I was reading this article http://www.msnbc.msn.com/id/5287945/ on where space begins and it mentions that escape velocity is 7 miles per second. To reach low earth orbit is 5 mps, and for SpaceShipOne's flight to 62000 feet it is 1 mps. He sort of hints that going further may be a power problem.

My question is, with the same power source, say a rocket, as you get further away from the earth and its gravity has less of an effect a ship isn't that same rocket going to keep accelerrating you? Ummm... what I mean is, if SpaceShipOne got to 1 mps with it's rocket, if it could have burned the rocket twice as long wouldn't it have kept accelerrating? The article makes it sound like that's not the case.

They also compare it to the power of a space shuttle rocket. But given enough space isn't it still a power to weight ratio problem and not simply a velocity problem (given that one is going to keep accelerrating as they get further away from earth's gravity)?

Hmmm... I've said it three different ways and I'm not sure I even understand my question. Oh well, I'll wait for some responses.

Thanks!

I'll have a read of the article first, then come back if no-one else has replied in the meantime

Rob.

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OK, I think I can clarify a few things for you here mesaba. The general confusion you have is probably from these two phrases within the article:

Then later on:

First, a comment about gravity. Strictly speaking, you’re never beyond the gravity of Earth, it pervades the whole of space, although it does fall off with distance from Earth. Just that, after a certain distance, the gravity of Earth is negligible in comparison to that of nearer and more massive objects. However, at 62,000 metres the gravity field of Earth is still more than 98% of it’s value at ground level, and even at 200 miles (the typical altitude of a Space Shuttle orbit) the gravitational strength is 90% of that on the ground. It falls off very slowly indeed. (This is why scientists get narked whenever news commentators talk about the “Zero-gravity” environment of the Space Station, or the Space Shuttle being “beyond the earth’s gravity”.)

The main difference in the terms in the article here is between energy and power. Power is basically the rate of energy output per unit time, so for example, a 1000-Watt electric drill puts out twice a much energy, every second, as does a 500-watt one. But if I ran my 500-watt drill for a full minute, it would put out more energy in total than if I ran my 1000-Watt one for only 15 seconds. Both would make short work of drilling through my fence, but the 1000-Watt drill would do the job faster.

What’s this got to do with spaceships and aircraft? Well, in spaceflight, once you are above the significant parts of the earth’s atmosphere, getting anywhere is all about the energy you need to climb out of the earth’s gravity well, which in energy terms, is the same as climbing out of a 4,000 mile deep hole in the ground.

Now, you do in fact have the option of doing this quickly or slowly in terms of time, and the option taken depends of the rocket technology you’ve got. The Apollo Spacecraft, using chemical fuels and putting out a lot of power, could get to escape velocity within minutes. However, the Europaean SMART-1 spacecraft, using Ion engines, it going to take months to climb out to the Moon, with it’s engine running pretty much all the time – it has lots of energy, but not as much power.

In conventional chemical rockets like the Space Shuttle, it actually makes economical sense to have lots of power, as the chemical fuel itself has a lot of mass, and it is best to burn as much as you can quickly, pick up lots of speed (and what Physicists call kinetic energy) and get rid of that mass of fuel as close to earth as you can, rather than haul most of it with you up the gravity well, which all the time is trying to pull you back down, and which would actually require even more fuel. But in theory, you could take your time.

So, why does the article go on about power being significant in aircraft? Because, aviation (as opposed to spaceflight) is all about being within the earth’s atmosphere, and there the main problems are aerodynamic drag and aerodynamic lift. Air drag is not like gravity – the force of air drag gets worse the faster you try to go, and it’s literally in your face tending to stop you. All this means that to go twice as fast in an aircraft, you need roughly eight times as much power. (I can hear aeronautical engineers choking on their cornflakes here, but that is the general principle. Power is in proportion to the cube of the speed.) And you have to go a certain minmum speed, otherwise your aircraft will ‘stall’ as it can’t sustain enough lift under the wings, and it would tumble back to Earth somewhat untidily.

In summary then, getting around in space is primarily an energy problem (not necessarily velocity), and whereas power/weight ratios are important for aircraft, they are less important for chemical rockets and not even essential if you have other technologies at your disposal and are starting out above the atmosphere of Earth.

I hope this has clarified some of your questions, please come back if you have others.

Rob.

[edited to tidy up some loose bits of my reply, and to correct the height of Spaceship one to 62000 metres, not 62000 feet. ]

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"We need rigidly defined areas of doubt and uncertainty!"

I'll just mention a couple of possible sources of confusion. Unlike an airplane, a rocket will use (almost) all of its fuel in a couple of minutes. To run the engine twice as long, you need to bring along about twice as much fuel, requiring a much larger (and more importantly, heavier) rocket. As a result of the weight of the bigger rocket and the extra fuel, the same engine will not get you to 1 mps nearly as quickly if you put it on a rocket with twice as much fuel. Its a matter of diminshing returns. You need a much bigger rocket to get a little bit faster. The way the apollo rockets did it was with stages. When the fuel was exhausted , you reduce your weight by jettising the part of the rocket that held the fuel. Then you have a smaller rocket with its own fuel to continue the trip. IIRC, the apollo was a 3 stage rocket. The space shuttle uses the solid rocket boosters for the first stage and also drops the external fuel tank when it is empty. The Rocket/glider that went above 62 km is designed to be reusable, so the idea of a multi-stage rocket that is completely reusabe adds a lot of complexity and expense to the project. Their solution is to use a plane to get above as much of the atmosphere as possible (a sort of pseudo first stage), and then use the rocket for the rest of the trip.

I think you mean the Saturn V rockets used for the apollo missions.

But other than that, this has been well covered, I can add no more.

Just to pick a nit...

SS1's last flight went to 100 km, or 100,000 meters, or 62.5 miles, not 62,000 meters. Remember what happened to one of the Mars orbiters when units got mixed.

Fred

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#-o Good job I don't work for NASA.

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"We need rigidly defined areas of doubt and uncertainty!"

Real scientists don't.

Earth escape velocity is the same fixed 7 mi/s, no matter what your mass is. That number is a velocity and it falls out of some basic physics (I remember doing all the calculations in college physics, but I can't do it "off the top of my head" 25 years later -- i'd have to look it up). Someone else on this board probably knows, right now.

Keep in mind that escape velocity (25,200 mph) is far greater than minimum orbit velocity (~17,000 mph).

If SS1 had more fuel and continued to fire its engine, it would continue to accelerate, gaining velocity.

If it had enough fuel it could achieve orbit.

If it had enough fuel it could "escape" Earth and travel to the moon.

If it had enough fuel it would be huge, probably look more like a Delta/Proton/Ariane type rocket and cost almost as much.

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The formula for escape velocity is: v = sqrt(2GM/r)

Where,
G is the gravitational constant.
M is the mass of the object in kilograms
r is the radius of the object in meters.

For the Earth, M = 5.972E24 kg and r = 6.378E6 m

Plug these into the formula and you get: v = 11,178.75 m/s or a little more than 11 km/s.

I must confess that I had to look up the formula.

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In fact, escape velocity at any altitude is sqrt(2) * orbit velocity at that altitude.

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i just had a thought isn´t this formula only correct assuming that the object has equal density fron the centre to the surface?
I might be wrong I admit just speculation on my part :-?

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if it's a symetrical distribution, it'll hold easily . I.e. if variations are the same everywhere, so density at 100 miles from the center is the same, in a "shell" about the core.

Now, if the northern hemisphere is much denser thanthe souther...you may get a problem. Complications at least.

So no, it doesn't require uniform density, just symetrical.

I was thinking more in the line of an object the size of say Betelgeuse which has a very low density outer shell (not sure if that´s the right word)

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As long as you are outside the object, it doesn't matter.

If you're inside, well, then you've got to whip out some calculus and integrate that equation (or a similar one, i'd have to really look at the problem).

Ok thank you that pretty much disproves my "theory" I conceed defeat #-o

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This is a case where not only does size matter, but shape does as well. Take for example a dumbbell shaped object. Assume you start at the midpoint of the bar. The gravitational pull is zero assumming both ends are symmetric. Depending on which direction you go, the gravitational pull will be different. Eventually when you are far enough away that the dimension of the object is small compared to the distance, the object can be treated as a point mass with all the mass concentrated at the object's center of mass.

Back to the real world: If you have a homogeneous spherical mass, the only mass that contributes to the gravitational pull that you feel is the mass that is closer to the center of mass than you are. So, as you dig deeper into the Earth, the gravitational pull decreases. If the mass is non-homogeneous, then you need to know know how the density of the mass is distributed and use some calculus. To correctly do calculations regarding the Earth, you must take the oblateness of the Earth into consideration. Sometimes you can get a good estimate by assuming that the Earth is spherical but surrounded at the equator by a ring. Then you separately calculate the gravity due to the sperical Earth, and the gravity due to the mass in the equatorial ring, add them together (using Vector arithmetic of course), and you have a very good approximation.

of course, since the difference is something like .01%....

I don't usually jump through the hoops.

And notice I said in an earlier post...that as long as the thing is symetrical .
I probably should have specified mostly spherical.

I think one of the issues the creator of this topic wanted to adress was:

what if a ship like SS1 didn't achieve escape velocity but continued in its velocity, say, 3000km/h, into space indefinitely. When would it reach a point where it would no longer fall back to earth? At the geostationary orbit?

(just edited the original mistake 3000k/s to what I meant: 3000km/h

at 3000 km/s it would be far beyond escape velocity...

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