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There are huge tidal forces at work around a black hole. But the larger the hole, the smaller the tidal effects near the horizon. So the differences between nearby local forces is small.

What about the force itself? How strong must a rocket be to stay just above the horizon? The horizon grows linearly with M, the mass of the hole. This means that for a black hole containing the mass of the universe the Schwarzschild radius would be greater than the universe itself.

How big would the force be just above a black hole with such a large radius? Is there a formula that relates the force we need to let a rocket stay stationary above its horizon? Can this force be smaller than the force the rocket would need to hoover just above the Earth? If so, why can't light from inside the hole escape?

Qmechanic
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MatterGauge
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The force required to hover at radius $r$ goes to infinity as $r$ approaches the Schwarzschild radius

Dale
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  • That's what I thought too. But I can't imagine that for a hole with the mass of the universe. Wouldn't the force be small on such a large distance? – MatterGauge Feb 22 '22 at 04:25
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    The tidal force would be small, not the force – Dale Feb 22 '22 at 04:29
  • But say we have such a big hole. Say with mass of the universe. And say we are at a distance equal to the radius of the universe (larger even). How can the force be that big? Does time play a role? – MatterGauge Feb 22 '22 at 04:33
  • @Felicia, your question says you are located just beyond the Schwarzschild radius. The acceleration at that point is the same for all Schwarzschild black holes. – BowlOfRed Feb 22 '22 at 04:52
  • @BowlOfRed But what's the formula for that force? I mean, you cant calculate it with Newton. That would give an incredibly low value. – MatterGauge Feb 22 '22 at 07:46
  • @Felicia, this might be close to what you're asking about: https://physics.stackexchange.com/questions/47379/what-is-the-weight-equation-through-general-relativity – BowlOfRed Feb 22 '22 at 08:08
  • @BowlOfRed Thanks for the link. – MatterGauge Feb 22 '22 at 09:45
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Is there a formula that relates the force we need to let a rocket stay stationary above its horizon?

Yes. The gravitational acceleration is given by $a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$ which means that the force you are mentioning approaches infinity as $r$ approaches the Schwarzschild radius $r_s$ as already stated by @Dale.

timm
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  • Ha! I was just looking to a linked answer. All clear. It seems strange though that Newton gives almost zero for such a big hole. But the formula tells it. Thanks! – MatterGauge Feb 22 '22 at 09:42