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What's your point? The math was applied correctly and it reflects the mainstream literature I quoted. I never claimed otherwise, nor is the discussion about that. It is about where the "spaghettification" occurs: outside the EH vs. inside.

True for freefall. Not true for the realistic case of approaching the BH in aby type of spaceship. The electromagnetic force of the ship will work against your enormous weight.

Because you would collapse under your own weight. Imagine that you weighed 10 tonnes. Your skieleton would not be able to support you.

I was trying to keep things simple, since some of you argued for using the gravitational acceleration GRADIENT instead of the acceleration proper (acceleration proper makes you into a "pancake", not a "spaghetti"), here is the more complex treatment for "spaghetti": [math]a=\frac{2GMd}{r^3}[/math] where d is the length of the object and [math]a[/math] is the acceleration gradient over the distance d. Re-doiing the calculations: [math]r=(\frac{r_sc^2d}{k*g})^{1/3}[/math] The points I made stand, depending on the characteristics of the BH spaghettification can occur inside the EH or way outside it, you need to compare [math](\frac{r_sc^2d}{k*g})^{1/3}[/math] with [math]r_s[/math] . correct

We don't know what happens inside the EH. We will NEVER know (because no information escapes from beyond the EH). Correct. To put it in math form, let's say the spaghettification occurs for an acceleration of [math]a=k*g[/math] where [math]k[/math] is a value greater than 100 and [math]g=\frac{GM}{r^2}[/math]. On the other hand, the EH is at [math]r_s=\frac{2GM}{c^2}[/math] So, spaghettification happens at a distance [math]r[/math] from the singularity: [math]k*g=\frac{r_s c^2}{2r^2}[/math] i.e. [math]r=c \sqrt{\frac{r_s}{2k*g}}[/math] It is possible to have [math]r<r_s[/math] if : [math]r_s>\frac{c^2}{2k*g}[/math] The above condition can be fulfilled for supermassive BHs. For BH with [math]r_s<\frac{c^2}{2k*g}[/math] the spaghettification occurs way outside the EH. The main lesson from this thread is that the language of physics is MATH. We need to learn to express ourselves in mathematical terms.

You completely miss the point, a BH of Sun's mass will spaghettify an object at 100 km distance.

[math]\frac{G}{c^2}=10^{-27}[/math] See the answer to problem 3

The radius of the EH (also known as the Schwarzschild radius) is extremely small (for the Sun is 3km, as an example (the Earth is 9mm). The reason is the [math]c^2[/math] in the [math]r_s=\frac{2GM}{c^2}[/math] The "G" is not helping either.

Actually, if you listened to him to the end , you would have found out that we can never find out what happened to Bob near or at the EH. Though ANY experiment. Which is exactly what I said earlier. What we DO know is that the tidal forces WAY OUT of a BH (far from the EH) spaghettify Bob. Tidal forces are REAL, they are not a coordinate "artifact".

The theory gives us the "internal solution" for the EFEs in the form of metric that is quite different from the :external solution". The metric IS the "theory", it predicts how test probes would move inside the EH. We can only say that the predictions are quite different from the ones made by the "external" solution. We will never know because we will never be able to test it. To your second question, I only know of the internal Schwarzschild solution, I do not know about the internal solutions for Kerr, Riessner, etc.

This question is unanswerable since no information can escape from beyond the EH. We cannot know what happens, we can only theorize.

There is no such thing as "the force of the Singularity", so your questions are ill-posed. Did you get the part where the "spaghettification" due to tidal forces happens OUTSIDE the event horizon?

yes

The link is fully relevant, he's trying to find "a" as a function of "b". You can't do that. This is a DIFFERENT problem. Do you see the difference?

The "ripping apart" happens BEFORE the object crosses the event horizon. A picture will help.

This is because: [latex]\frac{355}{113}=3+\frac{16}{113}=3+\frac{1}{7+\frac{1}{16}}[/latex] So, you replaced one division with two inversions (which are divisions themselves). The net effect is more error and more calculations.

You do not understand the notion of "noise". Try studying it.

Transcendental equations

It isn't "silent", there is noise in the 13 Mhz range. See here

You need to read the paper I linked, all the notions are explained on pages 3 and 4.

yes

You should kept on reading.

Yes, it has a closed solution.

It is more complicated than that. The problem has a rigorous mathematical formalism, it is known as the three-body problem

you are welcome