## [2015-05-29] Horrifying indeed!

Blame Quintushalls for this.

Moderator: Kimra

### [2015-05-29] Horrifying indeed!

So if I understand correctly, the Teacher in this comic uses horrifying physics to teach the joys of killing koalas, right?

Specific Potential Energy of an object at the surface of Pheibos:
-64 m²/s²
Specific Orbital Energy of an object on a martian orbit around the Sun:
-291.113.810 m²/s²

Specific Potential Energy of an object at the surface of Luna:
-1.411.053 m²/s²
Specific Orbital Energy of an object on a terran orbit around the Sun:
-443.562.776 m²/s²

Bringing an Object of 1 Kg mass from Pheibos to Earth:
1 Kg * [(443.562.776-291.113.810) + 64] m²/s² = 152.5 MJ
Bringing an Object of 1 Kg mass from Luna to Earth:
1 Kg * 1.411.053 m²/s² = 1.4 MJ

So in short: Not actually more. Not even close.

Also the main reason why people in the space industry think its funny when politicians start discussing Asteroid mining (>200 MJ/Kg not even counting the transport of vehicles and tools), as you can recycle most Metals for 5 to 20 MJ/Kg.
ShieTar

Posts: 4
Joined: Fri May 29, 2015 8:31 pm

### Re: [2015-05-29] Horrifying indeed!

Actually... Neither of you guys are correct. ShieTar, you've made two errors: First, the gravity of the sun HELPS rather than hinders moving mass from Mars to Earth. It's essentially a sign error. Second, for some reason all your values are half of what they ought to be. U=-GMm/r, so specific energy is = -GM/r.

As to your comments about asteroid mining, the energy required to move an asteroid orbiting the sun to an earth orbit is primarily governed by how much you would have to alter the velocity of said asteroid in order to change the orbit.

Zach made an error as well: Neglecting the gravity of the planet that the moons are orbiting. So the energy required to move the mass from the moon to the earth is determined by how far out of the moon's gravity well the object would have to climb, but the energy required for a transit from Phobos to earth has to climb out of not only Phobos's gravity well but also Mars's. Since Phobos is one of the nearest moons to its planet (only 9376 km from the center of mars) Mars contributes significantly to the energy requirement. The specific energy required to exit Martian gravity well is 4.56 MJ/kg, as opposed to only 2.82 MJ/kg to escape lunar gravity (It actually is slightly less since the earth is only about 3.8*10^5 km from the moon instead of approximately infinitely distant, it's about 2.80 MJ/kg). It's still easier to move from the more distant Deimos to Earth; at 23463 km from Mars it would only take 1.83 MJ/kg to move mass from Deimos out of the Martian gravity well to earth.

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novak
Guest

### Re: [2015-05-29] Horrifying indeed!

-GM/r, or -µ/r for simplicity, is the potential energy of an object, yes.
+GM/2r is the kinetic energy of an object required to stay in a stable circular orbit with that amount of potential energy.
Thus the specific energy of an object in orbit is effectively -µ/2r, exactly half the potential energy of a stationary object at the same position.

But you are right for the stationary objects on the moons surfaces, they don't have orbital kinetic energy of course and thus should be double of what I have written down. I've just been in the groove of retyping the same formula into the calculator. Doesn't really change the overall conclusion though.

Your point of missing Mars/Earth potential energy in the original lecture in the comic is correct though. But again, any object on a moon has also the same specific kinetic energy as the moon itself, so an object on Phobos is in an Mars-Orbit with a specific energy of -2.28 MJ/kg.
Your own math of on the gravity well of earth seems incorrect though. The standard gravitiational paramter of earth is 398600 km³/s², so specific potential energy at a lunar distance of 385000 km is still about 1MJ/kg, or 0.5MJ/kg at a lunar orbit.

Guest wrote:As to your comments about asteroid mining, the energy required to move an asteroid orbiting the sun to an earth orbit is primarily governed by how much you would have to alter the velocity of said asteroid in order to change the orbit.

Exactly, and the enrgy required to change from one circular orbit to annother is exactly the difference in specific energy between the two orbits. In general you do that by a http://en.wikipedia.org/wiki/Hohmann_transfer_orbit , in short you first change the kinetic energy on one point of the orbit to match the kinetic energy of the target orbit, then wait for half an (now eliptical) orbit until you have reached the correct orbital distance. At this point the falling towards the central body has built up new additional kinetic energy, so you have to need to change it again to match the target orbit.

You could in theory just spend half the difference to put the asteroid on an eliptical orbit, make it intersect the earth orbit with asteroid orbit kinetic energy at its perihelion.
The remaining half of the energy differential would then be converted to heat when your Asteroid enters the Earth atmosphere. Just warn the dinosaurs before you do that

And as for the gravity of the sun helping: It's not that easy, the sign makes no real difference here. There are no hybrid engines in space, where you could just store excess energy: You always have to off-set it by spending additional energy. Granted, instead of burning fuel you can use planetary swing-by to put the energy into the planetary orbit itself. But again, that does not depend on the sign of the energy difference, you can also use swing-bys to accelerate when leaving the solar system.

Here's a video of the swing-by dance that our forth-coming Bepi Colombo mission will have to go through just to decelerate to mercury velocity:

And all of that means the transport from Earth to Mercury is only possible on a very specific date every few years.

Also, Earth does not care about the swing-by of a single 1-ton satellite, but bring in an asteroid with a mass in excess of a billion tons? You talk about investing 2x10^20 J of kinetic energy into the earth. The rotational energy of the earths daily rotation is 2x10^29 J. It'll only make the day 40µseconds shorter (or longer), which may not effect you or me directly, but it will throw of presiccion calculations like those used for GPS & Co.

I would also expect some rather spectecular flood waves during the initial swing-by.
ShieTar

Posts: 4
Joined: Fri May 29, 2015 8:31 pm

### Re: [2015-05-29] Horrifying indeed!

Ok, I suppose the energy calculations point is reasonable, I was only looking at potential energy. Planetary eccentricities in question are pretty low so that's a fairly valid approximation.

The whole point of talking about specific energy requirements is to talk about what's possible, not a single transfer method. Yes, you can do a Hohmann transfer, and you probably would if you care about TOF, but the whole point of the comic (the science part) is that technically, it is possible to find a route which uses less energy, even over a very long distance in space. This is impossible in the two-body problem with centrobaric bodies but not so much in the real solar system. It might take a decade or three, but you really could go from Mars's sun orbit (if you were far enough from Mars) to earth expending less energy than going from the moon to the earth. That's the point the teacher is making (trying to make?) in the comic- how potential energy works.

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novak
Guest

### Re: [2015-05-29] Horrifying indeed!

Guest wrote: It might take a decade or three, but you really could go from Mars's sun orbit (if you were far enough from Mars) to earth expending less energy than going from the moon to the earth. That's the point the teacher is making (trying to make?) in the comic- how potential energy works.

But you do not actually "expend" less energy, you merely use a different reservoir of energy, e.g. the orbital energy of one of the planets instead of fossil fuel. It may seem to be "free energy", at least in an economical sense, but that equation would change if you bring Earth on a collision course with Mars after a million years of asteroid catching, or if you wipe out the human race by changing the day/night cycle to a duration we (or our food-stock) can no longer cope with.

So in a way the teacher is telling you that you can constraint a physical problem to just the one aspect, without checking or even defining the validity of your assumption.

I'm beginning to think that you and that Weiner guy are just paid cronies of the Martian dead Koala industry!
ShieTar

Posts: 4
Joined: Fri May 29, 2015 8:31 pm

### Re: [2015-05-29] Horrifying indeed!

The mouseover text in this comic mentions Big Bird, and it reminded me of a story.
At one point a few decades ago, NASA was thinking that the children of America were not interested enough in space, and they wanted to do something to bring out interest so that kids would be excited about science.

And they decided to approach Sesame Street and Carol Spinney to ask if they could find a way to do "Big Bird in Space". Spinney was really excited about the idea, and they tried to get it to work, but there was no way to fit an eight-foot tall costume into the shuttle. So they came up with a different plan.

They went with a teacher named Christa McAuliffe to go up in the space shuttle Challenger.

Honestly -- "dead Big Bird"? We came uncomfortably close to "lost Big Bird in the shuttle disaster." If that had happened, I think there wouldn't be a person of my generation who even looked up into the sky, let alone was interested in space science. Astronomy, rocketry, the space program -- those would all be The Evil Stuff That Killed Big Bird.
Ian Osmond

### Re: [2015-05-29] Horrifying indeed!

I'm beginning to think that you and that Weiner guy are just paid cronies of the Martian dead Koala industry!

You got me!

So in a way the teacher is telling you that you can constraint a physical problem to just the one aspect, without checking or even defining the validity of your assumption.

I guess... In much the same way that physics teachers often explain about blocks on inclined planes without accounting for the curvature of the earth, or explain the parabolic motion of a ball without using relativistic equations. A sack of dead koalas isn't going to deorbit anything, though I hope the Australian space program doesn't take that statement as some kind of challenge.

Also, interestingly, if you say, take an asteroid from approximately Mars orbit and move it in to approximately earth orbit, you'd be adding energy to the rest of the solar system, not reducing it, as the system is conservative and you've reduced the energy of the asteroid. Probably, you would increase the orbital energy of the earth and decrease that of Mars, as the asteroid would be dragging Mars inward and the earth outward. It's difficult to say in such simple terms, but at any rate, no energy would have to be lost by the system until you exert a non-conservative force such as a rocket engine.

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novak
Guest

### Re: [2015-05-29] Horrifying indeed!

ShieTar wrote:
Guest wrote: But you do not actually "expend" less energy, you merely use a different reservoir of energy, e.g. the orbital energy of one of the planets instead of fossil fuel. It may seem to be "free energy", at least in an economical sense, but that equation would change if you bring Earth on a collision course with Mars after a million years of asteroid catching, or if you wipe out the human race by changing the day/night cycle to a duration we (or our food-stock) can no longer cope with.

The same could be said about Solar Power. It's not free energy; instead of heat that would have radiated to the atmosphere, it is instead converted to electrical energy and then used elsewhere. Eventually we will create a global weather shift due to Solar Power... Of course, as you pointed out, the "eventually" is very, very far out. Also (assuming we don't use it for rockets) the system we are in is closed for the most part.
guest2

### Re: [2015-05-29] Horrifying indeed!

Guest wrote:I guess... In much the same way that physics teachers often explain about blocks on inclined planes without accounting for the curvature of the earth, or explain the parabolic motion of a ball without using relativistic equations.

Well, but the teachers usually do the experiment and verify that their planes are not actually affected by the curvature of earth, and the do throw their balls slow enough so that the non-relativistic equations remain sufficiently accurate.

The teacher in the comic on the other hand talks compares the small gravity wells of the moons while ignoring the fact that they are themselves at different positions of the larger gravity wells of their planets, which in turn are also at different positions of the even larger gravity well of the sun. In this case, the moons are actually the negligible part of the problem, not the important one. Its like saying it takes more work to get from a 10m deep hole in the Netherlands to the peak of Mt. Everest than it takes from a 20cm deep ditch in the North Base Camp, because the hole is deeper. Its just very much not true, unless you make some assumption like "You only need to get out of the hole yourself, the rest of the way is done by helicopter". Yes, such assumptions exist and can even be realistic, but you can not just leave them out of your teaching.
Guest wrote:Also, interestingly, if you say, take an asteroid from approximately Mars orbit and move it in to approximately earth orbit, you'd be adding energy to the rest of the solar system, not reducing it, as the system is conservative and you've reduced the energy of the asteroid. Probably, you would increase the orbital energy of the earth and decrease that of Mars, as the asteroid would be dragging Mars inward and the earth outward. It's difficult to say in such simple terms, but at any rate, no energy would have to be lost by the system until you exert a non-conservative force such as a rocket engine.

Fair enough, though reducing and increasing are rather irrelevant terms regarding the energy of a system, as the 0 is a completely arbitrary definition (Do you have 0 kinetic Energy because you just sit in your chair, or do you have a lot because you travel quickly around the sun?). Also as long as you change the sign on all your definitions, the physics remains exactly the same.
But granted, the general definition will state that the Asteroid now has less energy because it is on a lower orbit around the sun. So depending on where and how you have done your swing-bys, either Mars or Earth or both will now have a larger orbit about the sun with more energy, or maybe the Earth-Moon Distance is now larger. Dragging Mars inward would not help you, because that would push the asteroids out, not pull them in.

But where you are incorrect here is by calling the rocket fuel "non-conservative". It is not, if you use the rocket engine to push the asteroid in, you push the rocket fuel into a higher solar orbit. So the entire solar system, including the now interplanetary fuel cloud, would still keep the same energy. Some of it will be transferred from chemical potential energy to heat due to the way effective rockets work, but there is still no overall loss of energy. There never is.

Unless you push earth a tiny bit to far out. Our solar system is "dynamically full". If the distances between the planets get smaller, they start to interact with each other (much like a swing-by), and there is a high chance that one of the two planets will eventually get pushed on an increasingly eccentric orbit, and finally get removed entirely from the solar system. I believe I read that Mercury is destined to be kicked out in a few billion years, as its still a tiny bit too close to Venus for its own good.

guest2 wrote:The same could be said about Solar Power. It's not free energy; instead of heat that would have radiated to the atmosphere, it is instead converted to electrical energy and then used elsewhere. Eventually we will create a global weather shift due to Solar Power... Of course, as you pointed out, the "eventually" is very, very far out. Also (assuming we don't use it for rockets) the system we are in is closed for the most part.

Absolutely true. Firstly though, rocket fuel is also solar power, just photochemical rather than photo-voltaic. Secondly, I would not necessarily agree that Solar Power will be very slow to change the weather. When you remove heat from a sufficiently large area, you create artificial low pressure centers. Our complete weather system is created by nothing but local differences in ground absorption, water evaporation and wind speed resistance. The sun itself would only quiet homogeneously heat up half the planet at a time, so all the weather effects are created by local differences in cooling and convection. Play around with that, and you may get some nasty surprises quiet quickly. Or you're lucky and improve your locale climate, it's a gamble because it is extremely difficult to understand which human decision may be connected to which effect, because of the chaos theory and the lack of thousands of copies of the Earth for statistically relevant experiments.
ShieTar

Posts: 4
Joined: Fri May 29, 2015 8:31 pm

### Re: [2015-05-29] Horrifying indeed!

Well, but the teachers usually do the experiment and verify that their planes are not actually affected by the curvature of earth, and the do throw their balls slow enough so that the non-relativistic equations remain sufficiently accurate.

The teacher in the comic on the other hand talks compares the small gravity wells of the moons while ignoring the fact that they are themselves at different positions of the larger gravity wells of their planets, which in turn are also at different positions of the even larger gravity well of the sun. In this case, the moons are actually the negligible part of the problem, not the important one. Its like saying it takes more work to get from a 10m deep hole in the Netherlands to the peak of Mt. Everest than it takes from a 20cm deep ditch in the North Base Camp, because the hole is deeper. Its just very much not true, unless you make some assumption like "You only need to get out of the hole yourself, the rest of the way is done by helicopter". Yes, such assumptions exist and can even be realistic, but you can not just leave them out of your teaching.

Close, but backwards. It is more like the teacher is saying that it technically requires less energy to get out of a 20cm hole on top of mount Everest and to the bottom of death valley than it is to get get there from a 10m hole right next to death valley. As you point out, this is unlikely to be true if dead koalas ever spoil (and the on earth version is even less likely to be true as there is more friction and fewer gradually sloping routes). But his point is still totally true. You don't need a helicopter, though it's unclear how long it would take to make the journey or if it is even possible given earth's real geometry.

Fair enough, though reducing and increasing are rather irrelevant terms regarding the energy of a system

I agree, of course. But common sense will tell you that a more energetic orbit is one that has more energy relative to the object orbited. Not that this is a particularly useful definition, just that it's what I was trying to say without devolving into coordinate frame arguments (typically I suppose this would actually be less energy because the potential energy is taken to be negative, as you did in your math- the point is, the magnitude would be greater).

But where you are incorrect here is by calling the rocket fuel "non-conservative". It is not, if you use the rocket engine to push the asteroid in, you push the rocket fuel into a higher solar orbit. So the entire solar system, including the now interplanetary fuel cloud, would still keep the same energy. Some of it will be transferred from chemical potential energy to heat due to the way effective rockets work, but there is still no overall loss of energy. There never is.

I think you're missing what I was trying to say, maybe a slight vocabulary confusion. As you say, "there is never any overall loss of energy," and this is true in as far as no mass is created or destroyed (which actually does happen a bit during chemical reactions...). But I'm not saying that firing a rocket engine destroys or loses mass from the solar system. A force is considered "conservative" when the work it performs is independent of path. This is true of something like gravity or electrical fields where you can assign a potential energy to every point. It is not true of something like a rocket engine, where the force is not a function of only position. If you are at a given distance from the sun you know what your potential energy is relative to the sun, and what it would be if you were to move to another position- and based on this, you know the minimum amount of work required to get from where you are to that other point. This is the teacher's whole point. Your point appears to be that the teacher's point is completely invalid if you use any of our primary means of interplanetary travel- which are not conservative.

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novak
Guest

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