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# Help with BlackHole question: Schwarzschild radius

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The Astronaut is 1000 km away from the Schwarzschild radius of a Black hole of a solar mass (??⊙). Astronaut B is 1000 km away from the Schwarzschild radius of a black hole one billion solar masses (??9?⊙). Both astronauts weigh 80 kg. Which of the two astronauts feel an gravitational attraction more intense ?

a The Astronaut A
b Astronaut B
c Impossible to know without the information of the radius of the black holes
d The two feel the same gravitational attraction
And why? Its important and i cant figure out the right alternative.

Thanks for any help.

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Since the 1000 km height is larger than the Schwarzschild radius of the Sun, but smaller than the Schwarzschild radius of the billion solar mass object, this is not easy to predict with arguments like: this goes like this, and that goes like that, and therefore the answer scales as ...

I have derived an expression on my board (probably with mistakes), but I haven't run the numbers.

Anyone interested in seeing the math?

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Me. But, if you have to chose one of the alternatives, what is the one who matches better  ? (C)?

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5 hours ago, Andre Gazoto said:

The Astronaut is 1000 km away from the Schwarzschild radius of a Black hole of a solar mass (??⊙). Astronaut B is 1000 km away from the Schwarzschild radius of a black hole one billion solar masses (??9?⊙). Both astronauts weigh 80 kg. Which of the two astronauts feel an gravitational attraction more intense ?

a The Astronaut A
b Astronaut B
c Impossible to know without the information of the radius of the black holes
d The two feel the same gravitational attraction
And why? Its important and i cant figure out the right alternative.

Thanks for any help.

Trick question - both the same, you've already said they both weigh 80kg.

Now if they both had a mass of 80kg...

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They're repeating the Swallowed by a Black Hole program ATM on EDEN + 1

Dave

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3 hours ago, Stub Mandrel said:

Trick question - both the same, you've already said they both weigh 80kg.

Now if they both had a mass of 80kg...

Its a miss translation, or a error in the question, the astronaut have a mass of 80 kg.

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7 hours ago, Andre Gazoto said:

Its a miss translation, or a error in the question, the astronaut have a mass of 80 kg.

It's a strange question, an astronaut will only 'feel a force' if he has a surface to react against.

The force = G x M1 x M2/R^2

As G, the mass of the astronaut are the same we only need to compare the mass of the black hole/R^2.

Mass of earth = 1 unit

R~=1000km

Force for earth-sized black hole is proportional to 1/1,000,000 in my arbitary units

Force = 1 / 1000 x 1000

1 x 10^-6

A billion suns weigh about 300000 earths, so big hole mass = 3 x 10^15

Black hole radius is proportional to mass (at least something about black holes is simple)

Big hole schwartzchild radius is 3x10^15 cm

1000 km is 100,000 cm, negligible compared to this figure.

So I my arbitary units.

Force = 3 x 10^15 / (3 x 10^15  x 3 x 10^15) = 1/ 3 x 10^15 = 3 x 10^-15

So the force from the big hole is vastly smaller.

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Imagine that that an astronaut uses a powerful rocket to hover at constant height (here, 1000 km) above the black hole. Then, the astronaut will have an (apparent) weight.

Newtonian gravity works well for the solar mass black hole, but it looks like Newtonian gravity is off by several orders of magnitude for the large black hole, and Einstein's gravity must be used to find the weight in this case. Still, the weight of the astronaut hovering above the large black hole seems to be much, much smaller than the weight of the astronaut hovering above the solar mass black hole.

I have to go shovel 10 cm of snow off the driveway, but I hope to post a little math later.

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7 hours ago, Stub Mandrel said:

It's a strange question, an astronaut will only 'feel a force' if he has a surface to react against.

The force = G x M1 x M2/R^2

As G, the mass of the astronaut are the same we only need to compare the mass of the black hole/R^2.

Mass of earth = 1 unit

R~=1000km

Force for earth-sized black hole is proportional to 1/1,000,000 in my arbitary units

Force = 1 / 1000 x 1000

1 x 10^-6

A billion suns weigh about 300000 earths, so big hole mass = 3 x 10^15

Black hole radius is proportional to mass (at least something about black holes is simple)

Big hole schwartzchild radius is 3x10^15 cm

1000 km is 100,000 cm, negligible compared to this figure.

So I my arbitary units.

Force = 3 x 10^15 / (3 x 10^15  x 3 x 10^15) = 1/ 3 x 10^15 = 3 x 10^-15

So the force from the big hole is vastly smaller.

You are right. The astronaut A feels the stronger force.

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This has got me thinking the entire universe is a black hole, but everything lies within its schwarzchild radius.

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Thanks for the help guys

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The weight felt by the hovering astronaut is

W = WN * sqrt( 1 + (Rs) / h ) ,

where Rs = 2 * G* M / c^2 is the Schwarzschild radiaus, WN =G * M * m /(Rs + h)^2 is the astronaut's Newtonian weight, and h is the height above the Schwarzschild radius.

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On 10/14/2016 at 10:01, Andre Gazoto said:

Quote
On 10/14/2016 at 10:01, Andre Gazoto said:

The Astronaut is 1000 km away from the Schwarzschild radius of a Black hole of a solar mass (??⊙). Astronaut B is 1000 km away from the Schwarzschild radius of a black hole one billion solar masses (??9?⊙). Both astronauts weigh 80 kg. Which of the two astronauts feel an gravitational attraction more intense ?

Astronaut B has the most force applied to him in a GR solution.

One of the deferences between Newton's theory and GR is the location of infinite forces. In Newtons theory it is at R=0. The GR location is at R=(Rs), where all objects fall at the speed of light. This is why GR equations written in Newton form may have the term (R-Rs) or its equivalent.

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I disagree,

Whether by GR or Newtownian calculations, at the Schwarzchild radius the escape velocity is, by definition, c, the speed of light. An object travelling at C will orbit at the Schwarzchild radius.

The larger the back hole, the larger this radius, and therefore the longer the orbital period, and therefore the smaller the force experienced by the object to keep it in orbit.

In fact for a really massive black hole, the force experienced at the SR could be tiny, so a photo arcing away from the black hole might take millennia or even aeons to return.

This is the same question as to whether or not the universe is 'closed'.

It also raises the counter-intuitive fact that something at the SR doesn't get 'ripped apart by the intolerable forces of gravity' - as far as it's concerned everything is perfectly normal...

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Yes, the escape velocity at Rs is ,c, the velocity of light. In GR an orbital velocity of c is at R=1.5(Rs) while the orbital velocity at Rs is infinite.

Question, how much force is required to accelerate a 1 gm mass to c?

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On ‎10‎/‎14‎/‎2016 at 18:01, Andre Gazoto said:

The Astronaut is 1000 km away from the Schwarzschild radius of a Black hole of a solar mass (??⊙). Astronaut B is 1000 km away from the Schwarzschild radius of a black hole one billion solar masses (??9?⊙). Both astronauts weigh 80 kg. Which of the two astronauts feel an gravitational attraction more intense ?

A is approx. 1003 km distance from COG of M=1 hole

B is approx.3,000,001,000 km distance from COG of M=10^9 hole

'A' therefore feels about 90,000 times more force than 'B'.

Force is proportional to M/R^2 and Schwarzschild radius =2M = 3km approx.

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On 10/23/2016 at 07:31, Tiki said:

A is approx. 1003 km distance from COG of M=1 hole

B is approx.3,000,001,000 km distance from COG of M=10^9 hole

'A' therefore feels about 90,000 times more force than 'B'.

Force is proportional to M/R^2 and Schwarzschild radius =2M = 3km appro

Force has both time and velocity components. Newtonian physics will give reasonable results when time is absolute and velocity is linear. GR must be used when a distant observer's rate of time differs from the local observer and\or where velocity is not linear. In the following, I am assuming both 'A' and 'B' are falling at negative escape velocity.

Is 'A's clock running at the same rate as 'B's clock?

Time slows down near a black hole by 1/√(1-(2GM/c^2R)). For every time unit 'A's clock shows, a distant observers clock would show 1.002 time units. Time also slows down do to velocity. 'A' has a velocity of .0547c. The time dilatation factor for velocity is √(1-v^2). For every time unit 'A's clock shows, a distant observers clock would show 1.001 time units.  Time in this case is almost passing at the same rate as the distant observer so Newtonian physics should give reasonable results.

For every time unit 'B's clock shows, the same distant observer would show 1732 time units. In this case, GR must be used to get reliable ansers. If a distant observer could do spooky action and apply a g of force to 'B' for 1732 seconds, 'B' would experience 1732g for 1 second. 'B's velocity would be .9999998c. The time correction would be 1581 seconds for the distant observer to 'B's 1 second. Multiple the two corrections times each other gives 273,800 as a correction factor for time.

Which is the bigger change of velocity, .05470000c to .05470001c or .99999980c to .99999981c?

(.05470001-.0547)/(1-.05470001•.0547)= .00000001

(.99999981-.9999998)/(1-.99999981•.9999998)=.0256

Close to Rs small changes in Newton velocity will be large changes in real velocity.

Edited by Absolute zero
Correct typo

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My point about orbital velocity is that the larger a black hole, the less the force required to keep an object in orbit at any given velocity. c is a convenient velocity as the orbital radius at c = the schwarzchild* radius (by definition so independent of GR or Newtonian physics).

Trying to imagine a photon on the SR but travelling angled slightly outwards, it will go a lot further before being pulled back in to a large black hole. But of course a photo can't be travelling in that way as it would have to reach the SR from inside which can't be done!

Horrors - thinking about this I though the only way would be if a pair of complementary photons popped into existence right on the SR - one would be trapped an the other COULD escape.

Blow me! I've discovered Hawking radiation :-)

But coming back to my point, an arbitrarily huge black hole would have an arbitrarily large SR.

GR means that to a particle on the SR the orbit IS a straight line (though we outside see it as a circle in curved space).

Even with GR a larger orbit (critical point, in the question the astronaut is OUTSIDE the SR) is parallel to the SR.

For a small black hole that orbit is clearly going to be circular, in the first example about 1000 and a bit km across. The force to keep an object in an orbit of 1000kn radius is 'a fair bit'.

For an arbitrarily huge black hole, any modest length of the orbit would be virtually indistinguishable from a straight line and the force required to keep an object following it would be infinitesimally small.

For the second Black Hole in the example the size of the orbit is clearly between 1000 and a bit km and arbitrarily huge.

The force required to maintain that orbit the force experienced by an astronaut at that distance has to be less than 'a fair bit' and greater than 'infinitesimally small' so the bigger the Black Hole the less the force felt at any given distance from the SR.

QED

*The Spill Chicken wants me to use 'schoolchild' which creates the disturbing mental image of astronauts orbiting massively obese youngsters.

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18 hours ago, Stub Mandrel said:

My point about orbital velocity is that the larger a black hole, the less the force required to keep an object in orbit at any given velocity. c is a convenient velocity as the orbital radius at c = the schwarzchild* radius (by definition so independent of GR or Newtonian physics).

*The Spill Chicken wants me to use 'schoolchild' which creates the disturbing mental image of astronauts orbiting massively obese youngsters.

From Wikipedia, (sorry I have not figured out linking here)

## Orbital velocity in general relativityEdit

In Schwarzschild metric, the orbital velocity for a circular orbit with radius 'r' is given by the following formula:

v=√(GM/(r-rs))

where  rs=2GM/c^2 is the Schwarzschild radius of the central body.

So when r=rs, the orbital velocity is infinite or undefined. At r=1.5rs, the orbital velocity is c.

Script did not transfer correctly
Edited by Absolute zero

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It seems you are right!

I'm not sure it invalidates the argument as you can just transfer it to 1.5rs where the same logic applies. The force at rs will be 2.25 times that at 1.5rs in each case (there may be a modest adjustment for GR), so the proportionality argument still applies.

My Hawking radiation idea isn't relevant - photons can escape at the schwarzhild radius as they are massless, not because of their speed.

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7 hours ago, Stub Mandrel said:

My Hawking radiation idea isn't relevant - photons can escape at the schwarzhild radius as they are massless, not because of their speed.

If Hawking radiation exists it is predicted to consist of photons, neutrinos, and to a lesser extent all sorts of massive particles. As they form outside the event horizon the escape velocity is <c.

Regards Andrew

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On ‎10‎/‎24‎/‎2016 at 09:04, Stub Mandrel said:

For a small black hole that orbit is clearly going to be circular, in the first example about 1000 and a bit km across. The force to keep an object in an orbit of 1000kn radius is 'a fair bit'.

For an arbitrarily huge black hole, any modest length of the orbit would be virtually indistinguishable from a straight line and the force required to keep an object following it would be infinitesimally small.

I like this argument. The least curved path is of course a consequence of a lesser force.

Orbits near black holes are in some ways very different from their Newtonian analogues. For instance in a Schwarzschild space-time there are no stable circular orbits for r<6M and no circular orbits at all for r<3M.

On ‎10‎/‎24‎/‎2016 at 03:15, Absolute zero said:

Force has both time and velocity components. Newtonian physics will give reasonable results when time is absolute and velocity is linear. GR must be used when a distant observer's rate of time differs from the local observer and\or where velocity is not linear.

OP's original question can be answered using purely Newtonian gravity. In fact, a distant observer would be able to deduce Kepler's third law if he were to observe a body in circular orbit around a distant black-hole.

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On 10/25/2016 at 01:11, Stub Mandrel said:

It seems you are right!

I'm not sure it invalidates the argument as you can just transfer it to 1.5rs where the same logic applies. The force at rs will be 2.25 times that at 1.5rs in each case (there may be a modest adjustment for GR), so the proportionality argument still applies.

I am not sure what you are referring to as being right. Is it v=√(GM/(r-rs)) is the correct equation for orbital velocity? Or could it be v=√(GM/r) is just a Newton approximation, good whenever r is much larger then rs?

Until someone does a full GR evaluation, I will assume the equation, GM/(r-rs)^2 , is the correct one to answer the op's question with.

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14 hours ago, Absolute zero said:

Until someone does a full GR evaluation, I will assume the equation, GM/(r-rs)^2 , is the correct one to answer the op's question with.

I posted the results of a full GR evaluation above, i.e.,

On 2016-10-15 at 13:52, George Jones said:

The weight felt by the hovering astronaut is

W = WN * sqrt( 1 + (Rs) / h ) ,

where Rs = 2 * G* M / c^2 is the Schwarzschild radiaus, WN =G * M * m /(Rs + h)^2 is the astronaut's Newtonian weight, and h is the height above the Schwarzschild radius.

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On 10/29/2016 at 11:11, George Jones said:

I posted the results of a full GR evaluation above, i.e.,

The conversion of Newtonian orbital velocity to a Schwarzschild metric orbital velocity would use

√(1+(rs/h)) or more familiar expression to me √(r/(r-rs)) or we could mix and match √(r/h).

It would be odd for the same conversion factor to work for Newtonian weight to Schwarzschild metric weight. If you would be so kind, I would like to see the equation for Schwarzschild metric weight or where it can be found.

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