Delta-V

[Baythan]

The point is that if you are high above a planetary body in a roughly circular orbit and want to escape, you can burn retrograde and drastically drop your periapsis (which decreases your orbital energy), and then burn for escape at your new lower periapsis.

This strategy will typically provide you with a substantial dV savings due to the oberth effect.

Now THAT is exactly what I need to know! I would not have thought to lower my Pe first before making an escape burn, I'd have just done it at the current Pe.

Thanks!

[rkman]

maccollo posted an interesting example of the Oberth effect in action

On the one hand intuitively it makes sense to me that capture at lower Pe is more efficient. But the example seems to defy the Oberth effect in that the more efficient burn takes place at a lower speed (246m/s, req Dv=75m/s) than the less efficient burn (360m/s, req Dv=284m/s).

Another way to look at it is that energy is a force F applied along a distance s. For a fixed burn time t and force (thrust) F, the distance

s along which the force is applied is longer if the craft moves faster, which increases the energy.

That makes sense to me, sort of.

Accelerating your car from 100 km/h to 120 km/h takes 22% more energy than from 80 to 100. The same is true for rockets.

From a kinematic point of view, this is everything you need to know to understand the Oberth effect. Note that this has nothing to do with

orbital mechanics, gravitational drag or what have you.

But this not; if achieving a certain amount of delta-v at a high speed requires more energy than achieving the same delta-v at lower speed, but you put in the same amount of energy (same burn time and thrust), then would the result not be a lower achieved delta-v?

Edited May 2, 2014 by rkman

[Red Iron Crown]

But this not; if achieving a certain amount of delta-v at a high speed requires more energy than achieving the same delta-v at lower speed, but you put in the same amount of energy (same burn time and thrust), then would the result not be a lower achieved delta-v?

This is the part that throws intuition for a loop: The energy contained in the fuel changes depending on how fast the ship is moving. Its chemical energy stays the same, but the amount of kinetic energy that can be extracted is greater when the ship is moving faster.

This leads to the unusual circumstance of a rocket engine producing, say, 100KW at one speed and 200KW at a higher speed.

Delta-V is not a measure of energy. A m/s of dV contains differing amounts of energy depending on speed.

Edit to add:

On the one hand intuitively it makes sense to me that capture at lower Pe is more efficient. But the example seems to defy the Oberth effect in that the more efficient burn takes place at a lower speed (246m/s, req Dv=75m/s) than the less efficient burn (360m/s, req Dv=284m/s).

Not sure where you're getting those speeds. Don't look at the speed on the navball, that's current speed and not what it will be when at the maneuver node.

Edited May 2, 2014 by Red Iron Crown

[eypandabear]

This is the part that throws intuition for a loop: The energy contained in the fuel changes depending on how fast the ship is moving. Its chemical energy stays the same, but the amount of kinetic energy that can be extracted is greater when the ship is moving faster.

This is correct. More precisely, the relative amount of kinetic energy *imparted on the rocket* is higher. The chemical energy of the fuel is converted into kinetic energy, and this energy is distributed between the rocket and the exhaust.

For example, let's take a rocket that is not moving relative to us. The rocket engine is fired for just a split second, resulting in

a very small delta_v. The rocket gains the kinetic energy (1/2) m delta_v^2, neglecting the lost mass of the propellant, delta_m. The propellant gains the energy (1/2) delta_m w^2, where w is the exhaust velocity.

Now let's do the same burn, but with the rocket already moving at v=v_0. The rocket gains (1/2) m [(v_0 + delta_v)^2 - v_0^2]. The propellant gains (1/2) delta_m [(v_0 - w)^2 - v_0^2] = (1/2) delta_m [w * (w - 2 v_0)].

But.. wait, why?

Because, prior to the burn, the propellant was already moving together with the rocket. Because it is expelled backwards,

it needs to first be decelerated to zero before it can be accelerated again in the direction of the thrust. Therefore the energy

that it gets is less than it was in the case of v_0 = 0. You cannot really see from the above equations that this "missing" energy is exactly the same amount as that which the rocket gets more because I have neglected the transfer of mass, but it gives you an idea of what is happening.

An interesting thing happens if the rocket's initial velocity v_0 is more than twice as high as the exhaust velocity w. In this case, the propellant energy gain becomes negative! The rocket actually gains additional energy from the fuel! This is because, seen from the outsider observer relative to which the rocket is moving, the fuel is now moving in the same direction as the rocket, but slowed down.

Like I said, the above calculation is an approximation, but the gist is that a slow rocket has to put energy into the fuel to push it backwards, whereas a fast rocket steals stored kinetic energy from the fuel by pushing itself away from it (so to speak).