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Atlantis shuttle and satellite orbits

  1. Oct 16, 2004 #1
    Dear Forum friends,

    Here's a problem I can't solve: The space shuttle Atlantis tries to put a satellite in a circular orbit with vo velocity. But they make a mistake and the actual velocity is vo(1+e).

    I have to find the eccentricity of the orbit when e>0 and e<0, considering |e|<<1.

    Can you guys help me with this?Thanks!
  2. jcsd
  3. Oct 16, 2004 #2


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    A quick trial and error method would give you the rough estimate (i.e. the limit).

    The shuttle normally has a radius (r) of about 6675 km. The velocity for a circular orbit is found by:


    The specific energy [tex]\epsilon[/tex] of an orbit is:


    The semi-major axis (a) of an orbit is:

    [tex]a=\frac{-\mu}{2 \epsilon}[/tex]

    The circular orbit has been turned into an elliptical orbit. If the velocity increased, the satellite was at perigee when placed into orbit. The radius of perigee for an elliptical orbit is:


    Don't confuse my e, which is eccentricity, with the e you used. Some books use a different system of symbols, since there is not as much standardization as there really should be. I suspect your book uses different symbols for eccentricity and specific energy than I use - in fact, you probably use epsilon for eccentricity and maybe an E for specific energy? (That's a less common system of symbols, but common enough)

    If you slow the satellite down, the effect is reversed. Your point is apogee instead of perigee:


    You can combine those through substitution (or just do it on a spreadsheet, which let's you compare the values you get for different values).

    Anyway, close to zero you will notice a definite pattern.

    If you go the opposite way, you'll notice your e can only reach a certain number before the the orbit turns into an escape trajectory (either a parabola or a hyperbola). If you know that the specific energy has to be negative for a closed orbit, you'll also know that a secific energy of zero means a parabolic escape orbit. You can rearrange the equation to find your escape velocity. If you compare your equation for circular velocity to your rearranged specific energy equation, you'll notice the maximum absolute value for your e before the orbit turns into an escape trajectory.
  4. Oct 16, 2004 #3
    Thanks a lot for your answer. That has helped me. However I didn't quite understand how you prove that when v is increased the satellite is at perigee and viceversa.

    I didn't give all data for this problem. Actually we know radius R for the wanted circular orbit, but the error is of 1m/s. What height can the satellite have? I've tried to work it out but I don't know why the book tells me to consider |e|<<1 (where e is MY e).
  5. Oct 16, 2004 #4


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    I'm thinking the e in your equation is the error in the velocity (for example, a 5% error would make e = .05)

    Whenever you maneuver a satellite, it's new orbit has to include the maneuver point (In other words, if you turn at Main and First Street, your new path would have to include Main and First Street, right? A turn at Main and First doesn't suddenly teleport you to Broadway and Fifth Avenue).

    If you increase the velocity, you increase the distance that the satellite will move away from the Earth before gravity overcomes its velocity and starts to pull it back in.

    The mathematical answer lies in the equation for specific energy and the equation that uses specific energy to calculate the semi-major axis. For any given orbit, the total energy (and the specific energy per unit of mass) stays constant. You add kinetic energy by increasing the velocity, the total energy (and the specific energy) increases, meaning you're going to have a bigger orbit. (If you look at the equation for specific enery per unit of mass, it's just kinetic energy and potential energy, each with the mass divided out since you're concerned with the resulting motion of the satellite vs. the forces acting on the satellite)

    You have a bigger orbit, but the satellite has to pass through the point where you made your change in velocity. Being a circular orbit, there were no points closer to the Earth than your maneuver point. Since the orbit is made bigger, there are still no points in the orbit closer than your maneuver point.

    (If you started with an elliptical orbit and threw in the possibility that you didn't maneuver at either the closest point or the furthest point, the problem would get more complicated.)
  6. Oct 16, 2004 #5


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    If you work out the equations for an object in an orbit around a central force in polar coordinates, r, [tex]\theta[/tex], where r is the distance of the satellite to the central body, and [tex]\theta[/tex] is the angle, you will get the well known result

    [tex] m \frac{d^2r}{dt^2} - m r (\frac{d\theta}{dt})^2 = f(r) [/tex]

    (which can be interpreted as saying that the total radial accleration is given by the difference of the centripetial acceleration needed to maintain constant r and the applied radial force)

    plus another equation that is the conservation of angular momentum

    [tex] m r^2 \frac{d\theta}{dt} = L [/tex]

    where L is a constant, the angular momentum of the orbit.

    When you combine these two equations, you get the one dimensional equation

    m \frac{d^2r}{dt^2} -\frac{L^2}{m r^3} = f(r)

    which yields a conserved quantity, the energy of the body in the orbit

    E = \frac{1}{2} m [ (\frac{dr}{dt})^2 + r^2(\frac{d\theta}{dt})^2 ] + V(r) = \frac{1}{2} m [ ( \frac {dr}{dt})^2 + \frac{L^2}{m r^2} ] + V(r)

    This is often described as the "one dimensional equivalent problem", because the above equations of motion are the motion of a body of mass m with a constant energy E in an effective potential

    [tex] Veff = V(r) - \frac{L^2}{2 m r^2} [/tex]

    For an inverse square law force due to gravity, we can specifically write out f(r) and V(r)

    f(r) = -GmM/r^2
    V(r) = -GmM/r

    So you can find the maximum and minimum "turning points" of the orbit by solving for the radius r where dr/dt = 0 in the expression for the energy

    \frac{1}{2} m [ (\frac {dr}{dt}) ^2 + \frac{L^2}{m r^2} ] -GmM/r = E
    Last edited: Oct 17, 2004
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