# How does one calculate the escape velocity starting at the center of a planet?

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#### jeffreyH

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##### How does one calculate the escape velocity starting at the center of a planet?
« on: 05/02/2014 10:05:05 »
I am looking for an equation that will always result in a value of 1 above a certain value. So if we have a range between 0-n, n will be the value above which the result will always be 1. Am I asking too much?
« Last Edit: 07/02/2014 23:10:58 by CliffordK »

#### RD

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##### Re: An impossible equation?
« Reply #1 on: 05/02/2014 10:41:18 »
A limit of a sequence could converge to 1 as n increases ...
e.g.   n sin(1/n)

Quote
n      n sin(1/n)
1       0.841471
2       0.958851
...
10      0.998334
...
100    0.999983
http://en.wikipedia.org/wiki/Limit_of_a_sequence

n=1000000 ,    n sin(1/n)=0.999999999999833

[ the formula uses sin in radians not degrees ]
« Last Edit: 05/02/2014 11:00:57 by RD »

#### CliffordK

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##### Re: An impossible equation?
« Reply #2 on: 05/02/2014 11:03:22 »
All of your limit equations approach 1, but never quite equal 1.  Is that what you want?

Did you say your equation has to equal anything other than 1?

Y = 1x

One could also have an equation like:

Y = 1 + |x| - x

If you want to shift the point where it becomes 1, then you just add in another factor:
Y = 1 + |x-n| - (x-n)

#### jeffreyH

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##### Re: An impossible equation?
« Reply #3 on: 05/02/2014 11:16:54 »
I don't think what I am asking is possible. At least not without horrendous mathematics. I have worked round the issue. The jpeg attached shows what I was doing. The tipping point is at the spherical surface. The figures used are rough estimates of values without spherical components. I am working on that part of the calculation. So don't take these as correct.
« Last Edit: 05/02/2014 11:18:50 by jeffreyH »

#### evan_au

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##### Re: An impossible equation?
« Reply #4 on: 05/02/2014 11:17:20 »
Software notation gives us a very compact notation for describing a function like this.

Depending on the computer language, it could be as simple as "(n>N)", where:
• n is an integer variable (or it could be a floating-point variable)
• N is an integer constant
• This expression returns 0 if n<=N
• This expression returns 1 if n>N

#### jeffreyH

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##### Re: An impossible equation?
« Reply #5 on: 05/02/2014 11:30:44 »
Internally this function bears a remarkable similarity to the atomic asymtotic freedom principle. The plateau effect nearing the surface is also interesting. If we take this as an analogy of the atom then electron orbitals would congregate around this plateau.
« Last Edit: 05/02/2014 11:33:16 by jeffreyH »

#### CliffordK

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##### Re: An impossible equation?
« Reply #6 on: 05/02/2014 12:01:39 »
I think Evan is right, some kind of if-then-else statement would be easiest.

My absolute value equation above will give you a nice linear progression that you could use, but it would just make all of your equations unnecessarily complicated.

As far as your escape velocity calculation, I think there is a problem.
Say you start 1000 miles below the surface of the Earth.  The gravity at the starting point is less than at the surface, but to actually escape from the planet, one has to travel up the hole to the surface, and continue through the atmosphere and into space.  Thus, starting at the bottom of the hole, the escape velocity must be greater than it would be starting at the surface.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #7 on: 05/02/2014 12:26:46 »
I think Evan is right, some kind of if-then-else statement would be easiest.

My absolute value equation above will give you a nice linear progression that you could use, but it would just make all of your equations unnecessarily complicated.

As far as your escape velocity calculation, I think there is a problem.
Say you start 1000 miles below the surface of the Earth.  The gravity at the starting point is less than at the surface, but to actually escape from the planet, one has to travel up the hole to the surface, and continue through the atmosphere and into space.  Thus, starting at the bottom of the hole, the escape velocity must be greater than it would be starting at the surface.

If gravity cancels at the centre then movement upwards is easiest surely and increases in an exponential manner from there. You have to take into account the partial gravitational cancellations all the way up to the surface. This is still a work in progress and once I have the proper spherical terms the profile may look rather different so any conclusions will have to wait. I have no idea how to test this but if true it would show that gravitation has the same properties as some of the atomic mechanisms. Of course I may also be talking rubbish.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #8 on: 05/02/2014 12:38:30 »
Take point P0 to be at the centre of gravitation. Take point P1 to be n diatance from Po. As long as n < r then it should get harder to move from P0 the nearer we get to distance n. If n > r then this situation is revered and assumes an inverse square formula. So to get to P1 from P0 will be easier and require less energy than moving from P1 to Pn. Again Pn must reside at a distance less than r from the centre.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #9 on: 05/02/2014 12:42:47 »
I think Evan is right, some kind of if-then-else statement would be easiest.

My absolute value equation above will give you a nice linear progression that you could use, but it would just make all of your equations unnecessarily complicated.

As far as your escape velocity calculation, I think there is a problem.
Say you start 1000 miles below the surface of the Earth.  The gravity at the starting point is less than at the surface, but to actually escape from the planet, one has to travel up the hole to the surface, and continue through the atmosphere and into space.  Thus, starting at the bottom of the hole, the escape velocity must be greater than it would be starting at the surface.

Sorry I see your point. This is an escape velocity gradient. This is not saying that if you start at point P within the mass there will be a constant escape velocity. Internally this is an increasing vector function requiring more input of energy the further out one travels. This makes it HARDER to escape from an internal gravity well and not easier.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #10 on: 05/02/2014 12:54:14 »
Just out of interest the function I have worked out as 1 - f(n) where n approaches zero.

#### JP

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##### Re: An impossible equation?
« Reply #11 on: 05/02/2014 13:55:25 »
The function you're looking for is the Heaviside step function H(x).  Or to get the effect you want, 1-H(x).

http://en.wikipedia.org/wiki/Heaviside_step_function

#### jeffreyH

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##### Re: An impossible equation?
« Reply #12 on: 05/02/2014 16:33:27 »
The function you're looking for is the Heaviside step function H(x).  Or to get the effect you want, 1-H(x).

http://en.wikipedia.org/wiki/Heaviside_step_function

I may yet need this so thanks for the link.

#### jeffreyH

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##### The similarities between a model of gravity and the atom
« Reply #13 on: 05/02/2014 18:53:55 »
I will be posting some results here in advance of posting a paper to a uk.arxiv.org.

#### RD

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##### Re: An impossible equation?
« Reply #14 on: 05/02/2014 19:46:59 »
« Last Edit: 05/02/2014 19:51:05 by RD »

#### jeffreyH

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##### Re: An impossible equation?
« Reply #15 on: 05/02/2014 21:07:54 »
Snap ?

http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth

The plateau profile is more rounded on the wiki graph as I haven't added the spherical geometry calculations to mine. As the wiki graph relates to density then my original assertion that mass-energy density affects gravitational feedback must have some credence.

#### JP

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##### Re: An impossible equation?
« Reply #16 on: 06/02/2014 00:07:37 »
What do you mean by gravitational feedback?  The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).

#### jeffreyH

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##### Re: An impossible equation?
« Reply #17 on: 06/02/2014 00:36:49 »
What do you mean by gravitational feedback?  The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).

I am not saying they are not negligible.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #18 on: 06/02/2014 00:38:12 »
A new graph but with the spherical geometry. This plot doesn't start from the centre of gravity. It is off centre so I need to correct that. Very unhelpfully Excel seems to continue the line.

#### JP

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##### Re: An impossible equation?
« Reply #19 on: 06/02/2014 01:09:53 »
What do you mean by gravitational feedback?  The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).

I am not saying they are not negligible.

Ah, but they are.  Unless you're planning on replacing general relativity with your own theory.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #20 on: 06/02/2014 02:25:13 »
What do you mean by gravitational feedback?  The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).

I am not saying they are not negligible.

Ah, but they are.  Unless you're planning on replacing general relativity with your own theory.

Now that would be silly.

#### JP

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##### Re: An impossible equation?
« Reply #21 on: 06/02/2014 02:34:17 »
What do you mean by gravitational feedback?  The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).

I am not saying they are not negligible.

Ah, but they are.  Unless you're planning on replacing general relativity with your own theory.

Now that would be silly.

Indeed, and it would belong in the New Theories section of the forum.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #22 on: 06/02/2014 12:31:06 »
I have just saved a PDF of the PREM results. I am going to look at the core data to see what conclusions were reached. I did notice they mentioned one unexpected result. It would be better if my calculations were able to follow the PREM results. Oh Joy more fun mathematics.
« Last Edit: 06/02/2014 12:33:52 by jeffreyH »

#### jeffreyH

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##### Re: An impossible equation?
« Reply #23 on: 06/02/2014 17:16:19 »
There was an error in the other plots. I picked up the wrong value for the Gravitational Constant. OOPS! So here is a new plot. I have zoomed in to investigate nearer the centre of gravity but that is not shown in this plot as the number of errors due to rounding made the data questionable. The pink line on the attached plot shows the point where we reach the spherical surface. Here it translates to the usual inverse square relationship. There is a sharp drop in Ve towards the centre. My hunch is this will coincide with the Schwarzschild radius. I still have to find a method to confirm this. If I find a match with rs then this should validate the model.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #24 on: 06/02/2014 17:35:36 »
This is a first attempt at a zoom with less uncertainty. The x axis shows metres from the centre of gravity.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #25 on: 06/02/2014 18:23:14 »
Now these calculations are either right or wrong. If wrong then it's back to the drawing board. If right then the striking thing is the Schwarzschild radius is invariant and always resides at the same point within a mass. If the Ve figures are right then the internal escape velocity at the central well is not c for an uncompressed mass. It can be seen from this that mass compression would have to have an amplifying effect on the gravitational field in order for this value of Ve to increase to c at the appropriate limit. For the Planck mass the radius is 2 Planck lengths. This puts the ultimate Schwarzschild radius within the elementary particles themselves. This is akin to quark confinement.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #26 on: 06/02/2014 18:44:18 »
Woo Hoo and there is the Earth's Scwarzschild gravity well.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #27 on: 06/02/2014 19:01:31 »
This also means that it is an exponential energy input function that traps objects in the black hole. You need an exponentially increasing amount of energy to make progress outward from any point past the horizon.

#### CliffordK

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##### Re: An impossible equation?
« Reply #28 on: 06/02/2014 20:45:20 »
If you have the acceleration function, then the "escape velocity" (ignoring wind resistance) should be simply an integral.

A(x) = acceleration due to gravity at distance X from the center of the Earth.

$$\int_d^\infty A(x)dx$$  for a starting point d from the center of the earth.

If you have two distinct equations, then you can simply add them together.

$$\int_d^{surface} A_{subsurface}(x)dx$$  +  $$\int_{surface}^\infty A_{above surface}(x)dx$$

Are you using some kind of a numerical approximation of the area below the curve?

I'm not sure about the magnitude, but the shape of your curves are looking a bit more like what I would expect.

#### JP

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##### Re: An impossible equation?
« Reply #29 on: 06/02/2014 21:47:11 »
Woo Hoo and there is the Earth's Scwarzschild gravity well.

It's great that you've gotten excel plots, but what exactly are you plotting?  How did you calculate this?  What equations did you use?

#### jeffreyH

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##### Re: An impossible equation?
« Reply #30 on: 06/02/2014 21:50:49 »
I won't be discussing it in the open but I will send you a copy of the paper when it's done. I also have to investigate time dilation the same way and that should be testable. At that point the model can be verified or falsified easily. I also need to take back something I said in an earlier post on this forum. I said I believed Karl Schwarzschild made a mistake in his radius calculation. I did him a disservice. He was spot on. I now can even believe in the possibility of a singularity. Unless of course I can disprove it. Now that I am above the radar I might be posting a bit less frequently.

#### JP

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##### Re: An impossible equation?
« Reply #31 on: 06/02/2014 21:52:45 »
Ok, Jeffrey.  But if you're going to be publishing plots of your own calculations without telling us what you're doing, could you please keep it to the New Theories section?  This part of the forum is for open discussion of the science behind "mainstream" ideas in physics, such as Einstein's field equations or Newtonian gravity.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #32 on: 07/02/2014 05:22:14 »
Ok, Jeffrey.  But if you're going to be publishing plots of your own calculations without telling us what you're doing, could you please keep it to the New Theories section?  This part of the forum is for open discussion of the science behind "mainstream" ideas in physics, such as Einstein's field equations or Newtonian gravity.

OK then I'll post them. I suppose here is as good as anywhere. We want to find Ve both internally and externally to a mass. So first we have Ve = f(M)/r. At the surface and beyond f(M) will simply return M as we have no internal gravitational cancellations to take into account. Internally we have f(M) = M*s/v where s/v will give us our effective mass once we have cancelled equivalent opposing field strengths. v is the total mass volume so s = (v-vs)/v. vs is our cancelled mass.    First we get the volume behind the direction of travel = v0 = (pi/6)h(3c^2+h^2).
c = SQRT(h(2r - h)). vs then equals v - 2(v - v0). I can post the spreadsheet somewhere for any interested parties. This gives the earth's Schwarzschild gravity well at just under 1 cm which agrees with the rs calculation. This shows the internal gravitation for a non compressed mass to follow a logistical curve rather than exponential. The curve will then go exponential when the rs compression limit is reached.

The values for h and c can be found at this page.

http://keisan.casio.com/exec/system/1223382199
« Last Edit: 07/02/2014 06:49:16 by jeffreyH »

#### jeffreyH

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##### Re: An impossible equation?
« Reply #33 on: 07/02/2014 06:47:39 »
I've just done a time dilation calculation for the earth and for 99.99 % of the radius from the surface to the core there is less than 1 microsecond's worth of time dilation. Not easily testable which is disappointing. This is why planets never ignite as stars do.

Correction there are 8 microseconds difference.
« Last Edit: 07/02/2014 07:25:50 by jeffreyH »

#### evan_au

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##### Re: An impossible equation?
« Reply #34 on: 07/02/2014 08:10:30 »
The latter graph seems to show the escape velocity decreasing when you reach 10m from the center of the Earth.

Assuming a spherically symmetric Earth, the gravitational force would be the attraction of a 10m radius sphere (possibly with a significant content of gold, osmium or other dense materials). You would be effectively weightless (if you weren't instantly crushed by thousands of kilometers of iron and rock).

This 10m radius sphere does not have enough density to form a black hole.

I think the graphed reduction of escape velocity near the center of the Earth is not  a real effect - I think it might be some sort of underflow error.
« Last Edit: 07/02/2014 21:04:54 by evan_au »

#### jeffreyH

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##### Re: An impossible equation?
« Reply #35 on: 07/02/2014 08:40:08 »
The latter graph seems to show the escape velocity decreasing when you reach 10m from the center of the Earth.

Assuming a spherically symmetric Earth, the gravitational force would be the attraction of a 10m radius sphere (possibly with a significant content of gold, osmium or other dense materials). You would be effectively weightless (if you weren't instantly crushed by thousands of kilometers of iron and rock).

This 10m radius sphere does not have enough density to form a black hole.

I think the graphed reduction of escape velocity near the center of the Earth is a real effect - I think it might be some sort of underflow error.

Hi Evan This is a gradient equation. What it says is at any point P you can find the velocity needed to escape the gravitational field. It shouldn't be viewed as a representation of a 10m mass. This is 10m from the centre of the earth. The well at the centre is not an error and matches the event horizon if and only if the mass of the earth were concentrated at that point which it isn't in this equation. This iss a view zoomed in to the core. The mass is at normal density.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #36 on: 07/02/2014 18:38:20 »
I am going to modify the model to use the sun's mass and radius. I want to see if the Scwarzschild radius matches for that. If so I think I can be confident that the model is actually showing a gravity well at the centre and not a mathematical error. Can I attach zip files here? I want to post a couple of Excel spreadsheets. Otherwise I'll post them on a website.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #37 on: 07/02/2014 18:43:49 »
Can someone help with a time dilation equation with respect to escape velocity. I have thought since t = 1/SQRT(c^2/c^2) would be 1 second then 1/SQRT((c-Ve)^2/C^2) should then give me the dilation for a particular point within a mass. Or would it be g rather than Ve?
« Last Edit: 07/02/2014 18:45:37 by jeffreyH »

#### jeffreyH

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##### Re: An impossible equation?
« Reply #38 on: 07/02/2014 19:05:37 »
The central well looks like an artifact rather than real data. For the sun the Ve at the core is given as 756708 m/s. Where on earth can I check this value??

#### jeffreyH

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##### Re: An impossible equation?
« Reply #39 on: 07/02/2014 19:48:05 »
I have just modified the model to plot g down to the core and it is very interesting. The value of g at the earth's surface is 9.819297 so I know it works there but take a look at what happens internally. The image of the Excel plot is attached. This in combination with escape velocity would explain a lot.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #40 on: 07/02/2014 20:04:38 »
Notice that the g force at the centre of gravity is equivalent to escape velocity at the surface. Maybe I am doing something bizarrely wrong. At this point I just don't know.

#### alancalverd

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##### Re: An impossible equation?
« Reply #41 on: 07/02/2014 21:39:43 »
Escape speed  at a distance r from the center of mass of a sphere is

se= sqrt (2gr)

where g is the gravitational acceleration at r. At the center of a sphere both g and r are zero, and increase smoothly (if the sphere is homogeneous) to the surface. For the nonhomogeneities of the earth, see the plot of g versus r given in the Wikipedia reference, which takes account of the depth/density profile of the planet.
helping to stem the tide of ignorance

#### jeffreyH

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##### Re: An impossible equation?
« Reply #42 on: 07/02/2014 22:48:48 »
Escape speed  at a distance r from the center of mass of a sphere is

se= sqrt (2gr)

where g is the gravitational acceleration at r. At the center of a sphere both g and r are zero, and increase smoothly (if the sphere is homogeneous) to the surface. For the nonhomogeneities of the earth, see the plot of g versus r given in the Wikipedia reference, which takes account of the depth/density profile of the planet.

Have you a link for the wikipedia reference. I can't find it?

#### CliffordK

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##### Re: An impossible equation?
« Reply #43 on: 07/02/2014 22:54:46 »
RD posted a couple of models for the internal gravity of the Earth.  The gravity should go to zero at the middle of the body.

Were you looking at theoretical singularities with all the mass concentrated at a single point?  That is a completely different beast than a planet with the mass surrounding a person as one bores towards the center.

Your escape velocity should always increase as you move towards the center of the body.  I.E.  no jagged lines as you have on some of your graphs.

I threw this chart together last night.
Acceleration Due to Gravity.
I drew a linear line from the center of Earth to the surface, which I believe is indicative of a constant density of Earth.  Current theories indicate an iron core, and greater density in the center of Earth, but for the purposes of my estimate, this was good enough.

[attachment=18468]

Equations:  (d is the distance to the center of Earth).
From center to surface:
Acceleration due to gravity is:  9.8*d/(6371 km)

From the surface into space,
Acceleration due to gravity is:

$$6.67x10^{-11}\frac{m^3}{kg s^2}\frac{5.972 x 10^{24} kg}{(d m)^2}$$

I'll try to convert my "acceleration" to a "velocity" soon.

#### jeffreyH

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##### Re: An impossible equation?
« Reply #44 on: 07/02/2014 23:09:41 »
RD posted a couple of models for the internal gravity of the Earth.  The gravity should go to zero at the middle of the body.

Were you looking at theoretical singularities with all the mass concentrated at a single point?  That is a completely different beast than a planet with the mass surrounding a person as one bores towards the center.

Your escape velocity should always increase as you move towards the center of the body.  I.E.  no jagged lines as you have on some of your graphs.

I threw this chart together last night.
Acceleration Due to Gravity.
I drew a linear line from the center of Earth to the surface, which I believe is indicative of a constant density of Earth.  Current theories indicate an iron core, and greater density in the center of Earth, but for the purposes of my estimate, this was good enough.

[attachment=18468]

Equations:  (d is the distance to the center of Earth).
From center to surface:
Acceleration due to gravity is:  9.8*d/(6371 km)

From the surface into space,
Acceleration due to gravity is:

$$6.67x10^{-11}\frac{m^3}{kg s^2}\frac{5.972 x 10^{24} kg}{(d m)^2}$$

I'll try to convert my "acceleration" to a "velocity" soon.

I appreciate that. My view is that gravitational cancellation is a gradient. If it all cancels out at the centre then this cancellation should decrease going outward but it is not a straight line. It can't be. The straight line approach makes no sense when the external field is inverse square. The cancellation must describe a curve with a particular functional description.

Moreover I believe that the current view of internal forces is inadequate. The only way to say which position is true is via experimentation. This is a thorny problem.
« Last Edit: 07/02/2014 23:13:52 by jeffreyH »

#### CliffordK

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##### Re: How does one calculate the escape velocity starting at the center of a planet?
« Reply #45 on: 07/02/2014 23:44:34 »
Technically you can divide your sphere into cubes, and calculate the acceleration due to gravity to each sub-cube, or perhaps one could do it with concentric rings or shells.

According to RD's Notes, the linear line represents the acceleration due to gravity if the planet or body has a constant density.

Anyway, it should be good enough for a crude estimate.  Otherwise one would need well defined acceleration or density functions.

#### jeffreyH

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##### Re: How does one calculate the escape velocity starting at the center of a planet?
« Reply #46 on: 07/02/2014 23:54:00 »
Technically you can divide your sphere into cubes, and calculate the acceleration due to gravity to each sub-cube, or perhaps one could do it with concentric rings or shells.

According to RD's Notes, the linear line represents the acceleration due to gravity if the planet or body has a constant density.

Anyway, it should be good enough for a crude estimate.  Otherwise one would need well defined acceleration or density functions.

I'm just thinking how many times in physics things have been counter-intuitive. Heisenberg's uncertainty principle, general relativity etc. I think g will increase with depth and fall sharply at the very core. I know this runs counter to common sense but this is exactly the point. An experiment to determine g in a deep mine shaft would resolve the issue. If it decreases linearly then my position is wrong. If it increases, however, we have a whole new view of gravity that may aid in a quantum theory of gravitation. Isn't that worth determining. I was as surprised as anyone when the figures went the wrong way but it makes sense when you think about partial field cancellations.

#### jeffreyH

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• The graviton sucks
##### Re: How does one calculate the escape velocity starting at the center of a planet?
« Reply #47 on: 08/02/2014 00:16:29 »
Now this is interesting.

http://davidpratt.info/gravity.htm
" In 1981 a paper was published showing that measurements of G in deep mines, boreholes, and under the sea gave values about 1% higher than that currently accepted.4 Furthermore, the deeper the experiment, the greater the discrepancy. However, no one took much notice of these results until 1986, when E. Fischbach and his colleagues reanalyzed the data from a series of experiments by Eötvös in the 1920s, which were supposed to have shown that gravitational acceleration is independent of the mass or composition of the attracted body. Fischbach et al. found that there was a consistent anomaly hidden in the data that had been dismissed as random error. On the basis of these laboratory results and the observations from mines, they announced that they had found evidence of a short-range, composition-dependent ‘fifth force’. Their paper caused a great deal of controversy and generated a flurry of experimental activity in physics laboratories around the world.5"

Now does this 1% increase agree with my model? That is the interesting question.

"As mentioned above, measurements of gravity below the earth’s surface are consistently higher than predicted on the basis of Newton’s theory.13 Sceptics simply assume that hidden rocks of unusually high density must be present. However, measurements in mines where densities are very well known have given the same anomalous results, as have measurements to a depth of 1673 metres in a homogenous ice sheet in Greenland, well above the underlying rock. Harold Aspden points out that in some of these experiments Faraday cage-type enclosures are placed around the two metal spheres for electrical screening purposes. He argues that this could result in electric charge being induced and held on the spheres, which in turn could induce ‘vacuum’ (or rather ether) spin, producing an influx of ether energy that is shed as excess heat, resulting in errors of 1 or 2% in measurements of G.14"

#### JP

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##### Re: How does one calculate the escape velocity starting at the center of a planet?
« Reply #48 on: 08/02/2014 00:59:43 »
Jeffrey, while I appreciate that you're discussing gravity and making calculations using your model, we have a strict policy here that we don't allow new models to be proposed and discussed outside of the New Theories sub-forum.  A major reason for this is that this is mostly a science Q&A board and users who come here seeking answers (without any scientific background) will find the forum very confusing if new theories are mingled in with accepted theories.

I'd like to ask that you keep this thread to asking questions about Newtonian gravity, general relativity and perhaps some questions about peer-reviewed theories of quantum gravity.  Please keep posts discussing your own models and their results to a thread in the New Theories section.

Thanks!

#### jeffreyH

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##### Re: How does one calculate the escape velocity starting at the center of a planet?
« Reply #49 on: 08/02/2014 01:14:38 »
Jeffrey, while I appreciate that you're discussing gravity and making calculations using your model, we have a strict policy here that we don't allow new models to be proposed and discussed outside of the New Theories sub-forum.  A major reason for this is that this is mostly a science Q&A board and users who come here seeking answers (without any scientific background) will find the forum very confusing if new theories are mingled in with accepted theories.

I'd like to ask that you keep this thread to asking questions about Newtonian gravity, general relativity and perhaps some questions about peer-reviewed theories of quantum gravity.  Please keep posts discussing your own models and their results to a thread in the New Theories section.

Thanks!

Is it possible for this thread to be moved into new theories?