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Escape Velocity

Posted by: roygrif - Tue Oct 19, 2004 5:27 pm
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Escape Velocity 
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Post Escape Velocity   Posted on: Tue Oct 19, 2004 5:27 pm
As I recall, you need something like 18,000 mph to escape orbit.

My questions:

#1 How fast does SS1 travel?

#2 What happens if you are 1 mph below escape velocity?

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Post    Posted on: Tue Oct 19, 2004 5:43 pm
Heh, that statement is a gross over-simplification.

But lets answer question 1 first:

SpaceShipOne travels just a fraction over Mach 3 on it's way up (and a little faster on the way down). That means almost exactly 1000 meters per second, or 2250 miles per hour.

As for part two. Firstly, true escape velocity is 25000 miles per hour, or 11100 meters per second, although anyone in their right mind would use the moon as a slingshot, which would bring it down a lot (into your 18000 bracket, but it would entirely depend on what your object was, and where it was going).

At this point, I think you might actually mean "going into orbit" rather than "escape velocity". Escape velocity means your heading out to mars, orbit just means your going around the earth. The answer, as a general rule, if you go 1 mph less than is needed for escape velocity you will go into orbit around the Earth. The orbit will be very eliptical (not circular), and you might just crash back into the Earth, although it depends what direction you were travelling in.

Was there a particular thought you had in mind when you asked these questions?

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Post    Posted on: Tue Oct 19, 2004 6:58 pm
Just curious, if you would stay at a constant height if escape velocity wasn't quite achieved.

Also, it seems that SS1 needs a lot more thrust to reach escape velocity.

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Post    Posted on: Tue Oct 19, 2004 9:54 pm
Anything just under escape velocity would be simply a very high orbit.

The moon is in orbit, but there are Earth orbits which are higher, but they are fairly unstable due to nearing the moon every few months.

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Post    Posted on: Wed Oct 20, 2004 12:13 pm
I thought that there was no such thing as "escape velocity" per se, I thought it was just the velocity required to be handed over from earth's gravity to that of another heavenly body (or bodies). So from my perspective Sev's right in the sense that "it depends ..."

I mean if earth was the only thing floating around with a gravitational field then no matter how fast you went (FTL travel lunatics please put your hands down) you would never, ever truly escape earth orbit. Although you could generate a very, very, very large one.

Is what I'm saying utter crap or not?

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Post    Posted on: Wed Oct 20, 2004 12:41 pm
Dr_Keith_H wrote:
I thought that there was no such thing as "escape velocity" per se, I thought it was just the velocity required to be handed over from earth's gravity to that of another heavenly body (or bodies). So from my perspective Sev's right in the sense that "it depends ..."

I mean if earth was the only thing floating around with a gravitational field then no matter how fast you went (FTL travel lunatics please put your hands down) you would never, ever truly escape earth orbit. Although you could generate a very, very, very large one.

Is what I'm saying utter crap or not?

Well, I would not say crap, but not 100% accurate either. Escape velecity refers to the velecity that is equivalent to the energy that is required to get out of the influence of the earths gavitational field. You can express this as a velocity because both potential energy caused by gravity and the kinetic energy of the speed are dependent on the mass of the object in the same way.
You are right that you can never get completely away from earth's pull (and let's, for the sake of simplicity assume earth is the only object with a gravitational pull). However, that does not mean that you will allways fall back! Because the pull decreases with the square of the distance, the force deminishes asymptoticly with the distance. That means that the gravitational force becomes extremely small with large distances. Mathematicly, it can be shown that the integral over the force with the distance is finite. This integral (that is: the area under the graph that you get if you set out the gravitational force against the distance to earth) is equivalent to the potential (gravitational) energy of an object at infinate distance to earth, and thus, the kinetic energy you need to escape from earth.

So, although the gravitational pull of the earth never completely vanishes, you can get away from earth and there is such a thing as an escape velocity.

edit: another small remark:
Note that you actually don't need to reach this velocity at all in order to escape! It's the velocity that you need to give an object to escape without further propulsion.


Last edited by André on Wed Oct 20, 2004 1:18 pm, edited 1 time in total.



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Post    Posted on: Wed Oct 20, 2004 1:18 pm
Does the escape velocity depend on the altitude? Your answer is sounding like this a little bit at least.

In general 40.000 km/h is said to be the escape velocity of earth - is that valid only at or near the surface and is another (less) escape velocity valid in LEO, HEO, GEO and lunar orbits each?



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Post    Posted on: Wed Oct 20, 2004 1:22 pm
Yes, the escape velocity is different for different altitudes. I believe "the" escape velocity is for the surface (not taking into account air resistance in that case!), but I'm not sure about that. Note that it does not matter that much either. If you start from 100km, you're just a fraction further from the center of the earth (which is the reference point for the distance) than if you started from the surface.


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Post    Posted on: Wed Oct 20, 2004 1:44 pm
100 km is 1/127 of the earthian diameter and 2/127 of the radius. But 1000 km already are 1/12 to 1/13 of diameter or 1/6 to 2/13 of radius . a relevant fraction I suppose.

10,000 km are 5/6 to 100/127 of diameter and 1.66 times to 10/13 of radius and GEO is around three times the diameter or around six times the radius.

So the escape velocity at these altitudes is interesting.



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Post    Posted on: Wed Oct 20, 2004 3:36 pm
André wrote:
You are right that you can never get completely away from earth's pull (and let's, for the sake of simplicity assume earth is the only object with a gravitational pull). However, that does not mean that you will allways fall back! Because the pull decreases with the square of the distance, the force deminishes asymptoticly with the distance. That means that the gravitational force becomes extremely small with large distances. Mathematicly, it can be shown that the integral over the force with the distance is finite. This integral (that is: the area under the graph that you get if you set out the gravitational force against the distance to earth) is equivalent to the potential (gravitational) energy of an object at infinate distance to earth, and thus, the kinetic energy you need to escape from earth.

Hey André, thanks for the input. Just want you to clear up that business of the asymptote, the graph and the integral being finite. I thought if any open-ended graph displays an asymptotic relationship then the integral under it is always infinite. Do you have a link to something that would explain your answer to me in straightforward terms?

But hey, I'm a biologist ... what the hell would I know?

DKH

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Post    Posted on: Wed Oct 20, 2004 5:28 pm
I don't have a link for you, I'm sorry, but I'll try to demonstrate it in mathematically, if I can remember how to do it...

The force of gravity can be expressed as:

(1) F(r) = G * (m1 * m2) / r²

To keep things simple, we assume we are dealing with point masses (pretty reasonalbe, since the earth is approximatly symetric in density and form) and we are flying in a strait line from the earth. Also, we have no other masses to deal with (which is not very reasonable).

Our goal is to calculate the amount of work (energy) we need to perform to get from distance a to distance b of our planet. Because work is the force * the distance we act against that force, that comes down to calculating the integral from a to b over (1) integrated to r.

Because G, m1 and m2 are constant, I stow them away in a single constant K for now:

(2) F(r) = K * r^(-2)

We need to calculate (I use int(a, b) to indicate the integral from a to b) the integral:

(3) int(a, b) { F(r) }dr = int(a,b) { K * r^(-2) }dr = K * int(a,b) { r^(-2) }dr

Calculating the primitive W(r) of F(r) results in:

(4) W(r) = K * -1 r^(-1) + c = -K/r + c

The constant 'c' is a part of the integration process. Here, it is 0 (it has no physical relevance), so we drop it. Note that even if we kept it in, it would have disapeared in equation 5.
So, now that we have the primitive of F(r), we can calculate the differential we needed, simply by filling in our equation (4) for the distances b and a, and substracting the values.

Now, as we are interested in the escape velocity, we want b to be infinate. Note that if we use infinate for the value of r, that W(r) becomes 0! That means that the integral is finite when b goes to infinate!

(5) E(r) = 0 - (- K/r) = K/r = (G * m1 * m2) / r

Now, the last step is to calculate the escape velocity this equates to. The kinetic energy of a moving object can be expressed as:

(6) Ek(V) = 1/2 m * V²

We use the mass of th earth as m1 and the mass of our object as m2. Recognize that we have escape velocity if our kinetic energy is equal to the emergy required to escape, so:

(7) Ek(V) = E(r)

Now, we rewrite (6) to get an expression for V(r) using (7):

( 8 ) 1/2 * m2 * V² = (G * m1 * m2) / r <=> V² = (2 * G * m1 * m2) / (r * m2) <=>
V² = (2 * G * m1) / r

Notice how m2 (the mass of our object) disapears from the equation? Now, we take the square root to arrive at:

(9) V(r) = sqrt ( (2 * G * m1) / r )

This equation gives the escape velocity for any distance to the centre of the earth r, with G the gravitational constant and m1 the mass of the earth.

It's very possible that I made a mistake somewhere in this calculation, but it's the method that counts :-) I hope this explains clearly enough how this thing works.

One final note: our assumption that we are only dealing with the earth and the mass we wish to get out of it's gravity is of course nonsense. We are dealing with many masses in our solar system that are very significant. It is possible to use the other planets or the moon as a slingshot to get the momentum you need. Calculating such trajectories is a complicated business which I don't grasp, let alone can explain.

edit: fixed error in equations 8 and 9


Last edited by André on Thu Oct 21, 2004 8:48 am, edited 1 time in total.



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Post    Posted on: Thu Oct 21, 2004 8:29 am
Uhnn ... my eyes have started to bleed ... so I think I'm going to print it all out and read it over a nice hot cup of tea or something. Thanks very much, 'tis appreciated.

DKH

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Post    Posted on: Thu Oct 21, 2004 8:41 am
Dr_Keith_H wrote:
Uhnn ... my eyes have started to bleed ... so I think I'm going to print it all out and read it over a nice hot cup of tea or something. Thanks very much, 'tis appreciated.


Hold on... I just noticed that I must have made a mistake somewhere, as the result is not right. Now it sais that the escape velocity is lower with a lower r, that is not right. I'll try to correct. Done. The mistake was in 8, and therefore in 9.


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Post    Posted on: Sat Oct 23, 2004 7:54 pm
Excellent job, André.

By the way, Dr. Keith H, there are several such asymptotic-yet-finite integrals. One of them produces the value of e -- I believe it's the summation of 1/(n!) -- and another gives you 2.

The actual speed required to orbit the Earth is, I believe, 17600mi/h. If you were able to move 17600mi/h at a distance of 1in from the earth's surface (assuming you picked an orbit that didn't slam you into a mountain), you'd achieve orbit. Far easier to achieve if you don't have a nice thick pea-soup atmosphere in your way, though.

However, once you get far enough away from the Earth, the speed decreases (the moon moves much slower than that number).

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Post    Posted on: Mon Oct 25, 2004 9:11 am
I got a mathematician friend of mine to slowly (and painfully) explain how André is right ... in the beginning he even rolled his eyes at me as if to say this was a complete no-brainer.

In short, I get it now. Thanks André.

DKH

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