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Using the Schwarzschild radius formula, I approximated the Sun's Schwarzschild radius to be $3\text{ km}$. Now assuming I have a body (not a human body) which is at a distance of $10\text{ km}$ from this hypothetical black hole. Using:

$$\frac{GM}{r^2}$$

i.e the Newtonian equation for the gravitational field at that point, I get:

$$\frac{6.67\times 10^{-11} \times 2\times 10^{30}}{10^8} = 13.34\times 10^{11}\ \mathrm{ms^{-2}}$$

which is an extremely great amount of acceleration. What would happen to the body in this situation? Is this absurd acceleration (and other such scenarios) proof that Newton's law of gravitation is flawed? (and hence, GR) Is it possible for any body to experience this much acceleration? If so, what would be the motion of the body under this much acceleration?

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    $\begingroup$ Here's an oft-used recommendation: Do not, under any circumstances, use Newtonian physics in conjunction with general relativity. $\endgroup$ – HDE 226868 Nov 23 '14 at 15:42
  • $\begingroup$ Hritik: why do you think a body shouldn't be able to experience such a large acceleration? $\endgroup$ – David Z Nov 24 '14 at 6:35
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    $\begingroup$ I'm thinking of spaghetti now, anyone else? $\endgroup$ – Harshal Gajjar Nov 24 '14 at 6:45
  • $\begingroup$ The object would have its momentum increased till its velocity is almost c, and then its mass would exponentially increase too, right? $\endgroup$ – Hritik Narayan Nov 24 '14 at 12:04
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I don't really see the problem. The situation is not far from the gravitational acceleration felt at the surface of a neutron stars where $M \simeq 1.4M_{\odot}$ and $R \sim 10\ km$. The acceleration at the surface must be something like $2 \times 10^{12}\ ms^{-2}$. Not only do we know that neutron stars exist, but this size of gravitational acceleration is needed to stop them blowing themselves apart.

Of course we need GR to properly understand these extreme gravitational fields, but the size of the acceleration (recall that GR does not distinguish between local gravitational fields and acceleration) is not a problem at all and such accelerations are already present in more down-to-Earth phenomena. For instance, a typical cathode ray tube (remember those!) might accelerate electrons using electric fields of $E=30\ {\text k}{\text V}/{\text m}$. The acceleration experienced by the electrons is $e E/m_e = 5.3\times10^{15}\ {\text m}{\text s}^{-2}$.

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At the Event Horizon of Black Holes, the acceleration an object experiences is infinite (Here, I am NOT talking about freely falling objects; Freely Falling objects are in inertial motion in General Relativity).

As for you, you can't feel that on the Event Horizon of small Black Hole because you won't survive by then. Between 4 and 8 longitudinal gs will knock you out. 14g of lateral acceleration can tear your organs loose from one another. Head-to-foot motion, meanwhile, plunges all the blood to the feet.

Forward or backward acceleration appears to go easiest on the body, because they allow the head and heart to accelerate together. Military experiments in the 1940s and 1950s with a "human decelerator" suggest we can slow down at a rate of 45g. We'd probably turn into a bag of spare parts up around 50g, researchers estimate.

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  • $\begingroup$ The body is 10 kilometers away, not at the event horizon. $\endgroup$ – HDE 226868 Nov 23 '14 at 15:59
  • $\begingroup$ Is it possible for a body to experience acceleration as high as the one I calculated? $\endgroup$ – Hritik Narayan Nov 23 '14 at 16:01
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    $\begingroup$ @Hritik Yes, whynot. But, objects may not remain in one piece. Note: such acceleration isn't enough to break molecules or atoms. $\endgroup$ – Schrödinger's Cat Nov 23 '14 at 16:09
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    $\begingroup$ "At the surface of Black Holes, the acceleration an object experience is infinite." - This statement needs qualification; presumably you are talking about the proper acceleration needed to maintain a constant distance in Schwarzschild coordinates, right? But for someone who's used to Newtonian physics, it must be understood that the acceleration for a free-falling observer passing a given point is not the same as the proper acceleration (as measured by an accelerometer) of an observer who is hovering at that point. $\endgroup$ – Hypnosifl Nov 23 '14 at 17:30
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    $\begingroup$ (continued) For a free-falling observer, the proper acceleration at any given point in their body is zero, the coordinate acceleration varies depending on what coordinate system you use (but it actually goes to zero at the event horizon in Schwarzschild coordinates), and probably the most physical measure would be the tidal acceleration over a short distance $\Delta r$ (like the distance between ends of a person's body), given here as $2GM\Delta r / r^3$ where r is the radial Schwarzschild coordinate. $\endgroup$ – Hypnosifl Nov 23 '14 at 17:35

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