The calculator on the website https://planetcalc.com/1758/ cites Wikipedia's Gravitational acceleration and implements:

$$g = G \frac{M}{(R + h)^2}$$

For a black hole with mass $M$ equal to 5 solar masses with a diameter $R$ of 1 meter at a distance $h$ of 1500 km, the acceleration given is 296000 million m/s^2 which is nearly close to trap light.

So one might expect the event horizon to be at least close to that. Is this correct?

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    $\begingroup$ I adjusted your format a bit and added the website's equation using MathJax. Please feel free to edit further. Thanks and welcome to Stack Exchange! $\endgroup$
    – uhoh
    Aug 2, 2020 at 22:52
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    $\begingroup$ Since you're interested in black holes, you may enjoy playing with the Hawking radiation calculator. $\endgroup$
    – PM 2Ring
    Aug 3, 2020 at 9:11
  • $\begingroup$ @PM2Ring Thank you. $\endgroup$ Aug 3, 2020 at 12:12
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    $\begingroup$ You're mixing up acceleration and escape velocity. $\endgroup$ Aug 3, 2020 at 13:50
  • $\begingroup$ What do you mean by 1 meter? A black hole of 10 solar masses has a diameter of 60 km (Schwarzschild diameter). $\endgroup$ Aug 4, 2020 at 0:44

1 Answer 1


This is the Newtonian model of gravity. It is a very good model, it is used for accurate calculating the motion of objects in the solar system to a very high degree of accuracy.

However, for very strong gravitational fields you need to use Einstein's model, which accounts for things like the constant speed of light for all observers. I'm not going to go into mathematical details (not least because they are far to hard for me!)

Now you can calculate the gravity of a sphere and you find that at a radius of $r=2GM/c^2$ something strange seems to happen, nothing can travel from a radius of less than this to a radius of more. It is the event horizon. This can only happen in Einstein's theory of gravity. Newtonian theory has no "speed limit" so there can be no event horizon.

Now it so happens that this is the same radius that Newtonian gravity would predict to have an escape velocity $v_e = \sqrt{\frac{2GM}{r}}$ equal to the speed of light. That is mostly a coincidence(there was a recent question about this, that I haven't been able to find), but a neat one.

You seem to be comparing the Newtonian acceleration due to gravity with the speed of light. The units of acceleration and speed are different, so the comparison is not valid.

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    $\begingroup$ Still it is cool . I was able to calculate the acceleration of the Sun to Alpha Centauri. It is fun if you play with them. $\endgroup$ Aug 2, 2020 at 23:22
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    $\begingroup$ By the way, the separation of R and h is not really useful. The only value you need is the distance from the centre of the planet that is what I've called "r" in the answer $\endgroup$
    – James K
    Aug 2, 2020 at 23:54
  • $\begingroup$ @JamesK - that formula is what is on the web page, for what it's worth. $\endgroup$ Aug 3, 2020 at 14:04

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