Hmmm. A very thought-provoking scenario.

My first thought is that you've simply changed coordinate systems, the different numbers are simply due to the way the energy is designated in each.

The 'added' energy was, in fact, always there. The initial energy expended to place the probe at the L4 point included the potential energy needed to place the probe's internal fuel supply there, too. When the engine was fired, some of this potential energy was lost and it showed up as kinetic energy. Within its own reference frame the increased KE was 4.5 MJ/kg; within the Solar reference frame the increased KE was 95 MJ/kg.

Hi Usbport,

Interesting comment. You wrote: "When the engine was fired, some of this potential energy was lost and it showed up as kinetic energy."

Immediately after the engine was fired, the spent fuel (exhaust gas) is essentially still at the same gravitational potential - 150 million km from the Sun. So how do you reckon that potential energy is lost in the firing?

-J

I agree with USB. This is not a puzzle, just a coordinate change trick.

Where does the energy come from? From the rocket fuel, full-stop.

Hi Guest, not so fast!

There are a few more subtleties in this scenario than you seem to give it credit for. And I'm not referring to relativistic effects - 30 km/s is not very relativistic.

As an example, suppose we change the orientation of the probe so that the same 3 km/s burn is not quite lined up with the orbital movement. The change of energy in the Sun's frame will now be less than 95 MJ/kg. In the extreme case where the burn is pointing normal to the orbital movement, the energy change will be some 5 MJ/kg only. No change of frame, yet different results.

I'm still after a good 'mechanism' for explaining these differences.

-J

Yea, but that sort of arithmetic is true for any inertial frame that moves relative to another one. You do not need orbits to demonstrate that. Kinetic energy is heavily frame dependent. The mechanism is simply the way we define and add energy. I do not see your point. :-(

Hi Guest, you wrote: "I do not see your point."

I guess my point is contained in the first paragraph of the opening post. Orbits do have some interesting modes of behaviour that we are utilizing for our benefit. For solar system exploration, the heliocentric coordinate system is normally the one that counts. What the mechanical energy happens to be in other coordinate system does not quite help us getting to Pluto and beyond.

Before we go to the "slightly tougher variant of this puzzle", I would like to get some consensus on this one.

IMO, the potential energy loss mechanism suggested by USB is not quite correct, as I've shown in #2. Remember, the 'extra energy' is there immediately when the firing is completed, before anything has time to 'fall in' towards the Sun.

Secondly, although the rocket fuel is obviously the root source of the energy, not all the resultant energy is extracted from there. This will hopefully become clearer in the variant puzzle...

-J

Me brain hurts

Bazzer

Hi Jorrie,

When you increase the velocity of the object, the orbit diameter will increase. I believe that the object will actually slow down relative to the earth and sun. Is this heading in the right direction?

Hi Poison, you wrote: "I believe that the object will actually slow down relative to the earth and sun."

Generally yes, but in this case the rocket first increases the orbital speed by 3 km/s (to ~33 km/s) and then it moves farther out and slows down. After the 'burn', the probe is in free-fall and it exchanges kinetic energy for potential energy. At the aphelion of the new orbit (~230 million km from the Sun), it actually would have slowed down to ~21 km/s.

-J

No, it's not a frame of reference thing; it's a VECTOR thing. In these energy equations, vee is the magnitude of the vector shorn of its direction; thus, what is called 'v' (vee) is the probe's SPEED. The probe's VELOCITY vector (let's call it the V-vector), when force is applied to it, can be changed in both magnitude (speed) and DIRECTION. The actual change is a function of the force vector (speed and direction, which may change applied over time, depending on both the magnitude of the force, and its direction, integrated over time of application to the probe. Thus, we get different results speed/direction results depending on the direction of the thrust applied to the probe. For example, if the probe were in a circular orbit around the Sun and a force were applied to it behind perpendicularly to the radius of the probe's orbit, then the probe would be pushed to a higher but still circular orbit; in such a case, the work done on the probe would be added to the probe's original orbital (rotational) energy (around the Sun) to give the final energy. Similarly, force applied in the opposite direction (that would 'slow' the probe) would lead it to have lower rotational energy and a smaller orbit. In both cases, the rotational speed (not velocity) in polar coordinates would be calculated by using taking the rotational energy and the radius of the new orbit. The point is that the probe moves in vector space and the problem's equations are those for a scalar one. Cheers! DZ

Hi DreadZontar, yep, vectors do come into it as soon as you have radial and tangential speeds. But the problem as stated is approximately a scalar one with a scalar solution, as Trevor has pointed out.

-J

Hi Trevor, you wrote: "If we look at the before and after energies of the system from the ground reference though its 100000-52020-48020 = 40 Nm. These are big input figures with the same answer as when measured relative to the c of mass of the system - as one would expect."

Good answer :)

It is all about the reaction mass (exhaust gas) that decreases in total heliocentric energy in order to increase the total heliocentric energy of the payload. Add in the chemical energy of the fuel that was lost and the total energy is conserved. In your case 2 one also can say: 52020 + 48020 - 40 = 100000, to show the conservation.

-J

Using Trevor's scheme for my original question, we get the following results, assuming the reaction mass is the same as the payload mass, so that the effective "exhaust speed" is 3 km/s (remember, for simplicity, Trevor used an 'explosion').

The total 'burn energy' required is: 2 x 0.5 x 3^2 = 9 MJ/kg of payload mass, half of it for the payload and the other half for the reaction mass.

Immediately before separation, the original kinetic energy in the heliocentric coordinates was: 2 x 0.5 x 30^2 = 900 MJ/kg

Immediately after separation, the payload kinetic energy is: 0.5 x (30+3)^2 = 544.5 MJ/kg

The new kinetic energy of the reaction mass is: 0.5 x (30-3)^2 = 364.5 MJ/kg

The kinetic energy increase: 544.5 + 364.5 - 900 = 9 MJ/kg, the 'burn energy', so everything adds up.

In this exercise, we have ignored potential energy, because we assumed a circular orbit and that during separation potential energy stayed the same. In the next Blog entry, we will consider what happens in the real elliptical orbit, where the potential energy depends on where in the orbit the burn takes place.

-J

Hi Guest. GA.

Yes, the Oberth effect is a valid, alternative and interesting way of looking at the situation.

In the next variant, I would like to show that it is not only the kinetic energy, but also potential energy of the fuel that can be converted into extra kinetic energy for the probe. Still working on the writing of it...

-J

"If the probe was at rest in an inertial frame, we would have added 'only' 3²/2 = 4.5 MJ/kg of specific kinetic energy - that's the energy imparted directly onto the probe. However, immediately^(d) after the burn the probe has actually gained (30+3)²/2 – 30²/2 ~ 95 MJ/kg of extra kinetic energy (relative to the Sun). This looks like a vastly 'over-unity' process at work: 4.5 MJ/kg energy input and 95 MJ/kg of output - more than a factor twenty! How does this work?"

It's a trick question. The probe is NOT at rest in an inertial frame. The kinetic energy works according to the formulas shown. It's like saying I weighed 800 pounds on Jupiter but now I weigh 200 pounds, so I have lost 600 pounds.

Question: After the burn, the altitude increases and the orbital velocity slows, you said. Does the probe stay ahead of the Earth in the orbit, or does the Earth pass up the probe eventually?

"In this entry I want to look at the way we utilize this energy."

I think you failed to address this. So the probe has more energy, So what?

Hi S, you wrote: "It's a trick question. The probe is NOT at rest in an inertial frame."

An unpowered probe in orbit is always momentarily in its own personal inertial frame of reference, so we are allowed to make that calculation before and after the burn. The 0.5 x 3^2 = 4.5 GJ/kg gain is relative to the inertial frame before the burn.

"Does the probe stay ahead of the Earth in the orbit, or does the Earth pass up the probe eventually?"

No, it quickly falls behind L4 and eventually behind Earth as well. The aphelion of the ellipse is at ~230 km from the Sun and at ~21 km/s momentary orbit speed. It is only at the perihelion where it is faster, but its period is much longer than Earth's, about 522 days.

"So the probe has more energy, So what?"

The 'so what?' should become clearer in the variant. What Guest wrote in #15 (and his reference) sheds some light on it for now...

-J

"An unpowered probe in orbit is always momentarily in its own personal inertial frame of reference,"

I disagree, and agree with Guest (#5). I understand that there is a multiplication of energy with respect to a probe starting from zero velocity, but you defined it to be in motion the same as the earth. You cannot redefine what you have already defined.

The Oberth Effect relates to gravity assist, which you said you were not talking about: "Here I am not referring to gravity-assist maneuvers." This example takes advantage of the energy from the creation of the solar system, as someone has said.

In the link about Oberth Effect, it talks about parabolic and hyperbolic orbits. Explain if you can what would cause each of those orbits.

-S

Hi S, I must have explained poorly. Here's a retry.

"... but you defined it to be in motion the same as the earth. You cannot redefine what you have already defined."

The misunderstanding probably arises because we cannot have one inertial frame to cover the whole orbit of Earth. What we are allowed to do is set up one inertial frame for a particular moment that is instantaneously moving with the probe, preferably for the moment when the burn starts. The probe is instantaneously at rest in this frame, which is moving relative to the Sun at ~30 km/s (probe's 'own personal inertial frame').

After the burn, we can compare the speeds and energies of the probe and the reaction mass in this inertial frame and we'll get the 3 km/s and 4.5 MJ/kg for each. In order to know what the energies are in the important Sun-centric coordinates, we obviously then have to transform the "own frame" energies to the Sun's frame. That's where the 900 MJ/kg energy comes in. All very valid.

The Oberth effect is not directly coupled to gravity assist. The latter is energy scored during unpowered flybys, by 'robbing' orbital energy from the host planet. The Oberth effect happens in powered flight anywhere (def: "Oberth effect is an effect where the use of a rocket engine when travelling at high speed generates much more useful energy than one at low speed."

The Oberth effect is often used in conjunction with gravity assist, the latter usually used near a planet. Because the speed of an orbit is higher near a planet and this is also the place where the gravitational field is strongest, both effects can be used there (hence the confusion, perhaps...)

"In the link about Oberth Effect, it talks about parabolic and hyperbolic orbits."

Parabolic orbits are those where the escape speed or energy is just reached and the resulting (open) orbital trajectory is a pure parabola (eccentricity e=1). Hyperbolic orbits are those where the escape speed or energy is exceeded and follows open hyperbolic trajectories (e>1). They are sometimes called "hyper-escape" orbits.

-J

PS: I still prefer the reaction mass energy balance as the 'explaining mechanism' for the Oberth effect. An effect is not quite a mechanism...

Hi Jorrie,

Thanks for the explanations and the thread. In this thread, gravity assist is shown to be temporary as far as velocity is concerned, but it changes the direction. To make the velocity increase permanent you would need the Oberth Effect. Firing the rocket when the probe is at its closest to the 'assist mass' would yield the best result as you said. The lost mass of the fuel is very important too. I give you a GA for #14 and #20.

-S

Hi S, thanks.

But I think you interpret gravity assist incorrectly. Correctly applied, it boosts the solar-centric energy of a probe without any Oberth effect, i.e., no need for a rocket firing. The second figure of the link you quoted shows that. Notice the Vout > Vin for the same distance from Jupiter. That's orbital energy 'robbed' from Jupiter. In a sense, it's the direction change that 'creates' the extra orbital energy.

Sure, a rocket firing at the perihelion would have given a bigger boost to Vout, but that was not done here. Where NASA always performs both, is for orbit-insertions around a planet. They pass in front of the planet in its motion around the Sun and hence give some probe energy to the planet, while the probe lose the same amount of sun-centric energy. At the same time they fire the rocket and optimally (with the least fuel, using Oberth) scrub off more energy, so that the probe goes into an elongated elliptical orbit.

Gravity assist alone cannot make a probe with less than escape energy to escape from a planet and it can also not be used on its own for orbit insertion. They always use Oberth at the perihelion of an 'escape-parking-orbit' to optimize the burn energy for escape. Bottom line is, gravity assist for energy gain only works if a probe passes a planet at higher than escape speed. Oberth works at below escape speed as well.

-J

Hi Jorrie,

Thanks again. I had not understood the meaning of "two dimensional analysis" on the first figure. This Wiki link is much more clear with statements such as: "A gravity assist or slingshot maneuver around a planet changes a spacecraft's velocity relative to the Sun, even though it preserves the spacecraft's speed relative to the planet—as it must according to the law of conservation of energy." I should have gone there first. I think I am ready for your next thread.

-S