# How does escape velocity work?

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#### jeffreyH

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##### How does escape velocity work?
« on: 28/02/2014 02:41:01 »
I have attached a graph of the final version of the escape velocity from the centre of a planet or other mass. This is not steeply exponential as the previous estimate suggested.
« Last Edit: 01/03/2014 10:54:41 by chris »
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#### CliffordK

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##### Re: Does my calculation match up to other data?
« Reply #1 on: 28/02/2014 07:56:38 »
As mentioned earlier, the escape velocity can not go to zero at the center of the planet.  It has to always decrease with distance.

Here is a free-hand drawing of what I would expect to see.

[attachment=18563]

The change in escape velocity may be a curve somewhat like what you have, in which case you would integrate it from distance d from the center of the planet to infinity (area under the curve).

#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #2 on: 28/02/2014 08:34:18 »
As mentioned earlier, the escape velocity can not go to zero at the center of the planet.  It has to always decrease with distance.

Here is a free-hand drawing of what I would expect to see.

[attachment=18563]

The change in escape velocity may be a curve somewhat like what you have, in which case you would integrate it from distance d from the center of the planet to infinity (area under the curve).

That is exactly the type of curve I had during the early stages. However that is for escape velocity with regard to a mass ignoring the effects of gravitational cancellation. Ve in my graph above could really be considered to represent a gradient with respect to mass cancelled moving out from the centre of gravity which is different altogether. It represents freedom of vertical movement away from the COG.
« Last Edit: 28/02/2014 08:50:41 by jeffreyH »
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#### CliffordK

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##### Re: Does my calculation match up to other data?
« Reply #3 on: 28/02/2014 10:09:09 »
Ve in my graph above could really be considered to represent a gradient with respect to mass cancelled moving out from the centre of gravity which is different altogether. It represents freedom of vertical movement away from the COG.

So, what you're graphing is not the Escape Velocity, but rather the change in velocity due to gravity.  In other words, the Acceleration Due to Gravity.

So, you would be drawing an acceleration graph, units m/s2 rather than velocity, m/s.

See the acceleration due to gravity on Wikipedia.
http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth

With the the best match probably being the PREM curve in navy blue, varying from 0 m/s2 near the center of Earth to just over 10 m/s2  at the core/mantle boundary.

You're escape velocity is the integral (area under the line) from distance d to infinity of the acceleration function and gives you the roughly the S-shaped curve I've drawn above with little change near the center of the planet, the greatest change near the surface (or at the core/mantle boundary), and then decreasing change the further one goes away from the planet.

#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #4 on: 28/02/2014 11:48:51 »
Well I am using Ve = SQRT(2GM/r) which is the calculation for escape velocity. The equation you are suggesting is g = -GM/r^2. I am aware of the PREM data but that is not what I am modelling. I could just as easily have used g but I am looking into asymptotic freedom within a mass which requires the use of a varying velocity gradient which only Ve will provide. This is not a straightforward use of Ve. For any point within the mass Ve indicates the speed required to move away to gain freedom from THAT particular point. This version of Ve varies point to point up towards the surface. I shouldn't really call it Ve as it is really asymptotic freedom I am measuring which is entirely different.
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#### CliffordK

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##### Re: Does my calculation match up to other data?
« Reply #5 on: 28/02/2014 15:32:10 »
Well I am using Ve = SQRT(2GM/r) which is the calculation for escape velocity.

That equation is giving you an answer that doesn't make any sense.  Thus, one must conclude that it is not valid for subsurface acceleration due to gravity.

If A(x) is the acceleration due to gravity at distance (x) from the center of the planet, then the escape velocity would be:

v = $$\int_d^\infty A(x) dx$$  for d is distance from the center of Earth.

Above the surface, A(x) = GM/r2

Substituting it into the integral above, one gets:

v = $$\int_d^\infty \frac{GM}{x^2} dx$$ =  $$\frac{-GM}{x} |_d^\infty$$ = 0 -  $$\frac{-GM}{d}$$ =  $$\frac{GM}{d}$$

Oops, it looks like velocity is in time, and I calculated it with respect to distance...  See the derivation in Wikipedia here.

So it should be: (1/2)v2 = $$\frac{GM}{d}$$, and v = $$\sqrt{\frac{2GM}{d}}$$

Now, we know that the acceleration due to gravity function below the surface of earth is NOT  A(x) = GM/r2, so we should expect the escape velocity function to fail.

We can break the function down into two parts.

Ab(x) = Acceleration due to gravity below the surface of Earth.
As(x) = Acceleration due to gravity in space (above the surface of Earth)

Set the radius of Earth = 6,371 km.

And one gets the equation:

(1/2) v2 = $$\int_d^{6,371 km} A_b(x) dx$$ + $$\int_{6,371 km}^\infty A_s(x) dx$$

We've already solved the second half of the equation above, but you need a good approximation for Ab(x) for the acceleration due to gravity below the surface.

Consider:

Ab(x) = 9.8*x/(6,371 km)

That would give you the case for the constant density of Earth (straight line on the Wikipedia graph above).

Simply integrate it, and one gets:

(1/2) v2 = $$\int_d^{6,371 km} \frac{9.8*x dx}{6,371 km}$$ + $$\int_{6,371 km}^\infty \frac{GM}{x^2} dx$$ = $$(\frac{9.8\frac{m}{s^2}}{6,371 km})\frac{1}{2}x^2 |_d^{6,371 km} + \frac{GM}{6,371 km}$$ =  $$(\frac{9.8\frac{m}{s^2}}{6,371 km})(\frac{1}{2})({(6,371 km)}^2 - d^2) + \frac{GM}{6,371 km}$$

So, your escape velocity (subsurface) is

ve = $$\sqrt{2(\frac{9.8\frac{m}{s^2}}{6,371 km})(\frac{1}{2})({(6,371 km)}^2 - d^2) + \frac{2GM}{6,371 km}}$$

Ok, so I graphed the above equation, giving a nice curve, of the general shape of what I had scribbled by hand above.

[attachment=18569]

Note, this uses the constant density of Earth.  You'll probably want to substitute in your preferred subsurface density/gravitational acceleration function.
« Last Edit: 28/02/2014 16:05:18 by CliffordK »

#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #6 on: 01/03/2014 02:31:21 »
As I said this was what I originally got shown below. However all forces at the centre of gravity are in equilibrium and the g forces pull equally in all directions. What sends an object to the centre is the imbalance in the overall force at the surface. Once these equalize the acceleration stops. The forces don't go away. Disturbing this equilibrium and moving away is easier than moving away from the combined force at the surface. It can't be the case that escape velocity is increased at the centre with respect to an acceleration from that point. While it is not necessary to reach escape velocity to leave a gravitational field it must also be an acceleration profile. Otherwise it is a deceleration which makes no sense.
« Last Edit: 01/03/2014 02:33:05 by jeffreyH »
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#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #7 on: 01/03/2014 02:51:53 »
Here is a proposition that will turn things on their head. Earth gravity is 9.78 m/s². I contend that if we were to put two laser detectors in a very deep shaft 9.78 meters vertically apart and then drop and object down the shaft it will have slowed down by the time it reached the detectors. This depends upon the depth of course and may be hard to measure. This is because as you descend the mass now above the object is exerting a decelerating force which acts against the bulk of the mass accelerating the object towards equilibrium. Instead of hurtling through the centre of gravity to the opposite surface of the planet you would come to rest at the centre. We have no experimental evidence for either your or my case. If we did and could settle this through observational evidence then we would be able to determine the conditions inside a black hole. Mathematics without observational evidence is conjecture.

Some interesting answers can be found here.

http://physics.stackexchange.com/questions/18446/how-does-gravity-work-underground
« Last Edit: 01/03/2014 03:00:41 by jeffreyH »
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#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #8 on: 01/03/2014 03:54:28 »
An interesting discussion from another post.

http://www.thenakedscientists.com/forum/index.php?topic=45321.0

evan_au posted.

"The above discussion has been focusing on "weight", or "the acceleration due to gravity".

There is another measure of gravity: How far you are down a "gravitational well". This can be characterised by an Einstein red-shift in the energy of photons which are emitted out of the well.

I expect that the center of the earth is the lowest part of earth's gravitational well, and will show an Einstein red-shift in photons emitted towards the surface in any direction. In this sense, gravity gets stronger the further you go down the well, in terms of producing a greater Einstein red-shift."

Now this red shift is interesting. So from an equilibrium position at the COG we have a red shift as if light is moving away. The total opposing forces on the emitted light have to be the reason for this red shift. Now as we know the whole universe is exhibiting a gradient of red shift as according to Hubble. I make no more comment on this observation other than it is interesting.

BTW This red shift implies a gradient of acceleration away from the COG. As light travels at c and will be affected by gravity as it leaves the well this is a good indicator of this assumption.
« Last Edit: 01/03/2014 03:56:56 by jeffreyH »
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#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #9 on: 01/03/2014 04:14:42 »
Clifford one thing I need to comment on.

"Oops, it looks like velocity is in time, and I calculated it with respect to distance...  See the derivation in Wikipedia here."

Doesn't time become dominant in the interior of a black hole? This should be taken into account. It should start outside the event horizon with normal mass densities and become more pronounced under compression of a mass.
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#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #10 on: 01/03/2014 16:14:44 »
One last word on this. Here is an article that highlights the problem.

http://vixra.org/pdf/1311.0183v3.pdf

Of particular note and what I am banging on about is discussed in the paragraph below but the whole article should be read.

"Physics, on the other hand, solicits unambiguous prediction of any gravitational attribute that can be
measured. Outside of a static, spherically symmetric gravitational source of eﬀective mass M > 0, the inverse
of the square root of the of the 00-component of the static, spherically-symmetric empty-space Schwarzschild
metric-tensor solution of the Einstein equation is supposed to yield the dimensionless gravitational redshift
factor as a function of the radial distance from the center of the source [4]. Every smooth invertible remapping
R(r) of the radial coordinate, however, yields yet another static, spherically-symmetric “form” of such an
empty-space Schwarzschild metric-tensor solution of the Einstein equation. In the absence of any persuasive
principle which links the radial coordinate marking scheme that human scientiﬁc observers happen to actually
use to a speciﬁc such Schwarzschild metric-tensor solution “form”, we can only regard the generally covariant
Einstein equation as physically ambiguous"

This then is also applicable to within the mass itself.
« Last Edit: 01/03/2014 16:16:44 by jeffreyH »
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#### jeffreyH

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##### Re: Does my calculation match up to other data?
« Reply #11 on: 06/03/2014 03:55:12 »
Well I am using Ve = SQRT(2GM/r) which is the calculation for escape velocity.

That equation is giving you an answer that doesn't make any sense.  Thus, one must conclude that it is not valid for subsurface acceleration due to gravity.

If A(x) is the acceleration due to gravity at distance (x) from the center of the planet, then the escape velocity would be:

v = $$\int_d^\infty A(x) dx$$  for d is distance from the center of Earth.

Above the surface, A(x) = GM/r2

Substituting it into the integral above, one gets:

v = $$\int_d^\infty \frac{GM}{x^2} dx$$ =  $$\frac{-GM}{x} |_d^\infty$$ = 0 -  $$\frac{-GM}{d}$$ =  $$\frac{GM}{d}$$

Oops, it looks like velocity is in time, and I calculated it with respect to distance...  See the derivation in Wikipedia here.

So it should be: (1/2)v2 = $$\frac{GM}{d}$$, and v = $$\sqrt{\frac{2GM}{d}}$$

Now, we know that the acceleration due to gravity function below the surface of earth is NOT  A(x) = GM/r2, so we should expect the escape velocity function to fail.

We can break the function down into two parts.

Ab(x) = Acceleration due to gravity below the surface of Earth.
As(x) = Acceleration due to gravity in space (above the surface of Earth)

Set the radius of Earth = 6,371 km.

And one gets the equation:

(1/2) v2 = $$\int_d^{6,371 km} A_b(x) dx$$ + $$\int_{6,371 km}^\infty A_s(x) dx$$

We've already solved the second half of the equation above, but you need a good approximation for Ab(x) for the acceleration due to gravity below the surface.

Consider:

Ab(x) = 9.8*x/(6,371 km)

That would give you the case for the constant density of Earth (straight line on the Wikipedia graph above).

Simply integrate it, and one gets:

(1/2) v2 = $$\int_d^{6,371 km} \frac{9.8*x dx}{6,371 km}$$ + $$\int_{6,371 km}^\infty \frac{GM}{x^2} dx$$ = $$(\frac{9.8\frac{m}{s^2}}{6,371 km})\frac{1}{2}x^2 |_d^{6,371 km} + \frac{GM}{6,371 km}$$ =  $$(\frac{9.8\frac{m}{s^2}}{6,371 km})(\frac{1}{2})({(6,371 km)}^2 - d^2) + \frac{GM}{6,371 km}$$

So, your escape velocity (subsurface) is

ve = $$\sqrt{2(\frac{9.8\frac{m}{s^2}}{6,371 km})(\frac{1}{2})({(6,371 km)}^2 - d^2) + \frac{2GM}{6,371 km}}$$

Ok, so I graphed the above equation, giving a nice curve, of the general shape of what I had scribbled by hand above.

[attachment=18569]

Note, this uses the constant density of Earth.  You'll probably want to substitute in your preferred subsurface density/gravitational acceleration function.

Here is my adjusted plot of escape velocity. I posted the wrong one earlier. So yes I get what you mean but I think it is in error. The error lies in not cancelling the gravitational feedback. If you have to sum it in the first place to get 2GM then internally you have to remove the cancelled portion otherwise you get an upward trend. If you don't get this then you don't follow Einstein.
« Last Edit: 06/03/2014 04:02:15 by jeffreyH »
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#### CliffordK

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##### Re: How does escape velocity work?
« Reply #12 on: 06/03/2014 07:35:20 »
Your graph now looks like the one I made.  It never hurts to double check my calculations as my calculus is getting a bit rusty by now.

Your graph will vary a little if you change the density/acceleration function subsurface.

You can't really have a gravity function that pulls towards the surface.  I think I read that the gravity within a hollow shell is zero.  And, potentially if you had a hollow shell with a non-uniform density, then the gravity could pull towards one side or the other, but that is quite a special case that is not reflected on Earth.

As far as falling in a hole.  Most of these escape velocity calculations ignore wind resistance.

In a planet, since the gravity is always positive towards the center, falling into the hole, you will always experience positive acceleration as you approach the center...  and thus speeding up (or zero acceleration at the center).  Then once you pass the center and approach the opposite side, you would experience negative acceleration.  In an "ideal system", falling through the sphere, you would accelerate towards the center, slow down going away from the center and barely reach the opposite surface.

« Last Edit: 06/03/2014 07:37:59 by CliffordK »

#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #13 on: 07/03/2014 01:03:31 »
Your graph now looks like the one I made.  It never hurts to double check my calculations as my calculus is getting a bit rusty by now.

Your graph will vary a little if you change the density/acceleration function subsurface.

You can't really have a gravity function that pulls towards the surface.  I think I read that the gravity within a hollow shell is zero.  And, potentially if you had a hollow shell with a non-uniform density, then the gravity could pull towards one side or the other, but that is quite a special case that is not reflected on Earth.

As far as falling in a hole.  Most of these escape velocity calculations ignore wind resistance.

In a planet, since the gravity is always positive towards the center, falling into the hole, you will always experience positive acceleration as you approach the center...  and thus speeding up (or zero acceleration at the center).  Then once you pass the center and approach the opposite side, you would experience negative acceleration.  In an "ideal system", falling through the sphere, you would accelerate towards the center, slow down going away from the center and barely reach the opposite surface.

I think the jury is out until some method of experimental confirmation can be done. I am currently reading up on linear algebra. I also recently bought a copy of "The Absolute Differential Calculus" by Tullio Levi-Civito. I also need to get a book on numerical relativity which I may use to develop a very simplified computer model. This may give me some further insights. As always it is good to debate Clifford. I always find your posts informative.
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#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #14 on: 13/03/2014 15:27:30 »
I have come across an online PDF titled "On the Gravitational Inverse Problem".

http://www.m-hikari.com/ams/ams-2011/ams-57-60-2011/yurtseverAMS57-60-2011.pdf

This ties in with the escape velocity issue and is worth a read through.
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#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #15 on: 27/03/2014 18:22:03 »
I have worked through an equation for the gradient of gravitational feedback down to the centre of a planet and now have to concur with Clifford that Ve is greater at the centre. At least now I have a simpler equation for the cancellation factors.
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#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #16 on: 27/03/2014 22:38:40 »
The image below shows the internal gravity well when cancellations of the gravitational field have been taken into account. This resembles the classical depiction with the rubber mat model.
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#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #17 on: 10/04/2014 10:13:26 »
One thing I forgot to add. The above graph is based on a 1kg mass as far as I remember, but its pretty low in any case. This is why escape velocity appears so low. It can be considered as a sphere of 1 metre radius containing 1kg of gas with a particle leaving the centre of gravity.
« Last Edit: 10/04/2014 10:16:10 by jeffreyH »
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#### CliffordK

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##### Re: How does escape velocity work?
« Reply #18 on: 10/04/2014 11:19:30 »
How did you get such a steep asymptote at the center of the body?

It looks like you used the equation: $$V_e = \sqrt{\frac{2GM}{r}}$$

That would be find for a point source mass.  However, for a body with volume, the gravity goes to zero as one goes subsurface towards the center of mass.

#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #19 on: 10/04/2014 12:42:29 »
How did you get such a steep asymptote at the center of the body?

It looks like you used the equation: $$V_e = \sqrt{\frac{2GM}{r}}$$

That would be find for a point source mass.  However, for a body with volume, the gravity goes to zero as one goes subsurface towards the center of mass.

That wasn't quite what I did. It is all tied up in an excel spreadsheet at the moment so I will have to pick it apart and post an equation. If it is wrong then I would appreciate any advice on correcting it. It is a modification of the logistic curve that we both posted previously. This new plot attempts to take into account the portions of the gravitational field that cancel as a particle travels out from the centre of gravity.
« Last Edit: 10/04/2014 12:44:24 by jeffreyH »
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#### jeffreyH

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##### Re: How does escape velocity work?
« Reply #20 on: 10/04/2014 13:02:33 »
Actually that is not escape velocity at all. I have no idea why it was titled that. I think I must have copied and pasted a graph and forgot to change the title and axis labels. I'm going to have to review the data. So ignore the reference to Ve. I had actually confused myself in the post following on from the graph. Referring to it as escape velocity. The posts were some time apart.
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