# Gravity for a stellar black hole

• stevebd1
In summary: What it reads is what it reads. It's up to you to interpret it. If you want to interpret it as a reading of g, you have to use the formula I gave. If you want to interpret it as a measurement of the "coordinate acceleration", you have to use the formula I gave. If you want to interpret it as an indication of which way is "down", then you have to "parallel transport" it to the surface of the earth, and then see which way is "down" there.
stevebd1
Gold Member
I'm currently looking at the various gravity gradients (tidal stresses) for black holes and I've realized that in respect of small stellar black holes, as you approach the event horizon, not only does the gravity increase but it (appears) to exceed c. For example, for a 3 sol black hole, the gravity at the event horizon, using GM/r^2, works out at 5.0845x10^12 m/s^2. With the notion that this is the rate of acceleration in every second, is it acceptable to use a figure like this to express something which is considered to travel as the same as c? I also noticed this only occurs with neutron stars and small black holes (and any other compact star that might exist in between). Doing some research, it seems acceptable to state that the surface gravity for neutron stars can range from 2x10^11 to 3x10^12 Earth g (which are in the same ball park). Also, working backwards, you can work out at what point the gravity exceeds 299,792,500 m/s^2, r = (GM/c)^0.5, for a 3 sol black hole this would be at 1152.2 km radius. Does this radius denote something significant? I'd appreciate any feedback.

regards
Steve

P.S. the gravity gradient at the event horizon for a 3 sol mass black hole increases at 1.1490x10^9 m/s^2 per metre which in itself appears to be superluminal (I'm aware that any object within a gravity field of this magnitude would not exceed c but be dragged by the gravity field to speeds very close to c). I'm also aware there are theories that propose that the speed of gravity exceeds c but this seems to be met with considerable skepticism.

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Hopefully I'm not being bullish as I understand that not all questions are answered straight away but I'm only looking for a simple response (or maybe there isn't one) to the above, such as 1) I've missed something out in the calculations, 2) This is nothing to be concerned about, it's just the way it is, 3) There's something theoretical I'm missing. I'd appreciate any feedback.

Steve

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stevebd1 said:
I'm currently looking at the various gravity gradients (tidal stresses) for black holes and I've realized that in respect of small stellar black holes, as you approach the event horizon, not only does the gravity increase but it (appears) to exceed c. For example, for a 3 sol black hole, the gravity at the event horizon, using GM/r^2, works out at 5.0845x10^12 m/s^2.

I'm not sure how you got GM/r^2. Are you using Newtonian gravity? Or are you using coordinate acceleration according to general relativity of a hovering observer?

Are you saying that acceleration exceeds c? The magnitude of acceleration can be any non-negative real number.

Hi George and thanks for the reply. I'm using Newtonian gravity which seems to be an acceptable solution according to most sources I look at, correctly it would be expressed as

$$g = G\frac{m}{r^{2}}$$

For earth, this (obviously) means a gravity acceleration of 9.8207 m/s^2 at sea level (based on a mean Earth radius of 6,371,000 m and a mass of 5.9736x10^24 kg), an acceleration of 9.79 m/s^2 at 8850 m, the top of Everest (r would equal 6,379,850 m) and 8.85 m/s^2 at 340 km above sea level, the orbit of the ISS (r would equal 6,711,000 m).

Using this same equation and principle for a black hole with a mass of 3 suns (5.9673x10^30 kg) and a Schwarzschild radius of 8861 m, based on the Schwarzschild equation-

$$R_{s}=\frac{2GM}{c^{2}}$$

Using the Newtonian equation to get an idea of the gravity at the event horizon, I get 5.0715x10^12 m/s^2, which tells me for every second an object is falling, it should increase in velocity by 5.0715x10^12 m/s, which is obviously superluminal. I'm not expecting that matter would accelerate at this rate but I just seem to be having a problem with the fact that it is acceptable to produce a figure for gravity that suggests acceleration in excess of the speed of light when it's common knowledge that gravity itself travels at the speed of light.

Steve

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stevebd1 said:
Hi George and thanks for the reply. I'm using Newtonian gravity which seems to be an acceptable solution according to most sources I look at, correctly it would be expressed as

$$g = G\frac{m}{r^2}$$

Could you give a reference either to a URL or a book?

I don't think that Newtonian gravity is acceptable here. The GR (Schwarzschild) coordinate acceleration has the same expression as Newtonian acceleration, but an accelerometer carried by a hovering observer would actually read

$$\left(1 - \frac{2Gm}{c^2 r} \right)^{-\frac{1}{2}} G\frac{m}{r^2}$$

This makes things worse, and diverges at the event horizon!

Using the Newtonian equation to get an idea of the gravity at the event horizon, I get 5.0715x10^12 m/s^2, which tells me for every second an object is falling, it should increase in velocity by 5.0715x10^12 m/s, which is obviously superluminal.

In relativity, velocities no longer add in the usual way. For example, if you zip off to my left at a speed of 3c/4 with respect to me, and Bob zips off to my left at a speed of 3c/4 with respect to me, you see bob zipping off to your right with a speed 24c/25, not 3c/4 + 3c/4 = 3c/2.

If, with respect to me, you start from rest, and then you undergo acceleration $a$ for a time $\tau$ as measured by your watch, then you observe our relative speed to be

$$v = c \tanh \left(\frac{a \tau}{c} \right).$$

This never exceeds $c$ no matter how large $\tau$ is. For $x \ll 1$, $\tanh x \doteq x$, which gives that $v \doteq a \tau$ when $a \tau \ll c$.

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Thanks George, I see the addition of the second part to the Newtonian equation is the way to go. When you say 'accelerometer carried by a hovering observer would actually read' are you implying that this is the gravity as it would be read and experienced real time by an observer? (as apposed to say redshifted light which has been stretched and isn't an accurate representation of itself as it was when it started its journey). The gravity read by the accelerometer is the gravity as it is (as the equation) and not dilated in some way.

One thing I did find while looking on the web was a slightly different expression of the equation you've provided-

http://astro.pas.rochester.edu/~aquillen/ast142/Lecture/Lecture9.pdf pages 18

$$a=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-2GM/rc^{2}}}$$

This implies that gravity blows up towards infinity at the event horizon (as also stated in the paper). This is quite a difference from what the equation you've posted suggests which is that the gravity drops to 0 and inverts at the event horizon. Are there 2 different theories as to what happens at the event horizon in respect of gravity.

Would you also apply this addition to the Newtonian gravity equation for neutron stars and other compact stars also.

cheers
Steve

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stevebd1 said:
Thanks George, I see the addition of the second part to the Newtonian equation is the way to go. When you say 'accelerometer carried by a hovering observer would actually read' are you implying that this is the gravity as it would be read and experienced real time by an observer?

Yes, this is the "g-force" that the hovering observer would feel.

One thing I did find while looking on the web was a slightly different expression of the equation you've provided-

http://astro.pas.rochester.edu/~aquillen/ast142/Lecture/Lecture9.pdf pages 18

$$a=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-2GM/rc^{2}}}$$

This implies that gravity blows up towards infinity at the event horizon (as also stated in the paper). This is quite a difference from what the equation you've posted suggests which is that the gravity drops to 0 and inverts at the event horizon. Are there 2 different theories as to what happens at the event horizon in respect of gravity.

The two formulas are exactly the same.

I see now how the negative power works, thanks. I've noticed that this works perfectly with a Schwarzschild black hole but when applied to a rotating Kerr black hole, the infinite gravity occurs at the edge of the ergosphere (or the static limit) in the equatorial plane (where the Schwarzschild radius would be if it were a static black hole) and the event horizon is further back. Also, because the ergosphere is a flattened sphere, the infinite gravity would occur somewhere outside the event horizon/ergosphere at the poles (see attached picture). Is this the norm or is there a separate equation or a modification required for a Kerr black hole?

Steve

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I was looking for surface gravity for a Kerr black hole on the web and happened upon a section on wiki which made reference to something called the Killing horizon (named after Wilhelm Killing) that had the following equation-

$$k = \frac{\sqrt{M^{2}-Q^{2}-J^{2}/M^{2}}}{2M^{2}-Q^{2}+2M\sqrt{M^{2}-Q^{2}-J^{2}/M^{2}}}$$

Based on a 3 sol mass black hole with a rotation factor of 0.8, M = 5.9673x10^30 and J = 6.3408x10^42 and considering the charge (Q) as zero, the answer comes out at 4.196276x10^-32. What significance does this figure have?

Oddly enough, when I calculate k using the more simple Schwarzschild solution, I get a very similar answer of 4.1897x10^-32

$$k = \frac{1}{4M}$$

Again, what is the significance of this number (from what I understand, it appears to be a geometric number) and how might it relate to surface gravity of a black hole. I'd appreciate any feedback.

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

Steve

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Attached is an image I put together comparing a 3 sol mass static black hole and a 3 sol mass rotating black hole (a = 0.95). I've used the modified Newtonian equation to calculate gravity for both black holes which results in gravity reaching infinite at the Schwarzschild radius for both types (the Schwarzschild radius is only apparent on the rotating black hole but the point of infinite gravity is the same for both models). I'd appreciate any feedback regarding the properties of gravity on the rotating black hole between the Schwarzschild radius and the outer event horizon (areas marked with G?).

Equations for rotating black hole-

Event horizon-

$$R+=(GM/c^{2}) (1+\sqrt{(1-a^{2})})$$

Cauchy horizon-

$$R+=(GM/c^{2}) (1-\sqrt{(1-a^{2})})$$

Static limit-

$$Re=(GM/c^{2}) (1+\sqrt{(1-a^{2}\,sin^{2}\theta)})\,\,\,\,\,\,\,\,\theta=latitude$$

(equatorial edge equal to Schwarzschild radius, polar edge equal to event horizon)

Photon sphere-

$$Rps=3GM/c^{2}$$

cheers
Steve

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stevebd1 said:
Hi George and thanks for the reply. I'm using Newtonian gravity which seems to be an acceptable solution according to most sources I look at, correctly it would be expressed as

$$g = G\frac{m}{r^{2}}$$

For earth, this (obviously) means a gravity acceleration of 9.8207 m/s^2 at sea level (based on a mean Earth radius of 6,371,000 m and a mass of 5.9736x10^24 kg), an acceleration of 9.79 m/s^2 at 8850 m, the top of Everest (r would equal 6,379,850 m) and 8.85 m/s^2 at 340 km above sea level, the orbit of the ISS (r would equal 6,711,000 m).

Using this same equation and principle for a black hole with a mass of 3 suns (5.9673x10^30 kg) and a Schwarzschild radius of 8861 m, based on the Schwarzschild equation-

$$R_{s}=\frac{2GM}{c^{2}}$$

Using the Newtonian equation to get an idea of the gravity at the event horizon, I get 5.0715x10^12 m/s^2, which tells me for every second an object is falling, it should increase in velocity by 5.0715x10^12 m/s, which is obviously superluminal. I'm not expecting that matter would accelerate at this rate but I just seem to be having a problem with the fact that it is acceptable to produce a figure for gravity that suggests acceleration in excess of the speed of light when it's common knowledge that gravity itself travels at the speed of light.

Steve
according to ur calculations anybody traveling near the black hole vil tare apart due to its tidal forces and the remaining debris vil revolve around it.if its gravitational accleration is more than c then its tidal force vil tremendously increase therefore increasing its roche limit.so it would bcome dangerous as a black hole vil show a vicinity of 10 to 9 mpc.therefore v should get explotions near from a black hole which v don't get.

George Jones said:
Yes, this is the "g-force" that the hovering observer would feel.

The two formulas are exactly the same.

according to stevebd1 a body vil fall inside a black hole vit a velocity more than c.then it vil have increasing mass and finally its mass vil reach infinity. how can it b possible that a body having infinite mass cannot escape from a black hole ?

I've realized that the equation that modifies the Newtonian equation for gravity is based on the Schwarzschild radius which is for a static black hole, if I modify it to incorporate the the Kerr equation for the outer event horizon I get-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-\frac{GM(1+\sqrt{(1-a^{2})})}{rc^{2}}}}$$

Basically I replaced the 2 with (1+(1-a^2)^1/2), the equation that dictates the outer event horizon for a rotating black hole (while the 2 dictated the event horizon for a static black hole). The equation works for calculating gravity in close proximity to rotating black holes, taking into account the reduced event horizon and keeping the infinite gravity at the event horizon.

The source for the original rotating black hole equations-
http://www.engr.mun.ca/~ggeorge/astron/blackholes.html

If anyone sees anything at odds with the equation, I'd appreciate it if you could let me know.

regards
Steve

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It appears common practice for models of black holes that gravity diverges (or reaches infinity) at the event horizon and the tidal forces (or gravity gradients) continue to the singularity where they reach infinity. I've looked extensively on the web and these forums and haven't been able to answer the following questions-

How is it possible to have a gravity gradient within the event horizon when gravity already reached infinity at the event horizon?

How crucial is it that gravity diverges (reaches infinity) at the event horizon? Is it related in some way to the way light is effected by gravity? Would it not be a reasonable proposal to establish infinity gravity (or divergence) at the singularity?

I'd appreciate any comments or feedback.

regards
Steve

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stevebd1 said:
It appears common practice for models of black holes that gravity diverges (or reaches infinity) at the event horizon and the tidal forces (or gravity gradients) continue to the singularity where they reach infinity. I've looked extensively on the web and these forums and haven't been able to answer the following questions-

How is it possible to have a gravity gradient within the event horizon when gravity already reached infinity at the event horizon?

How crucial is it that gravity diverges (reaches infinity) at the event horizon? Is it related in some way to the way light is effected by gravity? Would it not be a reasonable proposal to establish infinity gravity (or divergence) at the singularity?

I'd appreciate any comments or feedback.

regards
Steve
i still don't get it dat how can gravity reach infinity at the event horizon. i accept that gravity and the tidal forces reach infinity at the singularity but how can it be infinite at the event horizon. if the tidal forces are infinite at the event horizon then any matter even near to the horizon would tare apart into small moleculesbefore falling inside . and if v consider gravity to be infinite the the mass of the blakhole(event horizon) can reach infinity and would tare the fabric of the universe bending it to an infinite point.

The following is just an observation and hopefully acceptable within Physics Forums terms of use-

I was looking at the GR (Schwarzschild) coordinate acceleration equation while considering how gravity might be calculated (or estimated to some degree) beyond the event horizon. Keeping in mind that space and time are suppose to flip (or invert) at the event horizon, I looked at the option of inverting the 2GM/rc^2 part of the equation in order to maintain the flow of gravity and this produced results shown plotted in the attached document. While this isn't strictly speaking gravity as we know it, it does seem to collaborate with the curve outside the event horizon and the tidal forces.

equation for gravitational acceleration-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-2GM/rc^{2}}}$$

Based on the principle that the original equation diverges (or inverts) at the event horizon, for gravity beyond the event horizon the following might apply-

$$\frac{2GM}{rc^{2}}$$ inverts to become $$\frac{1}{2GM/rc^{2}}$$ or $$\frac{rc^{2}}{2GM}$$

This provides the following equation-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-rc^{2}/2GM}}$$

Using the revised equation beyond the event horizon, gravity appears to pull back from infinity, picks up more or less on the same curve it had previously been following, then increases to infinity again as it approaches the singularity.

The equation may be more of a geometric curiosity but it does appear to provide worthwhile information.

regards
Steve

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stevebd1 said:
The following is just an observation and hopefully acceptable within Physics Forums terms of use-

I was looking at the GR (Schwarzschild) coordinate acceleration equation while considering how gravity might be calculated (or estimated to some degree) beyond the event horizon. Keeping in mind that space and time are suppose to flip (or invert) at the event horizon, I looked at the option of inverting the 2GM/rc^2 part of the equation in order to maintain the flow of gravity and this produced results shown plotted in the attached document. While this isn't strictly speaking gravity as we know it, it does seem to collaborate with the curve outside the event horizon and the tidal forces.

equation for gravitational acceleration-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-2GM/rc^{2}}}$$

Based on the principle that the original equation diverges (or inverts) at the event horizon, for gravity beyond the event horizon the following might apply-

$$\frac{2GM}{rc^{2}}$$ inverts to become $$\frac{1}{2GM/rc^{2}}$$ or $$\frac{rc^{2}}{2GM}$$

This provides the following equation-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-rc^{2}/2GM}}$$

Using the revised equation beyond the event horizon, gravity appears to pull back from infinity, picks up more or less on the same curve it had previously been following, then increases to infinity again as it approaches the singularity.

The equation may be more of a geometric curiosity but it does appear to provide worthwhile information.

regards
Steve

i can help u in it with an example. consider the black hole is a centrifuge.any object falling in a centrifuge acclerates and finally comes to the centre. consider the same for the black hole where u will get that the accleration gradient of a body keeps on increasing once it gets into the event horizon. so simply get the relation between the accleration gradient and the gravitational accleration considering that gravity is not infinite. i got a relation that mass is directly proportional to event horizon, while it is also directly proportional to the gravitational accleration of the black hole. using the law of proportionality we say that event horizon is directly proportional to gravitational accleration. therefore g$$\proptor.$$

pseudo said:
i can help u in it with an example. consider the black hole is a centrifuge.any object falling in a centrifuge acclerates and finally comes to the centre. consider the same for the black hole where u will get that the accleration gradient of a body keeps on increasing once it gets into the event horizon. so simply get the relation between the accleration gradient and the gravitational accleration considering that gravity is not infinite. i got a relation that mass is directly proportional to event horizon, while it is also directly proportional to the gravitational accleration of the black hole. using the law of proportionality we say that event horizon is directly proportional to gravitational accleration. therefore g\propto r

pseudo said:
pseudo said:
i can help u in it with an example. consider the black hole is a centrifuge.any object falling in a centrifuge acclerates and finally comes to the centre. consider the same for the black hole where u will get that the accleration gradient of a body keeps on increasing once it gets into the event horizon. so simply get the relation between the accleration gradient and the gravitational accleration considering that gravity is not infinite. i got a relation that mass is directly proportional to event horizon, while it is also directly proportional to the gravitational accleration of the black hole. using the law of proportionality we say that event horizon is directly proportional to gravitational accleration. therefore $$\proptopgr$$

pseudo said:
pseudo said:
pseudo said:
i can help u in it with an example. consider the black hole is a centrifuge.any object falling in a centrifuge acclerates and finally comes to the centre. consider the same for the black hole where u will get that the accleration gradient of a body keeps on increasing once it gets into the event horizon. so simply get the relation between the accleration gradient and the gravitational accleration considering that gravity is not infinite. i got a relation that mass is directly proportional to event horizon, while it is also directly proportional to the gravitational accleration of the black hole. using the law of proportionality we say that event horizon is directly proportional to gravitational accleration. therefore g $$\proptop$$r

pseudo said:
pseudo said:
pseudo said:
pseudo said:
i can help u in it with an example. consider the black hole is a centrifuge.any object falling in a centrifuge acclerates and finally comes to the centre. consider the same for the black hole where u will get that the accleration gradient of a body keeps on increasing once it gets into the event horizon. so simply get the relation between the accleration gradient and the gravitational accleration considering that gravity is not infinite. i got a relation that mass is directly proportional to event horizon, while it is also directly proportional to the gravitational accleration of the black hole. using the law of proportionality we say that event horizon is directly proportional to gravitational accleration. therefore [g]\proptop[/r]
. g=kr

I appreciate the response Pseudo though I can't fully appreciate your equation. It's also unclear as to whether you're saying you allow for gravity reaching infinite at the event horizon or not. Personally I think it's essential that what ever calculations are made, gravity approaching infinite at the event horizon needs to be included for as it works with current mathematics and seems to signify a change in the quality of space (supposedly space and time swapping properties at the event horizon). I'd be interested to see an elaboration of the equation you've supplied, g = kr.

regards
Steve

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stevebd1 said:
I appreciate the response Pseudo though I can't fully appreciate your equation. It's also unclear as to whether you're saying you allow for gravity reaching infinite at the event horizon or not. Personally I think it's essential that what ever calculations are made, gravity approaching infinite at the event horizon needs to be included for as it works with current mathematics and seems to signify a change in the quality of space (supposedly space and time swapping properties at the event horizon). I'd be interested to see an elaboration of the equation you've supplied, g = kr.

regards
Steve

i worked on the dimensions of my equation ,g=kr(considering no dimensions for k) and got a new physical quantity called as accleration gradient. so the equation bcomes g=kr/a' where a' is the accleration gradient i.e dr/da which decreases with radius(event horizon) and also increases with it. but the accleration is inversly proportional to the radius.afterall m not considering gravity as infinite on the event horizon. can u help me with the correct elaboration of my equation or i shall stop thinking about it coz it is gettimg hypothetical. HELP !

Could you provide a working example?

regards
Steve

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stevebd1 said:
Could you provide a working example?

regards
Steve

consider a centrifuge. at the circumference the radius is maximum. if the radius is considered to be rotated at high speed then the accleration of the particle at the edge(CIRCUMFERENCE) is maximum and it reduces as radius reduces towards the centre. i.e the particle undergoes retardation towards the centre.

Thanks for the response. It appears the equation deals with a rotating black hole only. While it's likely that most black holes are rotating (Kerr black holes) due to that fact that most stars rotate, I still feel that what ever occurs with gravity at the event horizon and beyond happens to both a rotating (Kerr) black hole and static (Schwarzschild) black hole regardless of rotation. There may be some way of factoring in retardation for a rotating black hole but there would be an equation, in principle, that would apply to both in respect of gravity.

regards
Steve

Gravity well for a 3 sol mass rotating black hole, a = 0.95

I kept the following basic in order to keep the latex to a minimum. Hopefully it makes sense (it appears to make sense on the graph attached). I've kept my justification to a minimum but will probably elaberate on this later.

Equation for gravitational acceleration up to the event horizon of a static black hole-

$$g=\frac{GM}{r^{2}}\left(1-\frac{2GM}{rc^{2}}\right)^{-\frac{1}{2}}$$

which can be expressed as

$$g=\frac{GM}{r^{2}}\left(1-\frac{R_{s}}{R}\right)^{-\frac{1}{2}}$$

where Rs is the Schawartzschild radius- $$R_{s}=2GM/c^{2}$$

Following equations established to obtain a simple estimation of gravity acceleration-

Gravitational acceleration past the event horizon of a static black hole (based on divergence at EH and gravity and space swapping properies)-

$$g=\frac{GM}{r^{2}}\left(1-\frac{rc^{2}}{2GM}\right)^{-\frac{1}{2}}$$

which can be expressed as

$$g=\frac{GM}{r^{2}}\left(1-\frac{R}{R_{s}}\right)^{-\frac{1}{2}}$$

Attached is a graph for a 3 sol mass rotating black hole (a = 0.95) put together using the equations below-

Gravitational acceleration up to the event horizon for a rotating black hole (where a represents angular momentum, 0-1)-

$$g=\frac{GM}{r^{2}}\left(1-\frac{GM(1+\sqrt{(1-a^{2})})}{rc^{2}}\right)^{-\frac{1}{2}}$$

which can be expressed as-

$$g=\frac{GM}{r^{2}}\left(1-\frac{R_{+}}{R}\right)^{-\frac{1}{2}}$$

where R+ is the event horizon- $$R_{+}=(GM/c^{2}) (1+\sqrt{(1-a^{2})})$$

Gravitational acceleration between cauchy horizon and event horizon of a rotating black hole-

$$g=\frac{GM}{r^{2}}\left(\left(1-\frac{R}{R_{+}}\right)^{-\frac{1}{2}}+\left(1-\frac{R_{-}}{R}\right)^{-\frac{1}{2}}\right)$$

where R_ is the Cauchy horizon- $$R_{-}=(GM/c^{2}) (1-\sqrt{(1-a^{2})})$$

Gravitational acceleration within Cauchy horizon up to ring singularity-

$$g=\frac{GM}{r^{2}}\left(1-\frac{R}{R_{-}}\right)^{-\frac{1}{2}}$$

It could be said that the gravity acceleration peaks of infinity signify a 'flip' as the properties of gravity and space swap properties.

Steve

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## 1. What is a stellar black hole?

A stellar black hole is a type of black hole that forms from the collapse of a massive star. It is an object with such strong gravitational force that even light cannot escape from it.

## 2. How does gravity work for a stellar black hole?

The gravity for a stellar black hole is extremely strong due to its massive size and density. It is said to have an infinite amount of gravity and anything that gets too close will be pulled in and never able to escape.

## 3. How is gravity different for a stellar black hole compared to other objects?

The gravity for a stellar black hole is much stronger than any other object in the universe. This is because it has a higher mass and density, which creates a stronger gravitational pull.

## 4. How is gravity measured for a stellar black hole?

The gravity of a stellar black hole is measured by observing the effects it has on its surroundings. This can include the orbits of nearby stars, the bending of light, and the emission of radiation from its accretion disk.

## 5. Can anything escape from the gravity of a stellar black hole?

No, once an object or matter gets too close to the event horizon of a stellar black hole, it cannot escape its gravitational pull. This is why it is often referred to as the "point of no return".

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