What is the minimum amount of energy required to travel between planets?

[Simetrical]

This came up in another thread, and I couldn't think of a satisfactory answer.

Suppose I want to travel from, say, high Earth orbit to high Mercury orbit. What implications do conservation laws and so on have for the most energy-efficient way to do this? Since I'm moving inward in the solar system, I'm revolving more quickly, and therefore have more kinetic energy and more momentum (both linear and angular). I have less potential energy than I did before, and this will more than outweigh the loss of kinetic energy, since I'm in an elliptic orbit: total energy is negative and goes like 1/r, so decreasing r will decrease total energy.

Since I'm in high orbit, being in the gravity well of the planet itself is irrelevant. It seems to me that as long as I have somewhere to get my extra momentum (say, a planet), it should be possible to travel from one planet to the other for zero energy cost, or indeed perhaps negative energy cost. So is there any minimum energy cost to getting the needed angular momentum? How could I get that without expending energy?

If there really is no minimum energy cost for this kind of travel, how about the energy cost of travel that's required to have an average speed of at least v (i.e., practical travel)? If this still isn't a lot of energy, what factors contribute to the enormous cost of space travel? Mainly manufacture of equipment and escaping gravity wells?

[Ummon]

The highest consumption is reaching orbit (mass of the rocket and gravity acceleration being fixed factors), of course, then there is the cost of accelerating up until the middle of the course, and decelerating in the second part.

But with the methods used today, they exploit planet gravity to sling ships and satellites to destination, thus it is mostly a passive trajectory, except maneuvering.

It all depends on relative positions of the bodies involved in their orbits, etc.

Overall, if you google "space economics" you will find links with relevant information, I think.

[Sphere]

or indeed perhaps negative energy cost

To clarify the possibility of negative energy costs:

Two objects in space, by having mass and by being seperated by a distance, have potential energy. Much like pulling two magnets apart, when then come back together they create a force that can move a mass (aka work).

Leaving one planet and slingshotting around another planet effectively moves the two planets a little closer to to eachother, and the reduction in potential energy is transfered to your craft.

As to Sim's question. (this is all off the top of my head so a gaurantee some of it is flawed.)

Existing an orbit: Minimum required energy:
If you want to leave orbit you need the same energy as as if you were rocketing straight away from the planet ie integrate Gm1m2/r^2 from r=inital radius of orbit to r=infinity. So....

Gm1m2/ro where ro=initial orbit.

Which makes sense, the larger ro the less energy required to leave orbit

Travelling:
If you used the minimum amount of energy to leave earth orbit, you shouldn't have any relative velocity. So you gotta get moving using 1/2mv^2 unless you gonna bust out some Einstein.

Entering Mercury orbit.
I am fuzzy on this, but with some initial velocity from the above part, you should be able to enter some orbit without any added energy, unless you are going to fast or too slow, in which case you would need to adjust your velocity to keep from cratering or flying off into space.

[Simetrical]

The highest consumption is reaching orbit (mass of the rocket and gravity acceleration being fixed factors)

Which is why I said suppose we start in orbit. This energy cost could be made as small as desired by harvesting energy from descent: just as getting into orbit requires energy input, so getting down from orbit can produce an equal amount of output. In practice you won't quite break even, but you could get as close as you liked if you had good enough technology, so there's no lower bound here.

of course, then there is the cost of accelerating up until the middle of the course, and decelerating in the second part.

But if you go slowly enough, this energy cost could be made as small as you like, so again, no lower bound.

But with the methods used today, they exploit planet gravity to sling ships and satellites to destination, thus it is mostly a passive trajectory, except maneuvering.

No, long-distance travel is still heavily dependent on engines. You have to get to whatever you're slingshotting around to begin with, after all. Slingshotting can reduce energy costs, but it doesn't eliminate them, and certainly isn't relevant to a pure conservation-of-energy question.

It all depends on relative positions of the bodies involved in their orbits, etc.

Only if you account for practical concerns like how long it takes to get there. If you were content with moving at one inch per second and arriving sometime after the expected death of the Sun, it doesn't really make a difference if you can shorten the distance traveled by half or whatever.

Overall, if you google "space economics" you will find links with relevant information, I think.

No, that's engineering. I'm asking a pure theoretical physics question.

Existing an orbit: Minimum required energy:
If you want to leave orbit you need the same energy as as if you were rocketing straight away from the planet ie integrate Gm1m2/r^2 from r=inital radius of orbit to r=infinity. So....

Gm1m2/ro where ro=initial orbit.

Which makes sense, the larger ro the less energy required to leave orbit

Entering Mercury orbit.
I am fuzzy on this, but with some initial velocity from the above part, you should be able to enter some orbit without any added energy, unless you are going to fast or too slow, in which case you would need to adjust your velocity to keep from cratering or flying off into space.

I was assuming that we were far enough away from the planet that this was negligible, since it's clear that there's no minimum energy cost to leave orbit and then reenter it (which is basically what you're doing, dependent on the planets' masses and so on). If you like, assume that you begin and end stationary relative to the planets in question, instead of orbiting them.

Travelling:
If you used the minimum amount of energy to leave earth orbit, you shouldn't have any relative velocity. So you gotta get moving using 1/2mv^2 unless you gonna bust out some Einstein.

Yes, but there's no lower bound on that. You could move as slowly as you like. So is the only cost (ignoring orbital stuff) really just given by acceleration and deceleration?

The problem I have is this: you could then get from Earth to Mercury with zero energy expenditure, or arbitrarily close. But in the process, you've lost a ton of angular momentum. Where does that go? I suppose it goes to your propellant. You'd have to have a huge amount of propellant and propel it away with enormous angular momentum but virtually no energy. And yet that requires that the propellant's kinetic energy mv²/2 is negligible, but its angular momentum |r × p| <= rp = rmv is enormous. And yet this is true when r is a constant. The ratio of kinetic energy to rmv is v/2r, which must be practically zero. So since r is fixed, v must be practically zero.

I guess that works. If v is nearly zero but m is enormous, say k/v for some large-but-not-too-large k, then kinetic energy would be something like kv/2, which is tiny, and angular momentum would be <= kr (plus units), which is not. You could make k (and therefore the angular momentum) as big as you like without increasing the energy, as long as you make sure you make v small enough.

The problem is that if m is too large, you have to take into account the energy required to separate it from the craft, due to gravity. So m can't be taken too large without running into a minimum energy requirement. For instance, if you take m = mass of Earth, the craft needs to escape Earth's gravity well.

Some more computation would probably give me a minimum bound on this indeed. I'll think about it.

[Sphere]

The problem I have is this: you could then get from Earth to Mercury with zero energy expenditure, or arbitrarily close. But in the process, you've lost a ton of angular momentum.

Momentum is not energy. Conservation of momentum and conservation of energy are two different things.

mv = momentum
1/2mv^2 = Energy

The other flaw in your scenario.

Since I'm moving inward in the solar system, I'm revolving more quickly, and therefore have more kinetic energy and more momentum

Being a tiny spacecraft far from the sun means you are not in solar orbit. You can get to mercury on minimal energy, but to "catch up" with the faster pace of mercuy's orbit, you will have to match its faster pace or it will leave you behind, which takes energy.

[Ummon]

The reason why I suggested that you google "space economics" is that the issue is treated in detail in NASA articles about the viability of space extractive processes. The fact that you asked about engineering, doesn't forbid me to suggest an alternate way to get the data. There, you may find more specific reasons for my comments as well.

Going slowly enhances the exposure to space radiation, therefore it is never viable for human expeditions, unless of course there is an efficient cosmic ray shield in place.

[Ummon]

You mean kinetic energy.

[Simetrical]

Going slowly enhances the exposure to space radiation, therefore it is never viable for human expeditions, unless of course there is an efficient cosmic ray shield in place.

You're completely missing the kind of question I'm asking. This is theoretical physics. I don't care about practicality for the purposes of this question. So maybe it's a robot, whatever.

Momentum is not energy. Conservation of momentum and conservation of energy are two different things.

mv = momentum
1/2mv^2 = Energy

I know that, as I should think the computations in my post demonstrated. But can you really gain or lose a lot of momentum in space without expending energy?

Being a tiny spacecraft far from the sun means you are not in solar orbit. You can get to mercury on minimal energy, but to "catch up" with the faster pace of mercuy's orbit, you will have to match its faster pace or it will leave you behind, which takes energy.

Surely that's wrong. Any object in the solar system that's under the influence of no force but the Sun's gravity (e.g., a spaceship that draws level with Mercury and then turns off its engines) is going to be in some type of solar orbit. This is just the result of the two-body problem, I still remember the derivation using Lagrangian mechanics pretty well from my second-semester mechanics class. You have to be in either an elliptic, parabolic, or hyperbolic orbit. Your exact orbit will depend only on your mass, total energy, and angular momentum. Accelerating toward the Sun will put you in a closer orbit, changing some of your potential energy to kinetic energy, but you'll still be orbiting. You don't need to use your engines to match Mercury's orbital speed ― that's where your gravitational potential energy goes. To stop orbiting, you'd have to have zero angular momentum, in which case you'd eventually fall into the Sun unless you have some other forces acting on you.

[Ummon]

You're completely missing the kind of question I'm asking. This is theoretical physics. I don't care about practicality for the purposes of this question. So maybe it's a robot, whatever.

Technical concerns are all about practical matters. In any case, if it's a robot, then fuel consumption doesn't need to be high. Infact I think the idea is explored in that article I mentioned. I will track it if you need it, saw it years ago.

[Viking Prince]

Not to put a damper on the 'high orbit' but you will be in the planetary draw from earth until you pass into the planetary draw from another body. There is no free parking out there. The most reasonable draw would be from the sun. So all you need to do is calculate the energy needed to lag out or earth orbit into a spiral towards the sun. The next material bit of energy to be consumed would be to be slow down enough to be captured by mercury -- that may be the bigger trick. There might be a way to use Venus to slow the craft but not be captured -- a bit of a anti-slingshot effect. Normally the slingshot is flipping the craft out form the solar sink by accelerating toward another sink and taking advantage of the rotation around the sun to avoid capture. I am not certain how to reverse the process, but it should be possible with very little energy expended.

[Simetrical]

Not to put a damper on the 'high orbit' but you will be in the planetary draw from earth until you pass into the planetary draw from another body. There is no free parking out there. The most reasonable draw would be from the sun. So all you need to do is calculate the energy needed to lag out or earth orbit into a spiral towards the sun. The next material bit of energy to be consumed would be to be slow down enough to be captured by mercury -- that may be the bigger trick. There might be a way to use Venus to slow the craft but not be captured -- a bit of a anti-slingshot effect. Normally the slingshot is flipping the craft out form the solar sink by accelerating toward another sink and taking advantage of the rotation around the sun to avoid capture. I am not certain how to reverse the process, but it should be possible with very little energy expended.

As I said, forget about the orbit. Suppose I'm orbiting the Sun, then, in the same orbit as Earth but a million miles away, and I want to pull myself into a closer orbit, the same orbital distance as Mercury. Is that theoretically possible while expending arbitrarily little energy? If so, what kind of theoretical device could do it?

Put another way, is there some way to lose an arbitrarily large amount of angular momentum while expending arbitrarily little energy? You can transfer angular momentum to a nearby body without expending significant energy, I believe, as I discussed in my second post in this thread, but that requires you to be adjacent to a large mass, so you'll have to expend energy to escape its gravity. Is there some way to transmit the angular momentum while expending almost no energy and not being near a large mass?

[Viking Prince]

Once again, your angular momentum plus the gravity well of the sun has you in a stable orbit. To change the orbit you need to expend energy to to reduce speed to spiral towards the gravity sink (sun). At some point in the process you will need to break the spiral, but that can be a well aimed spiral that will be captured by the new gavity well (Mercury).

Technically a reallly small amount is required to move the orbit from stable to a slight spiral. Your problem is all within the gravity sink of the sun and only angular momentum keeps the craft from falling into the gravity sink. However this process may well take billions of years in the case of only a slight spiral into the sink.

Any unforseen event of another object can rearrange the plan. If you are talking in billions of years, the probability of an unforseen event is quite large. So your question is really Two:

1) probability of success with nearly no energy expended and no unforseen event and
2) how much energy needs to be expended for the time period of travel desired.

This problem is also a one way trip since the reverse to counteract the gravity sink of the sun takes a clear amount of energy.

[Simetrical]

Once again, your angular momentum plus the gravity well of the sun has you in a stable orbit. To change the orbit you need to expend energy to to reduce speed to spiral towards the gravity sink (sun).

Why should you have to expend energy to slow down?

Technically a reallly small amount is required to move the orbit from stable to a slight spiral.

That's not true. As soon as you stop decelerating, you'll stop spiraling. Spirals are not conic sections and are not solutions to the two-body problem. To spiral into the Sun, you'd have to continuously decelerate yourself, such as by finding a source of friction.

Any unforseen event of another object can rearrange the plan. If you are talking in billions of years, the probability of an unforseen event is quite large.

I don't really care about things like that. Let's look at a pure question about two point masses with nothing else in the universe.

This problem is also a one way trip since the reverse to counteract the gravity sink of the sun takes a clear amount of energy.

That's obvious, yes. Moving to a higher orbit will increase your total energy, so you need energy input.

[Belgian General]

Nevermind.

[Ummon]

Inertia and gravity make so that you need to expend energy to slow down and modify course (to correct trajectory) when you make passages near planets to accelerate.

[Gaidin]

Inertia and gravity make so that you need to expend energy to slow down and modify course (to correct trajectory) when you make passages near planets to accelerate.

with careful calculations and execution you can launch yourself right into the orbit around the other planet, assuming you take into consideration other gravity sources, and not expend any energy to slow down, and let the pull of the target planet do it for you as you go into an orbit. Depending on what you mean by 'expend energy' of course.

Next to impossible to actually pull off? Probably. But theoretically calculable.

Though the odds of me missing some technicality are fairly good as I never got to take a class on this level of physics.

[Simetrical]

Inertia doesn't require you to expend energy to slow down. If I'm driving in a car and apply the brakes, that takes no energy (or negligible energy) no matter how fast I'm going. I can even charge up the battery by braking, producing energy. After all, I'm losing kinetic energy and it has to go somewhere: why not my batteries?

So why can't something like that be done with a spaceship? What physical principle prevents it?

[Ummon]

In space there is no attrition.

[Simetrical]

In space there is no attrition.

"Attrition" has no meaning in physics that I'm aware of.

[Ummon]

Neither have I, I provide merely amateurish banter on this.