that point of the orbit 1 sec after the ISS did and would miss the ISS by some 109 m.

    The two crossing points of the orbits will not be 180 degrees apart either. At the same 10 m/s, the crossing points will be 179.999999745 degrees apart. If we increase the radial speed to 100 m/s it decreases to 179.9997450651 degrees apart. At 1 km/sec they're 179.7471522645 degrees apart, at 5 km/sec they are 151.1279524189 degrees apart, At 7 km/sec they will be less than 90 degrees apart(and your orbital period would be almost 15 times longer than that of the ISS.).
  6. Feb 12, 2016 #5


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    This would only be true if the ISS and astronaut were at rest with respect to the Earth, but they are not, they are already moving at orbital speed for their altitude. This makes its a whole new scenario
    If you were to jump at a speed of 2.5 m/s( enough to get you to 1/3 meter off the ground at the surface of the Earth), your new orbit would take you 2.2 km further from the Earth than the ISS. It will take ~46.45 min for you to cross the ISS orbit again and you will miss hitting your take off post on the ISS by almost 7 meters.[
    Last edited: Feb 12, 2016
  7. Feb 13, 2016 #6
    The ISS is 109 meters wide, so at least our poor astronaut wouldn't miss it on the way back down (yay!).

    Thanks, for the numbers, this is just the sort of thing I was looking for. Now I want to see if I can solve for this myself, and see what different scenarios look like. I'm more than a little out of practice, but I'm a bit obsessed with orbital mechanics recently for some reason.
  8. Feb 13, 2016 #7
    I'm a little stuck. Here is what I have so far:

    I calculated the new velocity (Vf) from the impulse velocity (Vj) of 2.5 m/s using the following formula:

    Vf = √(Vi2+Vj2)

    Then I used that to get the new semi-major axis using the following:

    a= (μ*r)/(2μ-Vf2*r)

    I assumed the ISS orbit is circular, and I'm sure we rounded differently, but I ended up with a difference between the old and new semi-major axes of ~3111.23 meters. However, I'm not sure how to find the point of intersection between the two orbits.

    How did you find the points and time of intersection? Am I even on the right track?
  9. Feb 13, 2016 #8


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    Checking back over my spreadsheet, I found an error in one of the formulas. After fixing it, I came up with a difference of ~3277 meters between the ISS orbit and apogee of the astronaut's orbit. The difference between two semi-major axes came out to ~888m

    Here's what I did: starting with the assumption of a circular orbit with an orbital speed of 7660 m/s for the ISS, and using 3.987e14 for GM, I got an SMA of
    6794988 m
    I got the new velocity the same way you did, and with that and the vis-viva equation solved for the new SMA of 6795876 for the new SMA. This is a difference of 888 m

    This is what I also get if I plug my numbers in to your formula for a. So I think our difference comes in in either our assumption of orbital velocity or orbit radius.

    After that, I needed to find the eccentricity of the Orbit, I did this by using two equations for the Areal velocity.
    [tex]A= \frac{rv \cos t}{2}[/tex]
    (where t = atan(2.5/7600) or the angle difference between the ISS orbit and the New orbit at the intersection point)
    [tex]A= \frac{\pi a^2\sqrt{1-e^2}}{P}[/tex]

    Where P is the period of the new orbit.
    Equating and solving for e gives an answer of 0.0003515532 for the eccentricity
    Since the apogee of the new orbit is found by a(1+e), this puts it at 6798265 m and the difference between this and ISS orbit is ~3277 m

    We also have enough info to work out the intercept points of the orbits as measured as angle from perigee of the new orbit.
    For that we use
    [tex]r = \frac{a(1-e^2)}{1+ e \cos q}[/tex]

    We set r to the ISS orbital radius and solve for q.

    At this point, my earlier error came back to really bite me. It resulted in an answer for q that was very near 90 degrees, that combined with the too small total difference in orbital periods it also led to, made it so that I felt safe in estimating the Intersection time as being close enough to half the orbital period as to not make much of a difference.

    However, the corrected formula showed that the intersection points were actually only 68.2 degrees from perigee ( it also gave and larger difference in the orbital periods).

    with the corrected numbers, it takes a few more steps to get the intercept time.

    First we must convert the true anomaly(q) to the eccentric anomaly(E)
    (note from this point on all angles need to be in radians)

    [tex]\cos E = \frac{e+\cos q}{1+ e \cos q}[/tex]

    then we get the mean anomaly(M) from the eccentric anomaly

    [tex]M = E- \sin E[/tex]

    The time it takes to go from perigee to the intercept can then be found to by

    [tex]t= \frac{MP}{2 \pi}[/tex]
    [tex]t=M \sqrt{\frac{a^3}{\mu}}[/tex]

    finally, P-2t gives the intercept time of 3463.68 sec.
    The time it takes the ISS to cross the same angle is the ISS period times 223.6/360 which equals 3461.86 sec or 1.82 sec less than what the astronaut will take. Thus by the time the astronaut arrives, the ISS will be some 1.82 x 7660 = 13941.2 m further along its orbit and he will miss it by this much (14071m by the spreadsheet)

    A fair bit more than the 7 meters answer my goof led to.
    Last edited: Feb 14, 2016
  10. Feb 14, 2016 #9
    Unfortunately, it looks like the formulas got screwy right at the part I couldn't figure out, which is how to find the eccentricity. It looks like you are trying to do A = (rv*cos(t))/2 but that doesn't seem to work out right.


    Actually, that seems to be exactly what you meant. My software was rounding things off too much. I have it figured out now. Thanks.
    Last edited: Feb 14, 2016
  11. Feb 14, 2016 #10


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    Went back and fixed the formatting. Should be good now.
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