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Let's say we have an elliptical orbit that goes from LEO periapsis to apoapsis near Earth's Hill Sphere.

You can merely nudge your satellite at the apoapsis a bit higher (a tiny burn at periapsis), and it becomes Sun's satellite, leaving Earth for a long, long time, moving really slow.

Or you can spend lots of fuel (and energy) to circularize the wide orbit. Then, at any point you can speed up just a little (again, just enough to breach the Hill sphere) and you'll escape the planet - choosing the point and speed right, just at the same speed and in the same direction as in the prior case.

Where did all the energy used to circularize the orbit vanish to?

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  • $\begingroup$ As a side answer, how would this affect the orbit around the sun ? $\endgroup$ – Antzi Jun 18 '15 at 15:20
  • $\begingroup$ In the elliptical case, you will need less thrust (less propellent) to breach the Hill sphere because you will do more work at the periapsis than you will at the apoapsis for the same duration burn. However, you can't match the same velocity that you get with the apoapsis burn as you can't independently raise the height of apoapsis and change the magnitude of your velocity at apoapsis solely with a burn at periapsis. $\endgroup$ – Erik Jun 18 '15 at 19:43
  • $\begingroup$ @Erik: But you can match the velocity of escape from elliptic orbit, with escape from the circular orbit (by choosing the right point of escape; it may be offset to a side, but head in the same direction and have the same value; different departure point but same velocity vector. Plus if it's "just a nudge", even that won't be the case as both will be tangent to the Hill sphere surface, meaning same value, same direction and same origin minus whatever "infinitesimal" allowance we use for the "nudge".) $\endgroup$ – SF. Jun 19 '15 at 8:38
  • $\begingroup$ @Antzi: If the escape is "just a nudge" (as in my reply to Erik), it wouldn't affect it at all. Same point, same speed, same direction. $\endgroup$ – SF. Jun 19 '15 at 8:42
  • $\begingroup$ ...and as a bottom line, I think the most unintuitive thing is that gram for gram, burning the same amount of propellant in the same engine can produce much more energy depending on how fast you move. It's as if the car milleage was rising if you're speeding after rolling downhill, and dropping after you're slow at the top of the hill. $\endgroup$ – SF. Jun 19 '15 at 9:11
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Where did all the energy used to circularize the orbit vanish to?

The exhaust.

Suppose a spacecraft with mass $m$ performs an instantaneous (either impulsive or infinitesimal) $\Delta v$ at some instant along its orbit by ejecting some quantity of matter $\Delta m$ at some velocity $\vec u$ relative to the spacecraft. The quantity of ejected matter is finite in the case of an impulsive burn, infinitesimal in the case of an infinitesimal burn.

In the case of an infinitesimal burn, the change in velocity of the spacecraft is frame-independent, given by $\Delta \vec v = -\frac{\Delta m} m \vec u$. This ultimately leads to the ideal rocket equation. The change in energy of the spacecraft+exhaust cloud system is also frame-independent, $\Delta E_\text{sys} = \frac 1 2 m\,\Delta m \, u^2$. The quantity of chemical potential energy released by burning that fuel (less thermodynamic inefficiencies) dictates the change in mechanical energy of the spacecraft+exhaust cloud system. Note: The above applies to infinitesimal burns. In the case of an impulsive burn, the change in velocity and in system energy are also frame-independent, just a bit messier.

How this change in the total mechanical energy of the vehicle plus exhaust cloud system is partitioned between the vehicle and the exhaust cloud is frame-dependent. Comparing identical burns at periapsis and apoapsis, more energy gets transferred to the vehicle when the vehicle is at periapsis than when it is at apoapsis.

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    $\begingroup$ In short by firing your rocket at higher or lower orbits you leave the exhaust in higher or lower orbits. Lower orbits means less potential energy in the exhaust. $\endgroup$ – Aron Jun 22 '15 at 16:29
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Remember, work done is force times distance. At periapsis, one second of thrust is performed over a larger distance than one second of thrust at apoapsis. Thus, more work is done by firing at periapsis. This is the Oberth effect.

The work done (energy used) to circularize the orbit at apoapsis went into increasing the energy (potential + kinetic) of the object at all points in its orbit.

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  • $\begingroup$ Why did it take an Oberth and 200+ years to find it out? And particularly for a then exotic space application, when it was evident in Newton's equations to begin with? And applicable to all kinds of acceleration no matter whether space or Earth, or reaction engine or jumping on your own feet? $\endgroup$ – LocalFluff Jun 20 '15 at 19:09
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The effect of the Oberth largely comes from reducing something known as Gravity Drag. Let's ignore air resistance. If we launch a rocket directly into space, for every second we are thrusting, we are also dragged down. This effect can be overcome in large part by moving faster, as we are in the gravity well for a smaller amount of time. Thus, gravity can't have as much of an effect on us.

In the elliptical case, we accelerate slightly when at the peak velocity. Thus, the drag of gravity on the spacecraft is less as we are moving faster. In effect, the energy comes because gravity has less of a drag on the spacecraft. Basically, the pull of the gravity well has accelerated the object to a high speed, and moving away faster gives the gravity source less time to slow it down.

In the case of the highly circular orbit, there is not a large amount of speed, in fact the orbital velocity is quite low. That velocity will only increase by the rocket power directly, and will not have the additional effect of the gravity drag.

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    $\begingroup$ At every point in an orbit there are two directions you can fire that have zero gravity drag. At periapsis and apoapsis these are retrograde and prograde. There is no relation between the Oberth effect and gravity drag. $\endgroup$ – Erik Jun 18 '15 at 14:36
  • $\begingroup$ Actually, it is a plane of vectors you can fire in -- as long as your thrust is perpendicular to the gravity field. $\endgroup$ – Erik Jun 18 '15 at 14:36
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    $\begingroup$ Gravity drag might be the wrong term, but the same principal affects it, namely that moving faster away from the gravity source will reduce the amount of time, and thus the amount of energy lost, by gravity. $\endgroup$ – PearsonArtPhoto Jun 18 '15 at 15:04
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    $\begingroup$ You "lose" no energy if you burn perpendicular to the gravity field -- which is the case in all of the examples mentioned by the OP. $\endgroup$ – Erik Jun 18 '15 at 20:34
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Where did all the energy used to circularize the orbit vanish to?

  1. Not much of fuel/energy is needed to circularize at the edge of the SoI. So the difference is not that big.
  2. If the fuel is burned and left at periapsis, it has much less energy, then burned fuel left at the apoapsis. The energy spent during circularization goes mostly to both kinetic and potential energy of the used propelant.
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