Hey guys, least say thanks to Jorrie for posting this sort of thing. I myself at least want him to know I attempt to understand some of it.

Transc, I think Jorrie's stuff is perhaps a bit tough for most to comment on. It mostly is for my woolly head.

Nevertheless, Jorrie, I'm puzzled by what you said about the Lagrangian. "We cannot use the Lagrangian with its positive potential energy here, because it is not valid for elliptical orbits".

We were taught at school and at college that you can treat potential energy as positive, e.g. when a one pound weight is dropped 10 feet off the ground, it converts all its positive potential energy, mgh = 32.2 x 10 = 322 ft-lb_f, into kinetic energy ½mv², reaching a speed of v = ±√(2 x 322) = 25.4 ft/s at ground level.

In more advanced stuff, we also worked out how far a cannon ball would travel, using the same sort of positive potential energy to find the height it reached and time of travel. Surely, the cannon ball was in some part of an elliptical orbit around earth?

SL

Hi SL,

What we used at school is a very good approximation for what happens near Earth's surface. You surely recall that we used a 'fixed g' = -32.2 ft/s². In real orbits, that 'g' is not constant; it is actually g = -GM/r², where G is the universal gravitational constant, M the mass of Earth or the Sun, if applicable. Here r is not 'height', but the distance from the center of mass. And even this is still an approximation for a real star or planet that is not precisely spherical and homogeneous. It also does not take any relativistic effects into account, but all are negligible in this case.

I suppose some confusion comes from the fact that potential energy gmh = -GMmh/r, so that 0.5 v^2 + gmh is actually equivalent to the Hamiltonian (sum of kinetic and negative potential energy) and not the Lagrangian (although it is an approximation).

Jorrie, thanks, I think you are right about the value and sign of g.

But, if I put g negative into my equation, I get mgh = -32.2 x 10 = -322 =½mv². But how to now extract v = ±√(-2 x 322)???

Maybe I should put in a negative h as well?

SL

Hi SL, you asked: "Maybe I should put in a negative h as well?"

In standard use, sure you should, because the mass that you dropped have lost some h. It is really Δh = h_final - h_start that we must use.

-J

Ok, understood, thx.

Now back to your puzzle. I guess the challenge lies in the fact that in the perihelion boost the fuel has more kinetic energy, but less potential energy - by equal amounts?

If so the question is: why does fuel with more kinetic energy but less potential energy work better in transferring energy to the probe? It may be just because the boost is along the path of the probe and not in the direction which will give it more potential energy. What if we rotated the rocket by 90 degrees so that it boosted directly away from the sun. Would it not give the same gain in energy, but this time for the aphelion boost?

I have tried, but I do not know how to calculate this sort of scenario.

SL

Hi SL, you wrote: "I guess the challenge lies in the fact that in the perihelion boost the fuel has more kinetic energy, but less potential energy - by equal amounts?"

Yes, kinetic energy exchanges 1:1 with potential energy in any free fall orbit.

"What if we rotated the rocket by 90 degrees so that it boosted directly away from the sun. Would it not give the same gain in energy, but this time for the aphelion boost?"

No, a radial boost is about the least efficient of all possibilities and even at the aphelion it does not help much. As you have hinted, this scenario is much more difficult to analyze and I would prefer to let it stand over till later. If you want to think further, here's a hint - look at velocity vectors. :-)

-J

This was nicely done for classroom theory. Some things taken as constants are actually a little variable like the gravity constant mentioned above. The is also vaiability in the burn because of fuel pressure and expansion rates relative to external forces. This kind of equation is a good starting point for the first burn but expect that there will be a few adjusing burns to achieve near perfect orbit. There will still be forces acting on the satelite that will require future adjustments.

"The challenge is once again to find a physical mechanism (not just the math) for explaining the resultant 2.99 MJ/kg energy difference between the two post-burn case"

Well here goes. The energy is related to the velocity. The velocity is different because the probes follow the 'Law of the Elliptic'. So the energy comes from the law. Since God created the law, the energy comes from God.

Hi S, yes, you are almost there...

Another name for the 'Law of the Elliptic' is obviously Kepler's laws of planetary motion, stating:

"1. The orbit of every planet is an ellipse with the Sun at one of the two foci.

"2. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

"3. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."

Law 2 gives a clear hint that the mechanical energy is constant along a free fall orbit. It also hints at the mechanism for the perihelion boost being more efficient than the aphelion boost, but that's just a tad tricky to see. God knows trigonometry!

-J

Jorrie, I am somewhat confused by your final remark to StandardsGuy. As I understand it, Kepler's 2nd law has to do with the angular momentum (L=m r v_t) of the orbit being constant. The area swept out represents momentum and not energy. Isn't the energy being constant a separate, independent condition?

SL

Hi SL, you wrote: "The area swept out represents momentum and not energy. Isn't the energy being constant a separate, independent condition?"

Hmm, I think you may be right! But, I'm not sure that they are two independent constants. AFAIK, if you have a free-fall situation with constant angular momentum around the main mass, you must also have constant mechanical energy. Anyway, making the connection between a change in angular momentum and a change in kinetic energy is not that simple, so perhaps my comment to S was not all that helpful! Sorry S...

The best path is still through analyzing the energies of the probe and the fuel immediately after versus immediately before the burn.

-J

OK, I've given the fuel side a try. I think you have a mistake with the sign in your opening post, for pre-burn energy: Should surely be -443.53? Your post-burn answers look right. I've tried to calculate the fuel energies, assuming an explosive release of energy giving the reaction mass 3 km/s negative delta-v, like TrevorM and you have done before.

Before the burn, the fuel must have the same energy as the probe, so I reckon the total energy of probe plus fuel is twice -443.5 = -887 MJ.

Fuel post-burn, perihelion: 0.5 x (30.29-3)^2 - 902.13 = -530 MJ

Add this to the -348 that you got for the probe at perihelion and I get: -878 MJ total.

Fuel post-burn, aphhelion: 0.5 x (29.29-3)^2 - 872.47 = -527 MJ

Add this to the -351 you got for the probe at aphelion and I get: -878 MJ total also.

So, where is the benefit for a burn at the perihelion? I do not see what you meant by "The energy difference, gained from 'somewhere', is 2.99 MJ per kg of payload mass, as measured relative to the Sun."

SL

Hi SL, thanks, I corrected the signs in note (d).

You wrote: "So, where is the benefit for a burn at the perihelion?"

But you have solved the puzzle of the 'mechanism' without saying so!

Note that the fuel's post-burn energy at the perihelion (-530) is 3 MJ/kg less than the fuel's post-burn energy at the aphelion (-527). This 'loss of energy' of the fuel of the perihelion burn was transferred to the payload, which gained 3 MJ/kg of energy relative to an aphelion burn. The total mechanical energy of the system remained the same, if we ignore heat losses and so on.

Question: with no heat losses, which is unrealistic but OK for this exercise, how much fuel energy did the burn have to release?

-J

PS: Sorry to sound so pedantic!

Ah, I see. Extra energy robbed from the exhaust gas transferred to the probe. This presumably means a faster exhaust gas, losing more kinetic energy (sun-relative) would result in a more energy transferred to a payload (for the same amount of fuel)?

OK, I haven't determined the needed fuel energy yet. I guess it has to do with my -887 MJ pre-burn total energy versus the -878 MJ post-burn total energy. There is 9 MJ increase in total energy, which I reckon must come from the fuel's chemical energy, not so?

In both burns, the payload received 3 km/s velocity gain, which per kg is an energy of 4.5 MJ. So it seems this rocket needed 9 MJ chemical to add 4.5 MJ kiunetic to the probe, 50% efficiency. I can only guess that say a 30 km/s exhaust would have needed only a 3/30 =10% of the chemical energy for the same boost to the payload. Would that have been a 90% efficiency?

SL

PS. Me I don't like to get "homework", but this is at least somewhat educational.

Hi SL, yep, that's the solution...

"I can only guess that say a 30 km/s exhaust would have needed only a 3/30 =10% of the chemical energy for the same boost to the payload. Would that have been a 90% efficiency?"

30 km/s would be just about the best exhaust velocity from an energy point of view - approximately all of the considerable kinetic energy of the fuel (sun-relative) would be recovered and could be converted into payload energy. It does not mean 90% propulsion efficiency, though. Wikipedia gives the rocket propulsion efficiency as:

where v = rocket speed (sun relative) and c = exhaust speed (rocket relative). For v=c=30 km/s, this gives 100% propulsive efficiency. Of course there will be heat and other losses of at least 30%, so in practice, some 70% overall efficiency is achievable. I'm not sure what sort of 'explosion' would give 30 km/s 'exhaust speed' - probably a very unlikely number.

-J