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Escaping gravity of a planet

  1. Aug 31, 2004 #1
    when you are in orbit and want to escape the gravity of a planet completely, what is the impulse/direction which would require the least effort? would it be a push in the direction you are moving already, or one(p=mv) in the direction exactly opposite of the center of the planet? and would it be impulse=p=m*escape velocity then?
     
  2. jcsd
  3. Aug 31, 2004 #2
    it would be perpandicular to the tangent of orbit at a point.
     
  4. Aug 31, 2004 #3

    Tide

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    How do you know that?
     
  5. Aug 31, 2004 #4

    pervect

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    You need to achieve an escape velocity of Vesc to escape. You have a velocity, Vorb. Let's look at what happens when we do a vector addition of an additional velocity, Vburn.

    When our burn velocity points in the same direction as our orbital velocity, we simply add the vectors together, and we get

    Vburn = Vesc - Vorb

    When our burn velocity is perpendicular to our orbit, we write

    Vesc^2 = Vorb^2 + Vburn^2, so

    [tex]
    Vburn = \sqrt{Vesc^2 - Vorb^2} = (Vesc-Vorb) \sqrt{\frac{Vesc+Vorb}{Vesc-Vorb}}
    [/tex]

    Thus we can see that the Vburn required is higher in the second case because [tex] \sqrt{\frac{Vesc+Vorb}{Vesc-Vorb}} [/tex] is greater than 1.

    If you want a more formal proof, look at the triangle inequality

    http://mathworld.wolfram.com/TriangleInequality.html

    |Vesc - Vorb| >= |Vesc| - |Vorb| for any two vectors, and we have achieved the minimum already by boosting in the same direction as the orbit.
     
  6. Aug 31, 2004 #5

    Janus

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    Just as a side note:

    If you are in an eliptical orbit there is also a best time in which to add your velocity such that you need a minimum effort to reach escape velocity.
    That would be when you reach perapis, or your closest approach to the body you are orbiting.
     
  7. Aug 31, 2004 #6

    Tide

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    Pervect,

    So the best burn is parallel to the tangent of the orbit and not perpendicular to it as the original response stated?
     
  8. Aug 31, 2004 #7

    Tide

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    Same question as before - how do you know that?

    For an elliptical orbit the escape velocity is smaller at apogee than it is at perigee. One might argue that you would need a smaller delta v when the escape velocity is smaller.
     
  9. Aug 31, 2004 #8
    wait. pervect says it is easier to boost in the tangent direction? but nenad says its perpendicular...
    ??

    so if i had something of mass m orbiting with velocity v, around a planet of mass M and radius r, how much energy would i need to push that object out?
     
  10. Aug 31, 2004 #9

    Janus

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    While the escape velocity is smaller, so is your velocity at apogee, and we are looking for the difference between your velocity and escape velocity. This is greatest at apogee.

    Example:

    perigee(Rp): 10,000km
    apogee(Ra): 20,000km

    escape velocity at perigee:

    [tex]\sqrt{\frac{2GM}{R_{p}}}=8948 m/sec
    [/tex]

    velocity at perigee
    [tex]\sqrt{\frac{2GM}{R_p+R_a}\frac{R_a}{R_p}}= 7306 m/sec
    [/tex]

    delta v needed to reach escape velocity: 1642m/sec


    escape velocity at apogee:

    [tex]\sqrt{\frac{2GM}{R_{a}}}=6327m/sec
    [/tex]

    velocity at apogee:
    [tex]\sqrt{\frac{2GM}{R_p+R_a}\frac{R_p}{R_a}}= 3653m/sec
    [/tex]

    delta v needed to reach escape velocity: 2674m/sec

    Which is more than you need at perigee.
     
  11. Aug 31, 2004 #10

    enigma

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    Are you sure about that Janus?

    Sure, you'd be going faster at periapsis, but the farther away you are, the lower the requirement to acheive escape velocity.

    If you're in a highly elliptical orbit, I'm have a hunch you'd spend less if you do the burn at apoapsis.

    The burn is best done in-line with your current direction daveed (assuming your current direction won't take you into the planet/sun). All you need is a magnitude. The easiest way to increase magnitude is in-line so you take maximum advantage of your current speed.
     
  12. Aug 31, 2004 #11

    Hurkyl

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    Well, here's the problem.

    If you burn at apoapsis or periapsis, that point will remain as a vertex of your new trajectory. Since burning fuel increases your total energy, it follows that:

    If you burn at apoapsis, then that point remains fixed while the periapsis moves away from the source: your trajectory converges to a circle.

    If you burn at periapsis, then that point remains fixed while the periapsis moves away from the source: your trajectory converges to a parabola.


    In an "ideal" escape trajectory, your velocity at infinity would be zero: a circular orbit is the worst approximation possible! A parabolic orbit, however, is exactly what you want: your velocity at the "apoapsis" is zero. (if it was positive, you'd actually have a hyperbolic orbit)


    I'm curious, though, if you can't save yourself more energy with a "nonoptimal" burn that is designed to actually decrease the distance to periapsis.

    I strongly suspect this cannot be done by accelerating in the direction of travel. So the question is that, although burning in a different direction doesn't maximize your overall energy, can it sufficiently decrease the energy necessary to escape to infinity? (Because the orbit converges to a narrower parabola)
     
  13. Aug 31, 2004 #12
    so I was right
     
  14. Sep 1, 2004 #13

    pervect

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    Your velocity at infinity will always be zero. The conserved energy of the system will be

    E =m*(rdot^2)/2+L^2/(2*r^2*m)-G*m*M/r

    where L is the conserved angular momentum
    [tex]L=r^2 \dot \theta [/tex]

    and r is the radius, [tex]\theta[/tex] is the angle, rdot = dr/dt, m=mass of small body, G= gravitational constant, M=mass of large body.

    This can be rewritten as
    rdot^2 = 2*E/m + 2*G*M/r - (L/m)^2/r^2

    When E=0, and r=infinity, rdot=0. The transverse velocity [tex]r \dot \theta = L/r [/tex] will also be zero at r=infinity

    I'm not absolutely positive, but preliminary calculations don't look good for this approach.

    If you are in a circular orbit, you only have to increase your velocity by a factor of [tex]\sqrt{2}-1[/tex] to escape. So there is no sense in even trying an orbital burn upwards or backwards of more than .414*velocity, because you could just burn that hard and escape.

    Let's try a specific example

    G*M=1
    2E/m=-1
    L/m=1

    rdot^2 = -1 + 2/r -1/r^2

    which is a circular orbit with apogee and perigee of 1, the orbital velocity is 1 as well.

    Burning upwards for .2, we get
    2E/m = -.96
    L/m=1

    solving the quardratic for rdot=0, we find

    closest approach is .83333
    escape delta-v at .83333 is .349

    So we've had to burn .2+.349 = .549 rather than .414
     
  15. Sep 1, 2004 #14

    Hurkyl

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    Hrm... I spent more time thinking on this. Is it a true statement that no satellite in a closed orbit can have nonnegative total energy? (KE + GPE)

    It would seem to me, now, that it doesn't matter when you burn, as long as you always acceleration and velocity vectors point the same way.
     
  16. Sep 1, 2004 #15

    pervect

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    Yes, assuming the GPE is normalized so as to be zero at infinity. This can be more-or-less seen from looking at the "turning points" in the equivalent one-dimensional problem (i.e the solutions for rdot=0 in my previous post).

    rdot^2 = 2*E/m + 2*G*M/r - (L/m)^2/r^2

    It's the easiest to see when E=0. An object with E=0 will have only one turning point in its orbit, as the equation is not a quadratic. This occurs when the object makes it's closest approach to the planet. The case where E>0 is slightly trickier mathematically, the quadratic has two solutions but one of them is at a negative radius, which is unphysical.

    let Em = E/m, Lm = L/m then the solution for the the turning points is
    [tex]
    1/r = \frac{GM +/- \sqrt{(GM)^2+2*Lm^2 Em}}{Lm^2}
    [/tex]

    Lm^2 is always positive, so if Em is positive, one of the solutions for 1/r is negative.

    It's best to burn when your velocity is highest, as Janus pointed out. The object is to get your energy up to zero in order to reach infinity.

    The subtle point is that delta-E = m*v*delta-v, so that the higher your velocity is when you burn, the more energy you gain from the burn. You'll be moving the fastest when you're closest to the planet, so it's best to burn as close to the planet as you can.

    I haven't been able to formulate any cases where one can lower the delta-v by making more than one burn, but I can't rule it out (maybe someone else can).
     
  17. Sep 1, 2004 #16

    robphy

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    minor quibble: Since gravitation has infinite range, one can never "escape the gravity of a planet completely". Far away, it's small since it's inverse-square... but not zero.
     
  18. Sep 1, 2004 #17

    Hurkyl

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    Right, because it's not the force that matters, it's the net change in energy. Duh! *sigh*
     
  19. Sep 1, 2004 #18
    well you know what i mean, wen you want to escape a planet you send something at escape velocity, right? so i guess basically, in which direction should an impulse be sent to have the lowest impulse needed to eject a satellite? at first it seems like parallel to the tangent of travel, because then the velocities would be added. but then, it seems like gravity would just pull it in a bigger orbit. but when you send an impulse to the perpendicular, it would be that vector of velocity that gravity would be decellerating, but if it changed the momentum(p=mv) so that it were going at the escape velocity at that point, wouldn't that be enough?

    when someone talks about escape velocity, it should mean the velocity required to leave opposite of the planet, right?

    :confused:
     
  20. Sep 1, 2004 #19

    pervect

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    The direction doesn't matter, as long as you don't hit the planet, that is. If you have enough energy (as measured in the planet's frame) you'll escape.

    Good choice. Boosting along the tangent is the easiest way to get the energy needed to escape (the lowest delta-v). Also note the remarks about boosting when you're closest to the planet if you're in a non-circular orbit.
     
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