I wouldn't call it a "loss" at all. It's a diminished gain from the Oberth effect. If you put 3km/s worth of impulse, you still get at least 3km/s worth of delta-v from it. It's just that you don't get the big Oberth effect boost you would have gotten if you had applied all of the impulse near perigee.
I don't think failure to utilize the Oberth effect can be called a loss, if you use it you get a gain in delta V above what you get from the rocket equation, if you don't use it you still get all the delta v of the equation.
So as long as a vehicle in orbit applies thrust perpendicular to the local gravitational field there is no gravity loss.
AiUI, what's referred to here as gravity loss is due to the burn not being instantaneous. i.e. most of the burn is done progressively further away from the ideal point. The lower the thrust, the worse the effect.
Page 19 has a handy graph of ignition T/W vs TLI gravity losses :http://esamultimedia.esa.int/docs/exploration/ReferenceArchitecture/Final%20ReviewJan09/04_Human_moon_mission_version9esa120109.pdf
Actually, now that I think about it, it's obvious that there is a perfectly standard definition of gravity loss. It's simply the integral over time of the local acceleration of gravity times the sine of the angle between the flight path and the local vertical. This loss is an amount which needs to be subtracted (along with drag and steering losses) from the ideal delta-V computed according to the rocket equation to give the actual delta-V.
By this standard definition, a long escape burn at low-thrust deep in the Earth's gravity leads to a large loss. Now it's true that some of that gravity loss is really lost, because by raising the altitude of the escaping spacecraft, the long burn has reduce the velocity needed to escape. But it's still less efficient than an impulsive burn.
Actually, a long escape to parabolic at low thrust can be EXACTLY the same delta-v as a fast escape burn to parabolic, as long as the burns take place at perigee.
IssacKuo was right in his first comment that using a combination, a chemical rocket at perigee to capitalize on the Oberth effect, and then using the STP to accelerate further on a voyage to Mars could give you 15km/s rather than the mere 12 I was getting with just STP.
I don't think the idea of storing energy in the form of heat for STP would be practical, unless the heat sink had other essential functions, the usable energy stored would be small compared to that in rocket propellants.
Good point. A couple of errors in my definition. It should be the integral with respect to time over the powered portion of the trajectory of the local gravity times the sine of the angle between the flight path and the local horizontal. LMK what's wrong with that.
Quote from: IsaacKuo on 06/24/2011 06:20 amActually, a long escape to parabolic at low thrust can be EXACTLY the same delta-v as a fast escape burn to parabolic, as long as the burns take place at perigee.Yeah, I think we've been over that before, hence the previous discussion of passes through van Allen Belts, boil-off and mitigants.
The counterexample I gave of a lunar lander approaching the landing site in different ways still stands. Consider a lunar lander that spends some time flying horizontally to line up over the desired landing spot. According to the above definition, it suffers zero gravity loss, because the sin of the angle is zero. In reality, the lander racks up increasing gravity drag losses the longer it takes to land.Also, consider the classic gravity drag example of a rocket booster getting up to orbital speed. In the simplified case, the rocket booster starts off on top of a mountain on an otherwise spherical airless world. Thus, the rocket simply flies horizontally until it builds up orbital speed. Again, according to your definition it suffers zero gravity loss. But in reality, it suffers more gravity loss the longer it takes to reach orbital speed. Suppose gravity is 10m/s/s and the thruster provides enough thrust for 14.14m/s/s. For the first second of flight, the rocket thruster is pointed at a 45 degree angle in order to maintain altitude. Its horizontal acceleration is only 10m/s/s.So, for each second of flight, the thruster provides 14.14m/s worth of impulse but the ship only accelerates by 10m/s. This is 4.14m/s worth of gravity drag loss per second, for the first seconds of flight. As the rocket ship builds up to speed, it essentially starts to experience lift. This allows the thruster to be tilted more horizontally, increasing horizontal thrust while reducing vertical thrust. This reduces gravity drag losses until eventually the rocket ship reaches orbital speed. At orbital speed, the thruster no longer needs to be tilted downward at all in order to maintain altitude. All of the thrust goes into horizontal acceleration, meaning zero gravity losses from this point onward.
Those are steering losses (a component of thrust normal to the flight path) rather than gravity losses.
Quote from: Proponent on 06/24/2011 07:01 amThose are steering losses (a component of thrust normal to the flight path) rather than gravity losses.Whether you like it or not, it's the actual definition of gravity drag that everyone else other than you uses. The example of a rocket booster which must tilt its thrust downward to maintain altitude is the classic textbook example of gravity drag.
"If you don't use heat storage..."That's not a refutation of my point which was that you can't physically store enough heat energy for it to be worth a damn, maybe you're planning to run your car to town on a pile of hot bricks in the boot, if so you won't be getting very far.