I don't think so.

Time matters for power, not energy (though power is energy per unit time).

It takes the same amount of energy to lift an object slowly, vs quickly.

Now, there are some efficiency ratings to throw in there too, but that depends on the method, not the goal.

Then again, anyone more familiar with the subject, feel free to correct me.

I think earlier in the thread it was said quite well: the amount of energy expended to get a unit mass out of the atmosphere and into orbit is the same no matter how fast you do it.

The real question is the method -- again, earlier in the thread it was pointed out that since rocket fuel is heavy (as are the tanks it's in) yuo want to blast it out as fast as possible, hit the relevant velocity at 3g's of acceleration (or whatever) and go. If there was some way to build -- I don't know, I am making this up -- a low-power ion drive you could use on Earth somehow, thrusting at only a few m/s/s but doing it for days, you cuold go into orbit much more gently.

But in the latter case you need a way to keep the engine running for a long time, and there is no technology that offers that right now. (The fuel cost would be insanely high).

The problem here is a confusion between the concepts of work and energy. The amount of work is the same, but the amount of energy is not. If you already have enough thrust to overcome gravity, then any extra thrust will work to accelerate the rocket. If you spread the thrust over time, then more of it will be necessarily devoted to overcoming gravity resulting in lower acceleration.

An extreme example is a rocket that maintains a thrust equal to 1 g The actual thrust decreases over time as fuel is consumed to maintain a 1 g force. This rocket will hover until it runs out of fuel and then drop.
If instead, the rocket even momentarily produces a thrust in excess of 1 g, then it will rise to a point that will never be reached in the first scenario. The difference is that all of the thrust in the first example was wasted on gravity. In the second example, some of the thrust accelerated the rocket.

The Mouse on the Moon. You could do it this way, but unless you had a big supply of Pinot Grand Fenwick or some other fuel with essentially infinite Isp there's no practical way to carry that much fuel.

No, not really.

Let's take this in stages. Until your rocket hits orbital velocity, it's fighting gravity. In the worst case, imagine a rocket, hovering. It's expending tons of fuel, but isn't going anywhere. Eventually it runs out of fuel and crashes. In order to minimize waste, you need to hit orbital velocity as quickly as possible--you need a high thrust.

OK. Now you're in a parking orbit--you have a few choices as to how you can get to your final destination orbit. Your equipment has also been subjected to 2-5 g's on blastoff. There's no real choice here.

There's an advantage to doing all your burns as close to earth as you can--you don't have to lift the fuel, and a change in velocity close to the earth gives you a higher velocity when you're away from the earth than the same change in velocity farther away. So, for a given Isp, a high thrust engine gets you to your target with less energy and less fuel.

There's kind of a trade-off as far as Isp (exhaust velocity) is concerned. The lower your exhaust velocity, the less energy it takes blast it out the back, but the more fuel you have to carry. High Isps require more energy but less fuel. If you want to minimize the total energy required, it works out that you want your exhaust velocity to be about 60% of your total mission delta V (so your package is about 80% fuel). I could have made a mistake in my derivation; you might want to check my calculus.

Note still that you really want to apply all this boost all at once--while you're deep in the gravitational well. Note also that for all current drives energy isn't really an issue (power is for ion drives, mass ratio is for chemical rockets).

So, currently we have our choice of drives--high thrust low Isp chemical drives or low thrust high Isp ion drives. The ion drives allow you to get more delta V out of a given amount of fuel, so they can send probes farther out then chemical drives. Past a certain point, they can also get a probe there faster (the higher top speed can compensate for the much longer time in getting there). For most missions, the ion drives will actually use more energy than the chemical drives, but again, it's not energy that is the limiting factor.

OK. To summarize. Chemical drives are absolutely necessary to get to orbit, and for many missions (inner solar system) are going to be the fastest way there. Ion drives can take over once in parking orbit or once in escape orbit and will allow larger payloads or faster outer system missions or more complicated missions (orbiters instead of fly-by's, that sort of thing).

Escape velocity is altitude-dependant. You can escape without ever hitting 11 km/sec, but if you've escaped by definition you've reached escape velocity.

No, that's not the definition of excape velocity.

Really? You don't like sqrt(2GM/r)? What do you think the definition of escape velocity is?

You can go into lunar orbit without ever achieving escape velocity, just a high enough orbit around the Earth that when the Moon comes along it grabs you.

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Yikes. Maybe we need a definition of "escape."

It looks like daver is saying that escape means to achieve escape velocity. But then, a rocket a few (ok a big few) feet off the gound would have escaped. Is that it?

I asked a similar question a few years ago and came up with the following conclusions:

1) Escape velocity applies to unpowered vehicles
2) Escape velocity is misleading, escape energy is a better term.
3) Escape velocity is dependant on initial conditions, e.g. it's different from the top of MtEverest and the bottom of death valley.

4) Some people define it with a formula as seen in previous posts, others define it as the launch speed required in unpowered flight such that the object won't return back to planet it was launched from.

Point 4 is particularly important as it means that the true escape velocity would be affected by the aero-dynamics of the object etc since they all remove energy that would otherwise keep its velocity higher at any given point.

For powered flight the escape velocity is dependant on the fuel, the thrust, the efficiency of the engines and all losses. So, if I have enough fuel I can get far enough away to get to a point where my current velocity when I shut the engines down is greater than the velocity where I would be in orbit at that distance. :-?

All very confusing, this ones even more so. I could build a meter that indicates what my velocity needs to be at any point in my accent such that I can safely shut down my engines without fear of falling back to Earth. If I watched this meter it would read very high at launch and keep reducing as I got higher. If I did a great job designing such a device (I think I would, maybe :-? ) it would show a negative velocity if in the pull of another planet (or the Moon).

I could do a 3d contour map showing the escape velocity at any point in the solar system too, but, this would get confusing as then I'd have to consider all the other objects in space and their map would need to be superimposed on top. It gets worse too, what would be the escape velocity of Earth if I'm on the surface of the Moon?

All valid questions if planning space travel. If the term escape velocity was changed in the above paragraphs to escape energy it all becomes clear rather than non-sensical.

Regards,

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MrObvious

[quote="ToSeekYou can go into lunar orbit without ever achieving escape velocity, just a high enough orbit around the Earth that when the Moon comes along it grabs you.[/quote]

Of course you can get to the Moon without achieving Escape Velocity. The moon is still within the Earth's sphere of gravitational influence. That means that the Earth is the primary gravitational object. Now to get to Mars or Venus or something like that, you need to exceed Earth's escape velocity, but not Solar Escape velocity.

BTW: Escape velocity is the same whether you are dealing with powered or unpowered flight. Acceleration is nowhere a part of the formula for escape velocity of sqrt(2GM/r), If you are going faster than that, and you are out of the atmosphere,you will escape regardless of which direction you are pointing. Of course if your powered flight slows you down, you might not escape, but that is not the issue here. While I'm thinking of it, I'm not really sure any more what the issue is in the thread.

I don't think that's necessarily true, either. If you had your timing right on the button, you'd just need to achieve an orbit of Earth with a 30 million mile apogee, though I daresay the difference between doing that and achieving escape velocity is minuscule.

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daver, JFR...

Thanks for the tweak, I knew I was forgetting something. I'm to used to ballistic trajectory style problems.

You can take a hohmann transfer ellipse to the moon--time from LEO to lunar encounter is about 5 days, the change in velocity required is about 3.13 km/sec. You're about 0.1 km/sec short of escape velocity at perigee; quite a bit shorter at apogee (about 1.25 km/sec). You can cut the travel time appreciably at a miniscule extra cost in fuel by going into a hyperbolic rather than elliptical orbit.

I'm not sure where your 30 million mile apogee figure came from--the moon is only a quarter of a million miles out.

I think he's talking about Mars, but he might want to add ten or twenty percent.

Yes, and you're right, it would be more like 35 million miles.

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My apologies; somehow I missed the bottom part of your post.

But I'm still confused--Mars at closest approach is closer to 50 million miles, isn't it?

You need about 3 km/sec left over after leaving Earth's vicininty to get to Mars--running a smidgeon under escape velocity to get you to a 50 million mile apogee is still going to leave you in the vicinity of Earth's orbit around the sun, although perhaps nowhere near earth.

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