Obviously, a smart phone has very little gravity to it. And from what I've looked up, you could turn anything into a black hole if you get it dense enough. Let's assume my phone turned into a black hole, and I'm standing 5 feet away from it. The equation for gravity is F = G((m[1]m[2])/r^2) right??? So according to this, since the mass would be the exact same if it got super dense, my black hole phone, will have no more force exerted on me than just my phone... right? Am I being stupid? r is the distance between them right?

no you have it right. The mass remains unchanged so the amount of gravity will remain unchanged.

 

The Schwartzchild radius is what changes. Assuming the situation where the radius of your phone condenses below its Schwartzchild radius. It will become a BH however the strength of the gravity field from your phone to you remain unchanged as your still at the same radius from your phone.

You will not be attracted towards the phone with greater force. However, if you enter into your phone black hole, you need to reach speed c to get out of this since c is the escape velocity of that black hole.

There is also that pesky hawking radiation to worry about but int the scenario you have set up it should go off like an atomic bomb,,

As Moon said you might get a little warm. A black hole with a mass of 152grams (my mobile phone mass) would be 10^-28metres radius (almost a billion times smaller than a proton) - but in the 10^-19 seconds after creation it would burn off all its mass as radiation at a whopping 10^34watts - and that flash would generate enough energy to keep the entire world going for an hour or so. At five feet away I think it might get a bit toasty - and not sure you would have much time to think about gravity

You will not be attracted towards the phone with greater force. However, if you enter into your phone black hole, you need to reach speed c to get out of this since c is the escape velocity of that black hole.

It is true that the escape velocity at the event horizon is c, but that is not the reason you can't escape. Even though this is often given as the reason. After all, you can leave the surface of the Earth, temporarily, at less than the escape velocity but you can never leave a black hole. The reason is tha space is so curved that there are no paths that lead out of the event horizon.

Edited January 27, 20179 yr by Strange

How can a curve not crossed?

Obviously, a smart phone has very little gravity to it. And from what I've looked up, you could turn anything into a black hole if you get it dense enough. Let's assume my phone turned into a black hole, and I'm standing 5 feet away from it. The equation for gravity is F = G((m[1]m[2])/r^2) right??? So according to this, since the mass would be the exact same if it got super dense, my black hole phone, will have no more force exerted on me than just my phone... right? Am I being stupid? r is the distance between them right?

 

you would lose your phone hole to immediate evaporation.

a black hole this size evaporates much quicker than one that is the mass of a star.

It is true that the escape velocity at the event horizon is c, but that is not the reason you can't escape. Even though this is often given as the reason. After all, you can leave the surface of the Earth, temporarily, at less than the escape velocity but you can never leave a black hole. The reason is tha space is so curved that there are no paths that lead out of the event horizon.

 

That is a very interesting thought.

I undestand what you mean. You mean to say that the edges of the event horizon are perfectly curved in 360° so there is no straight way out. However, if this is the answer, how is there a way in?