Space travel question.

[Traab]

This one gets into some math/physics type questions so bear with me. If you had two spaceships, identical in every way, one takes off from the moon, one takes off from earth. How much further could the one leaving from the moon travel before running out of fuel? I would assume with less gravity, less atmosphere, and all that stuff, it would take far less fuel to break free of the moon than the earth, but I wondered just how much further you could propel yourself "on a full tank of gas" before you would be reduced to coasting along on your momentum.

*EDIT* Also, would that effect your top speed potential? And if so, do g forces still apply in space the same way as on earth?

[thubby]

that is an insanely involved question.

[Lentrax]

Yup. That it is.

[Traab]

You mean there isnt a general figure like, "It takes "x" amount of fuel to leave earth and "y" amount to leave the moon?" Then you say, "Ok, so moon rocket has "z" amount of fuel left, which amounts to "c" length of time burning. That length of time of rocket burning equates to "g" number of mph increased speed."

I realize there are dozens of factors that would effect how much fuel it takes to leave earth, but much like how if you want to complicate things, you can talk about how the speed of sound varies depending on altitude or whatever, or you can just say, "the speed of sound is 768 mph give or take" you could also say, "It takes about this much fuel to leave earth give or take."

[Zorg]

Assuming the rockets can achieve escape velocity from Earth, the rest is irrelevant as they can then both theoretically travel for infinity after that due to the basically negligible drag of hard vacuum.
The big question is "where are you going?", as orbital mechanics (ie how much fuel you need) hinges greatly on that. If these rockets were going to Mars it would be different than if they were trying to go to Mercury as the Lunar rocket may want to slingshot around Earth and so on.

In terms of standards, there really isn't one as it depends on the build of rocket, payload weight and if you need to get back. If you look at this chart (http://www.chartgeek.com/types-of-rockets/) and spot the Vostok type (green and white in the middle) it was capable of launching a probe to the moon. but you need the mighty Saturn V to get three dudes there and back, and most of it is jettisoned almost right away.

Top speed isn't as much of a thing so much as necessary speed. You don't really want to go faster than you have to or else you need to carry more fuel to slow down when you get where you're going. Also to be smart everything has the same top speed potential of just under the speed of light :smalltongue:
G forces are a way of measuring gravity and also of forces acting on the body in those units instead. So yes, they act the same in space so that if you have a rocket exerting 10 gees of force when it fires on maximum, you'd still feel 10 gees of force when you do the same in space.

[AMX]

This looks like a job for Atomic Rockets. (http://www.projectrho.com/public_html/rocket/)

The Mission (http://www.projectrho.com/public_html/rocket/mission.php) page in particular has some useful info.

[Traab]

Well really I didnt have a destination in mind, its just, how far can they go before they run out of fuel. So heading in a straight line, same fuel load for both, same space ship for both, how much farther would the moon rocket go before it ran out of fuel.

[Drumbum42]

This one gets into some math/physics type questions so bear with me. If you had two spaceships, identical in every way, one takes off from the moon, one takes off from earth. How much further could the one leaving from the moon travel before running out of fuel? I would assume with less gravity, less atmosphere, and all that stuff, it would take far less fuel to break free of the moon than the earth, but I wondered just how much further you could propel yourself "on a full tank of gas" before you would be reduced to coasting along on your momentum.

*EDIT* Also, would that effect your top speed potential? And if so, do g forces still apply in space the same way as on earth?

Ok, a few things... What is "A full tank of Gas?" Because space flight is not like driving a car. If you're talking about Liquid/Solid fuel then I direct you to the equation F=MA (Force=Mass*Acceleration). The problem here is Rocket fuel is a LOT of MASS, which requires a LOT of FORCE to get off the ground. The more fuel you have, the more is needed to get that fuel off the ground. This is why 400lbs of people are strapped onto 200 Tons of rocket, and that's just to get into orbit.

The moon on the other hand has a lot less gravity, which is good for conserving fuel because you go a lot faster. The earth's gravity is about 9.8m/ss, the moon is 1.6 m/ss, so you're going to have ~8.2m/ss better acceleration. On the down side, there's no atmosphere, which normally causes drag on the rocket slowing it down. If you were to use the same rocket ship for both earth and the moon, it my kill the passengers at liftoff. I would not suggest using the same vehicle.

BUT, assuming surviving was unimportant, the amount of fuel used is needed to find out how far you go. But in short you go much farther, much faster.

But XKCD can sum this up better then me. http://what-if.xkcd.com/24/
That is about using model rockets to launch into space, he covers some of the same topics MUCH better then I can. I'd suggest reading though all of his "What If" posts, they're general very interesting.


Edit: As Zorg said, yes in a vacuum you would not slow down, so if your destination was "deep space," you'd simply get there faster.

[Zorg]

Well really I didnt have a destination in mind, its just, how far can they go before they run out of fuel. So heading in a straight line, same fuel load for both, same space ship for both, how much farther would the moon rocket go before it ran out of fuel.


That's... not how space works.

The short version is that once you have a velocity outside the pull of a planet you will just keep going forever. Like the Voyager probes heading out of the solar system. The tircky part is having enough fuel to get out of the various pulls of gravity: Earth's to go to the moon, the sun's to leave the solar system and so on.
There's also things like using planets to slingshot you around them to save on fuel (like boosting down a valley in your car for extra momentum going up the other side), but that's situational on where you're going and when.

The long version requires a PHD :smalltongue:

[thubby]

if you're willing to ignore air resistance,assume the planets are static, and treat the moon as a point mass, and you could get by with a basic energy analysis.

the moon is 3.8×10^8 meters from earth
gravity of earth is 9.81
gravity of moon is 1.622
we need a mass for our space craft so im going to use the combined mass of the lunar lander and module
so... 10,149 kg+ 30,332 kg
lets go with 40,000kg

it takes 40k*9.81*3.8×10^8-40k*1.622*3.8x10^8= 124terajoules of energy to get to the moon.

so liquid hydrogen has an energy density of 120 MJ/kg

it takes a little over 1000 metric tons of rocket fuel to get to the moon.

disclaimer: these are very rough cut estimates. from the rounded numbers to the obviously faulty assumptions, this figure will be off, and should only be used as a ballpark estimate.

[Lentrax]

The long version requires a PHD :smalltongue:

And a lot of whiteboard space to do maths on.

[SaintRidley]

Once you get into space, you'll basically turn off your fuel and let inertia carry you until you get close enough to where you're going that you need fuel to slow down.

So unless you burn all of your fuel at takeoff, you aren't running out.

Also, you're going to need more fuel on Earth to do what you can on the moon simply due to the greater gravity. The heavier your ship, the more fuel you need to lift. The more fuel you have, the heavier your ship is. Unless the reduced gravity messes with the ability of fuel to do its job (it shouldn't, to my knowledge), then you'll more easily be able to take off from the moon because the reduced weight reduces the necessary amount of fuel to take off (and you can pack your ship with as much fuel as you would on earth without increasing your burden nearly as much, ensuring you have plenty of fuel).


And that's about all the knowledge I can dispense on this really, really tricky question.

[thubby]

*EDIT* Also, would that effect your top speed potential? And if so, do g forces still apply in space the same way as on earth?

G force is the result of acceleration. anything can, in theory, be moved at any sub-light speed safely so long as you accelerate gradually.

[Jimorian]

One way of expressing "how full" your fuel tank is Delta-V. The difference in velocity you can achieve going from a full tank to an empty tank.

The "easy" answer is that assuming they both started with the same delta V on the ground, the Moon rocket will have more delta-V left over after escaping the Moon's gravity well than the Earth rocket after escaping Earth.

But that's assuming a "straight up" trajectory, which is terribly inefficient compared to getting into an orbital trajectory first.

This is further complicated by something called the Oberth Effect (http://en.wikipedia.org/wiki/Oberth_effect), which means that it actually takes less delta-V for a rocket starting in low Earth orbit to reach another planet than one starting in Moon orbit, because the orbital velocity around Earth is faster than that around the Moon, meaning there's more kinetic energy already stored in the ship and fuel mass.

And I'm only touching on the just the first order complications involved in your question.

Start playing Kerbal Space Program (http://kerbalspaceprogram.com), and you'll start to understand a lot of this without having to ever look up equations or do any complicated math. :smallsmile:

[Drumbum42]

I think that this thread is a great example that "Rocket Science" is generally used to mean "it's really hard." And we haven't even gotten to the "hard" math yet, OR astrophysics such as space contracting and expanding to decrease/increase velocity relative to earth.

It's a really involved and complex thing to leave earth's gravity. (That and they explode if you make a mistake)

[hamishspence]

One factor is how much acceleration the crew can withstand. Assuming the same "burn rate" as the fuel gets used up, the acceleration will increase- until the fuel runs out. Thus, the burn rate may need to be throttled back toward the end.

[factotum]

Also, you're going to need more fuel on Earth to do what you can on the moon simply due to the greater gravity.

That's true, but the fact the Earth has an atmosphere makes things harder as well--you lose a *lot* of fuel fighting your way through that.

[Traab]

One way of expressing "how full" your fuel tank is Delta-V. The difference in velocity you can achieve going from a full tank to an empty tank.

The "easy" answer is that assuming they both started with the same delta V on the ground, the Moon rocket will have more delta-V left over after escaping the Moon's gravity well than the Earth rocket after escaping Earth.

But that's assuming a "straight up" trajectory, which is terribly inefficient compared to getting into an orbital trajectory first.

This is further complicated by something called the Oberth Effect (http://en.wikipedia.org/wiki/Oberth_effect), which means that it actually takes less delta-V for a rocket starting in low Earth orbit to reach another planet than one starting in Moon orbit, because the orbital velocity around Earth is faster than that around the Moon, meaning there's more kinetic energy already stored in the ship and fuel mass.

And I'm only touching on the just the first order complications involved in your question.

Start playing Kerbal Space Program (http://kerbalspaceprogram.com), and you'll start to understand a lot of this without having to ever look up equations or do any complicated math. :smallsmile:

Ah ok, I was sort of hoping there was a fairly simple set of numbers to work with, a general range of say, how much fuel is used in each example, and that the rest could be extrapolated from there. Guess its just one of those way too complicated to answer without writing your thesis for a phd and probably needing three related fields worth of knowledge to fully understand type of questions. :smalltongue: Cheerfully withdrawn.

[Eldan]

Play Kerbal Space Program. That bears repeating. There's a free demo and it's insane fun.

Okay. Basically, you have to consider the Rocket Equation, as a start. The basic principle as I, as a non-physicist understand it is this:

You have a payload that needs to get away from the planet. This requires a certain amount of energy. However, this energy comes in the form of fuel. This fuel has a mass. Which also needs to be lifted, which requires more energy. So you need more fuel to lift the fuel and then more fuel to lift that...
And, of course, your rocket actually gets lighter the more fuel you burn along the way. And it can detach empty tanks, which makes it even lighter.

Then, once you get to the moon, your fuel also has less weight, due to lower gravity.

And that's without complications.

But basically, just from my experience with Kerbal Space Program: by far most of your fuel is just to get away from the planet and into orbit. Once you're in orbit, you don't need all that much fuel anymore.

[Mando Knight]

PhD? No, that's the kind of problem you'd find in the homework for a junior or senior year Aerospace undergrad... the only difficult part is factoring in air resistance, which you won't be doing on paper for a rocket problem anyway.

You have the rocket equation, you know the characteristics of your rocket (or this problem would be silly), and you know how the acceleration due to gravity varies. Add together those equations for velocity changes and integrate with respect to time to get a distance change to get your burnout altitude equation.

Adding in air resistance is just adding in another force... but it's one that varies with respect to both velocity and altitude, so the integration gets messy and it's much easier to put the whole thing through MATLAB or something at that point, particularly since now you have to pay attention to your Mach number.

[nedz]

You can go a very long way indeed if you use the Interplanetary Transport Network (http://en.wikipedia.org/wiki/Interplanetary_Transport_Network), it's not very fast though.

[Ravens_cry]

That's true, but the fact the Earth has an atmosphere makes things harder as well--you lose a *lot* of fuel fighting your way through that.
All combined with the fact if you need more fuel you need more fuel to carry the fuel, and a rocket to get into Earth orbit is a heck of a lot bigger than what's needed to get into lunar orbit (http://www.youtube.com/watch?v=HyWVGa6I7Gk).

You can go a very long way indeed if you use the Interplanetary Transport Network (http://en.wikipedia.org/wiki/Interplanetary_Transport_Network), it's not very fast though.
That will come in handy when transporting bulk cargo that doesn't have a expiration date.

[factotum]

Guess its just one of those way too complicated to answer without writing your thesis for a phd and probably needing three related fields worth of knowledge to fully understand type of questions. :smalltongue: Cheerfully withdrawn.

It's more that the question, as posed, doesn't *have* an answer. For instance, if the rocket in question has sufficient fuel to start with, it could theoretically go an infinite distance starting from either location!

[Kizor]

I don't think the topic's too complicated. I've been learning about these things and I'd like to try to edit together the matters that have come up in this thread. (plus I like the sound of my own voice, so you needn't feel embarrassed that I'm still going at it after you withdrew your question.)

It'll still be wordy as #¤%&. I don't know how much you know, so please don't be offended if I seem condescending.

Your basic assumption is right. The moon's lower gravity and lack of atmospheric drag mean that a rocket will spend far less energy fighting them, and thus will get far more speed out of the same amount of fuel. A lunar rocket thrusting straight up will have gone further by the time it runs out of fuel - unless either one of them hits escape velocity. Escape velocity is the spot where gravity can no longer pull you back down. Gravity's effect on you is dropping faster than your speed. Hit that, and bang, you're gone, and unless you're pointed at the Sun or the Earth or something you can go a very long way indeed. But if neither of the rockets hits escape velocity, the one from the moon will have gone further when its fuel runs out. They'll keep on moving up, and the one from the moon will have coasted further by the time gravity halts its progress and drags it back down.
Side note: I'm assuming that gravity and drag are the big movers. I'm cheerfully ignoring other factors, like the fact that earthbound rockets need to have engines for Earth's atmosphere and for vacuum, all the little subtleties of maneuvering in an atmosphere, and constraits like not jellying your crew (as hamishspence said). Then there's the Oberth effect. I have no idea how that works, and even though spaceflight was apparently considered absurd without it, I recommend ignoring it unless you get down to the equations. One can land on the moon in KSP without knowing about it, after all.
We live at the bottom of a deep gravity well. I don't know if this is a serious term or scifi geekery, but it's true. Hauling anything into space is an enormous hassle, compared to moving around once you're there. Orbital or lunar industry could make spaceflight much easier, but we still have to get it there in the first place.
The moon's escape velocity is far smaller than the Earth's. If I remember right, you could escape from one of the dinky moons of Mars with a bicycle and a ramp.

The problem is that thrusting straight up is a poor measure of range. In practice spaceships fire their engines for seconds or minutes at a time, coasting for possibly hours or days or years afterward. Our rockets would probably establish orbits as swiftly as possible to stop spending fuel fighting gravity, and then use ninja tricks like swinging by planets to change direction and have the planets slingshot them to faster speeds. Point in fact, the Apollo missions never achieved escape velocity. They thrust in Earth orbit, elongating it into a spindly one that would've brought them back down if they hadn't been intercepted by the moon's gravity partway through. With spaceship range depending on patience as much as this (and on how long your astronauts can last), distance becomes a difficult metric. A better one is delta-V, as Jimor mentioned, with "delta" standing for change and "V" for velocity, so the ability to change velocity.

There are currently five probes (and one booster rocket) that have achieved solar escape velocity and are bound to leave the solar system. NASA can and has sent objects to travel light-years with 1970s tech.
I'm looking at some of these numbers, and apparently a ship can get from low Earth orbit to low Mars orbit for cheaper than from to go from low Earth orbit to low Mars orbit. Your lunar rocket would be able to do more or bigger things in space, but I can't tell by how much. An expert could probably give you some numbers about modern-day spacecraft performance and estimate the difference.

[Cikomyr]

Well really I didnt have a destination in mind, its just, how far can they go before they run out of fuel. So heading in a straight line, same fuel load for both, same space ship for both, how much farther would the moon rocket go before it ran out of fuel.

There is no way to travel in a "straight line" in space, as far as we know, unless you start from a gravitational-free location, which is, as far as we know, beyond our knowledge. Even galaxies are pulled to each others and agglomerate in clusters.

Movement in space is defined by ellipses. Any burst of acceleration you will use (consuming fuel) is going to alternate that ellipse. "Escape velocity" means you are at a point and speed in the theoretical ellipse where your speed allows you to "escape" the reference gravitational body. In the case that worries us at the moment; the Earth.

however, you will still be part of the Sun's gravitational influence, and will simply develop an orbit around it rather than the Earth/Moon.


People have suggested picking up Kerbal Space Program, and I agree. Even if it's simplified at bit, the tutorial makes a good case to make you understand space mechanics on an intuitive level. Hell, I've had astrophysics classes before and even I did not fully grasped the movement of Apollo 13 until I played the game..

It makes you understand just how wonderful our modern science is to manage to send probes to Uranus.

And yes, that joke was on purpose :D

[Mando Knight]

Movement in space is defined by ellipses. Any burst of acceleration you will use (consuming fuel) is going to alternate that ellipse. "Escape velocity" means you are at a point and speed in the theoretical ellipse where your speed allows you to "escape" the reference gravitational body. In the case that worries us at the moment; the Earth.

Conic sections, actually. Escape velocity is what you need to enter a parabolic "orbit" with your current position as the periapsis. In the absence of any other objects, the path would continue indefinitely.

[nedz]

There is no way to travel in a "straight line" in space, as far as we know, unless you start from a gravitational-free location, which is, as far as we know, beyond our knowledge. Even galaxies are pulled to each others and agglomerate in clusters.

Movement in space follows geodesics (http://en.wikipedia.org/wiki/Geodesic) — which are the equivalent of straight lines in complex gravitational fields. Of course if you are firing rocket motors then this will change, but if you are exhibiting zero delta-V then you will be moving along a "straight line" or geodesic.

[shawnhcorey]

Rockets are measured in their Δv or maximum change in velocity.

If their Δv is less than the escape velocity of the Moon, both will rise up and fallback down, the one on the moon rising higher and taking longer.

If their Δv is greater than the escape velocity of the Moon but less than the escape velocity of the Earth, the one from the Earth will, at some time, fall back to Earth while the other will go into an Earth-Moon orbit.

If their Δv is greater than the escape velocity of the Earth but less than the escape velocity of the Sun, they both will go into solar orbit.

If their Δv is greater than the escape velocity of the Sun but less than the escape velocity of the galaxy, they both will go into galactic orbit.

If their Δv is greater than the escape velocity of the galaxy, they both will (eventually) fly between the galaxies. :smallbiggrin:

[shawnhcorey]

BTW, it is possible for an object in Earth-Moon orbit to go into solar orbit and back. See Newly Discovered Object Could be a Leftover Apollo Rocket Stage. (http://neo.jpl.nasa.gov/news/news134.html)

[shawnhcorey]

Movement in space follows geodesics (http://en.wikipedia.org/wiki/Geodesic) — which are the equivalent of straight lines in complex gravitational fields. Of course if you are firing rocket motors then this will change, but if you are exhibiting zero delta-V then you will be moving along a "straight line" or geodesic.

Sorry, no. The N-th body problem has no solutions. The only way to determine where a body will go in an inverse-square field when there are more than two bodies, is to follow it. It does not follow any predictable course.

[Mando Knight]

Rockets are measured in their Δv or maximum change in velocity.

If their Δv is less than the escape velocity of the Moon, both will rise up and fallback down, the one on the moon rising higher and taking longer.

If their Δv is greater than the escape velocity of the Moon but less than the escape velocity of the Earth, the one from the Earth will, at some time, fall back to Earth while the other will go into an Earth-Moon orbit.

If their Δv is greater than the escape velocity of the Earth but less than the escape velocity of the Sun, they both will go into solar orbit.

If their Δv is greater than the escape velocity of the Sun but less than the escape velocity of the galaxy, they both will go into galactic orbit.

If their Δv is greater than the escape velocity of the galaxy, they both will (eventually) fly between the galaxies. :smallbiggrin:
Actually, it doesn't matter much how fast you go, if it's strictly radially outward from the gravitational body. You'll fall back eventually. Any escape trajectory has some component of velocity that's perpendicular to the position vector. Edit: escape velocity is actually just a speed, without reference to a vector. However, those wishing to engage in actual gravity-well-escaping will want to go perpendicular as quickly as possible, however, so as to waste less fuel getting to escape velocity.

Sorry, no. The N-th body problem has no solutions. The only way to determine where a body will go in an inverse-square field when there are more than two bodies, is to follow it. It does not follow any predictable course.
It has no explicit solutions. However, a constrained N-body problem (such as using predefined paths for the other bodies) is quite solvable (at least, numerically. There aren't any tidy exact solutions for an N-body problem).

[shawnhcorey]

Actually, it doesn't matter much how fast you go, if it's strictly radially outward from the gravitational body. You'll fall back eventually. Any escape trajectory has some component of velocity that's perpendicular to the position vector.

No, velocity is the only thing that counts. Any direction that will not hit the planet will work.

[hamishspence]

Yup- an object leaving the planet at escape velocity, will slow down if no further thrust is given, over time, but it won't reach zero speed (relative to the planet)- until it's infinitely far away.

Thus- it cannot ever fall back.

[Mando Knight]

On review, I see where I made the mistake. Oops.

[nedz]

Sorry, no. The N-th body problem has no solutions. The only way to determine where a body will go in an inverse-square field when there are more than two bodies, is to follow it. It does not follow any predictable course.

My statement was entirely correct. The fact that it's very hard to predict a priori where your vehicle will go is irrelevant — an unpowered object will follow a geodesic.

[shawnhcorey]

My statement was entirely correct. The fact that it's very hard to predict a priori where your vehicle will go is irrelevant — an unpowered object will follow a geodesic.

Absolutely not. The general case of the N-th body problem has been proven to be unsolvable. If it followed the geodesic, it would have a solution.

If you assume the other bodies are moving slowly, then you can assume that the spaceship will follow a path close to the geodesic.

[nedz]

Absolutely not. The general case of the N-th body problem has been proven to be unsolvable. If it followed the geodesic, it would have a solution.

If you assume the other bodies are moving slowly, then you can assume that the spaceship will follow a path close to the geodesic.

You are confusing the real world behaviour of objects with our ability to model them. We cannot accurately predict the weather yet weather happens non the less. I am not claiming that we can solve the n body problem, but the argument that bodies in space follow geodesics follows directly from GR.

[shawnhcorey]

You are confusing the real world behaviour of objects with our ability to model them. We cannot accurately predict the weather yet weather happens non the less. I am not claiming that we can solve the n body problem, but the argument that bodies in space follow geodesics follows directly from GR.

If you assume the bodies don't move. And yes, the model is consider accurate, that is that gravity is an inverse-square field. And since GR is proportional to gravity, it too is an inverse-square field. The general case of motion of 3 or more bodies in an inverse-square field is unsolvable. Bodies, or even light, do not follow the geodesics.

[Animastryfe]

Absolutely not. The general case of the N-th body problem has been proven to be unsolvable. If it followed the geodesic, it would have a solution.

If you assume the other bodies are moving slowly, then you can assume that the spaceship will follow a path close to the geodesic.

The n-body problem for n greater than 2 has no closed form solution, but solutions (http://en.wikipedia.org/wiki/N-body_problem#The_theoretical_solution) can be found in terms of Taylor series. Also, apparently the 3-body problem has exact solutions for special cases, although I do not know much about that.

[Kelb_Panthera]

In the most lay terms I can put it, both ships will go just as far as they want. Once you're outside of a celestial object's gravity well you just don't stop until you hit something or apply the brakes.

The difference between the two otherwise identical ships will be in how high a velocity they can reach before running out of fuel. The exact speed they can reach is based on a whole host of factors that make it impossible to give you any kind of exact figures without a lot more information on the ships. Ultimately, the ship leaving from the moon will hit a higher velocity before running out of fuel because it'll take less fuel for it to break free of the gravity well it started in.