Questions with Gravity

[davidivad]

It is my impression that gravity is a field.

 

I have heard conclusions that you could survive the event horizon of a black hole if it were supermassive.

This seems counter-intuitive considering the forces involved.

 

I tried a thought experiment involving the earth as a test bed.

In this experiment, i am sealed in a box of unobtanium at the center of the earth.

 

In the first version I am alive and well because the crushing force of gravity is transmitted through the over-all mass pushing at me is held off by my box.

 

In the second version, i am crushed because of the cumulative intensity of the gravitational field.

 

i considered the second version due to a couple of external observations.

 

1. dense objects like neutron stars have dense crusts displaying no need for mass on the outside to compress the crust.

2. black holes can cause star formation leading to the conclusion that the field of a black hole and a cloud add the intensity both fields.

 

I then refered back to my supermassive black hole.

the first thing I considered was an object's acceleration and the resulting ionization that would occur.

I then wondered if this was acceleration or simply the object following the path of einstein's terrain (an obect in orbit does not experience angular acceleration of it's orbital path).

 

my conscience tells me that the two bodies (an object and a black hole) must cancel each other out (every action requires an equal and opposite reaction). the result would be an acceleration to the point of the object's loss of entropy and structure through radiation before the event horizon is reached. einstein contradicts me by saying i do not feel acceleration when I fall.

 

my conclusion is that the gravitational field would crush you before you ever got there thus still giving up your entropy to heat ( just like the earth gives up heat to gravitational compression).

 

what is your opinion?

 

 

 

 

 

 

 

[swansont]

The gravity at the center of a uniformly distributed spherical mass is zero. (i.e. you'd be crushed, but not by the gravity). If the sphere were hollow, there would be no force in the hollow interior.

[davidivad]

thank you; i knew i was missing something. the field is there but the attraction is cancelled out at the center (two people playing tug of war).

[imatfaal]

thank you; i knew i was missing something. the field is there but the attraction is cancelled out at the center (two people playing tug of war).

 

The field cancels out anywhere within a hollow symmetrical shell. From the outside all the mass can be thought to be concentrated at the centre, and from the inside you feel no gravity. You can look up Newtons Shell Theorem or Gauss' Law of Gravity. The former is more obvious if you have a good grounding in mathematics the latter is more intuitive

[Spyman]

I have heard conclusions that you could survive the event horizon of a black hole if it were supermassive.

Also in addition to what swansont and imatfaal have said, the event horizon is not at the centre or inside, it's considered to be the surface of the black hole.

 

An observer in free fall can survive the gravitational field close to the event horizon of a supermassive black hole because the tidal forces at the more distant horizon weakens with increasing mass, such that very very massive black holes can have even smaller tidal forces than what Earth has on its surface.

Edited January 17, 2014 by Spyman

[davidivad]

Also in addition to what swansont and imatfaal have said, the event horizon is not at the centre or inside, it's considered to be the surface of the black hole.

 

An observer in free fall can survive the gravitational field close to the event horizon of a supermassive black hole because the tidal forces at the more distant horizon weakens with increasing mass, such that very very massive black holes can have even smaller tidal forces than what Earth has on its surface.

yep, and it all works out in the math. in theory, i should be able to go on about my buisness regardless of the fact that escape velocity exceeds the constant. in this sense, the schwartzchild radius is simply the escape velocity at this distance.