View Full Version : Black hole mass, radius

imported_WINSTON

2004-Nov-28, 04:30 AM

At this site one can enter values of mass and radius of a celestial body and the escape velocity will be calculated for you.

http://www.geocities.com/CapeCanaveral/Han...culatoresp.html (http://www.geocities.com/CapeCanaveral/Hangar/9494/calculatoresp.html)

The QUESTION is:What are some typical mass/radii of black holes?

I visited several mainstream science sites to get the typical mass and radius of a black hole. I plugged the values into the formulae[at a site SIMILAR to this one], and the escape velocity was revealed. The escape velocity was 300 times slower than the speed of light.

The escape velocity of a black hole should be more than c, by mainstream definition.

You can plug in values at this site to see if you get the same results.......and let me know.

Erimus

2004-Nov-28, 06:19 AM

Well, this question can be answered two ways: the hard way, and the easy way.

The hard way is to use the formula for the Schwarzchild Radius, which is:

2MG / c^2 = SR

where:

M = black hole mass

G = universal gravitational constant

c = the speed of light

The easy way is to use a much simpler formula for black hole radius, which is simply:

R in km = 3M

where:

M = black hole mass in solar masses

With this in mind, a one solar-mass black hole would have a radius of 3 km. Most stellar black holes are probably four solar masses and up. A billion solar-mass black hole would have a radius of a 3 billion kilometers, enough to engulf the Solar System out past the orbit of Uranus.

imported_WINSTON

2004-Nov-29, 04:46 AM

The Schwarzchild Radius, which is:

2MG / c^2 = SR

This formulae is based on the speed of light, and useless as independent confirmation. I was hoping some other method was used to determine the spec's of a black hole.

Why I was asking:

Why must the escape velocity of a black hole be exactly c? If you pour a few tons of debris into one, the escape velocity should increase.[ Remember, I believe in the abide-by-the-laws-of-physics dark star theory, not the hole-in-space-with-gravity-remembered black hole theory.] Likely, if a black hole is black because the escape velocity is >= c, likely it goes beyond c upon initial collapse. I know of no proven theory that limits collapse of matter based on escape velocity.

Once the escape velocity surpasses c, the "Hawking Radiation" concept of "pair splitting" doesn't work, as the speed required is beyond c plus the Heisenburg variation.

I believe, of course, that matter in black holes is transformed back into elementary MALON particles. Maintaining an equilibrium of matter and energy, these MALON eventually collapse into galactic neutriods, forming matter in pair production, spiralling outward to form stars in the spiral arms. Above and below galaxies, stars form from the collapsing matter [the difference in abundance of elements of the two kinds of stars is documented].

If a cannon were fired at 90% of the escape velocity of Earth, perpendicular from Earth's surface, it would travel how far before returning to Earth?

If a photon [theoretical]were emitted perpendicular to a black hole at 90% of the escape velocity of a black hole, same answer?

This would make our black hole visible at an uncomfortably close distance.......

Matthew

2004-Nov-29, 05:30 AM

Winston you are looking at black holes in a classical physics way. Its like saying F=ma at .99c when it obviously doesn't.

You need to use relativity to understand and explain black holes.

antoniseb

2004-Nov-29, 03:36 PM

Originally posted by WINSTON@Nov 29 2004, 04:46 AM

Why must the escape velocity of a black hole be exactly c? If you pour a few tons of debris into one, the escape velocity should increase.

The escape velocity from a black hole at the event horizon is 'c'. If you pour a few tons onto a black hole, the mass increases, and the radius of the event horizon increases. This is just like any gravitational sphere and any sub-c escape velocity.

The escape velocity from an orbit 4500 miles from the center of the Earth is about 24,000 miles per hour [the orbit acounts for about 18,000 miles per hour of that]. Increasing the mass of the Earth would increase the distance for which 24,000 mph is the escape velocity.

Matthew

2004-Nov-29, 10:58 PM

The escape velocity from a black hole at the event horizon is 'c'. If you pour a few tons onto a black hole, the mass increases, and the radius of the event horizon increases. This is just like any gravitational sphere and any sub-c escape velocity.

Actually the escape velocity is >c, thats why we don't see any light coming out of the black hole.

TheThorn

2004-Nov-30, 01:34 AM

Matthew, Anton is right. The escape velocity of a black hole, "at the event horizon" is c. That's the definition of the event horizon.

The escape velocity from the singularity at the heart of the black hole is infinite. From just outside the event horizon, it's just less than c, which is why we can still see things just outside the event horizon. From just inside the event horizon, it's > c, which is why we can't see anything from inside the event horizon.

spacepunk

2004-Nov-30, 02:26 AM

The QUESTION is:What are some typical mass/radii of black holes?

and

Once the escape velocity surpasses c, the "Hawking Radiation" concept of "pair splitting" doesn't work, as the speed required is beyond c plus the Heisenburg variation

To add to this topic and clarify this question, just what is a typical "Hawking" black hole, and is it within a specific mass range?

http://www.iop.org/news/783

zephyr46

2004-Nov-30, 03:13 AM

The First Black hole that comes to mind for me is Sag A*.

Our own galaxy, for instance, has a black hole with mass 2.6 million Msun at its center, 8000 parsecs away from us. Our own galactic black hole is actually fairly wimpy. The Andromeda Galaxy, 700,000 parsecs away, has a black hole with a mass of 30 million Msun.

and

The closest star to this massive object comes within 17 light-hours (120 A.U.) of its center. The only plausible way to cram 2.6 million solar masses of stuff into a sphere 120 A.U. in radius is to make it all into a single big black hole. (For instance, if you tried to make a cluster of a million neutron stars 120 A.U. in radius, the collisions between neutron stars would make them all merge into a single big black hole within a few hundred thousand years.)

Quotes from this article (http://www-astronomy.mps.ohio-state.edu/~ryden/ast162_6/notes26.html)

The Second is Cygnus X1. This article (http://www.jfmto.pwp.blueyonder.co.uk/cygnus-xone.htm) gives 3.6 miles for one solar mass.

This article (http://www.eso.org/outreach/eduoff/edu-prog/catchastar/CAS2002/cas-projects/austria_cygnus_1/) puts examines the argument for Cyg X1 being a black hole.

Sorry, I haven't read Hawkings last paper about black holes. I am sure there is discussion about it though.

Matthew

2004-Nov-30, 06:40 AM

Originally posted by TheThorn@Nov 30 2004, 12:34 PM

Matthew, Anton is right. The escape velocity of a black hole, "at the event horizon" is c. That's the definition of the event horizon.

The escape velocity from the singularity at the heart of the black hole is infinite. From just outside the event horizon, it's just less than c, which is why we can still see things just outside the event horizon. From just inside the event horizon, it's > c, which is why we can't see anything from inside the event horizon.

Ah.... sorry Anton.

ferg.c.

2004-Nov-30, 01:10 PM

Hi all,

I know I'm just an amateur here but I'm going to try to explain why relativity is needed to show why black holes do not get as large as Winston fears they may do, and in laymans terms!

The most important feature of relativity is the observance of the speed limit, i.e. the speed of light in a vacuum. Nothing can travel faster than the speed of light, not even relative to another object traveling at the speed of light in the other direction. This sounds paradoxical but the truth of the matter is that in order to maintain the speed limit other factors have to change. These other factors are time, mass and gravity. Time will accomodate two objects traveling towards each other, both having a velocity over half the speed of light by slowing down just enough to make the closing velocity the speed of light. Gravity works in a similar way, adjusting itself to acommodate time and speed. The mathematical relations connecting these factors all relate to the inviability of this speed limit. If time did not slow down in massive gravitational fields the material and energy inside black holes would cover much vaster expanses of space and if only Newtonian mechanics applied here then your assumptions would be right.

Cheers

Ferg :)

ferg.c.

2004-Dec-01, 02:46 PM

Originally posted by WINSTON@Nov 29 2004, 04:46 AM

If a cannon were fired at 90% of the escape velocity of Earth, perpendicular from Earth's surface, it would travel how far before returning to Earth?

Hi All,

This is really for a better ( and at this moment more sober) mathematician than me but since no one has answered I'll have a go.

Firstly I know that Ve = sqrt2 * Vc = sqrt(a*r) form angular acceleration.

s= ut +1/2at^2 or so you would think.

If gravity was constant your cannon ball would go about 15500 km directly up and come back down.

That was easy mechanics but not the right answer.

But as you know gravity decreases with the square of the increase of distance from the earth so this involves more complex math.

The equation is an implicit function

s = ut +1/2 (a/s^2)t^2

Differentiate this mother with respect to t and find it's solution for Zero. I tried this and it's a *****.

s= (90%Ve)t + 4.9t^2 * t^2s^-2

firstly differentiate p = t^2s^-2

remembering that s is a function of a function

dp/dt = (2ts^-2) + t^2(-2ds/dt * s^-3)

substituing in

ds/dt = u + 9.81t + 2ds/dt * t^2s^-3 * (2ts^-2) * -t^2

ds/dt = ((90%Ve) + 9.81t ) / t^2s^-3 * (2ts^-2) * -t^2

when

((90%Ve) + 9.81t ) / t^2s^-3 * (2ts^-2) * -t^2 = 0

then t is the time traveled. put this in

s = 90%Vet +1/2 (9.81/s^2)t^2

and a good calculator, which I don't have, will give you the height your cannon ball will reach.

PS don't do calculus when you've been drinking you can loose terms under the table! don't take this math as absolutely correct. I'm quite blootered!

Cheers guys

Fergy :blink:

wstevenbrown

2004-Dec-01, 06:20 PM

Quite aside from the time-and-distance math, to put it simply-- it wouldn't come straight down. Consider: at Earth's surface, the cannonball participated in Earth's rotation and revolution as angular accelerations. Upon leaving the surface for a long ballistic trajectory, these will resolve-- only the tangential component goes with the cannonball. In the case of rotation, the earth below will have rotated at ca.1000 mph at the equator-- at the surface, equal to 15deg/hr. While the cannonball is at a much further-out radius, the tangential component, 1000 mph, will not subtend nearly so large an angular arc. Net effect: the launchpad rotated out from under. Similarly, the Earth's revolution will have carried it somewhat inboard of the takeoff point-- there will be farther to fall. Are we having fun yet? ;) S

wstevenbrown

2004-Dec-01, 06:30 PM

Afterthought-- I seem to have assumed in the foregoing post that the direction of the launch was opposite the sun-- ok, now it's explicit. S

ferg.c.

2004-Dec-01, 10:24 PM

Hi Steve,

I can't beleive I posted that math when I was three sheets to the wind this afternoon! It looks quite impressive but the differentiation is not quite right (pretty damn good though for blootered). I'm on tour in the north at the moment and I've just finished a concert. So I'l wait till I get home to a pencil and some paper to get the real deal up there. The idea is right though, Steve and the straight up and down thing doesn't come into it if you observe the whole thing from a vantage point far eough away not to see the Earth spinning. Also Escape velocity is not effected by direction unless of course you point you cannon at the earth then you can calculate consevation of momentum with perpendicular vector components and what not as your cannonball ploughs into the ground!

Cheers Ferg :)

wstevenbrown

2004-Dec-01, 11:16 PM

I haven't put pencil to it either-- I'm holidaying w/family. My intuition says that .9 v-sub-e puts you in eccentric orbit with apogee inside the atmosphere. Time aloft is extreme. S

ferg.c.

2004-Dec-01, 11:31 PM

Hi steve,

I can't believe the apogee will be so close. the 15500km (The wrong calculation) is going to be the smaller of the two and thats already way outside any thing that's considered atmosphere in aerodynamics.

Ferg :)

wstevenbrown

2004-Dec-01, 11:54 PM

My bad-- perigee, da****all! :( S

ferg.c.

2004-Dec-02, 12:01 AM

Hi Steve,

Sorry, I still disagree.The perigee of a satelite launched from the surface of the earth is at the surface of the earth! This is theoretical because we don't take into account orbital decay due to air resistance. That's why we could have very low level sustained orbits round the moon on Apollo missions: no air resistance. Of course the orbit of our cannonball would decay on its first time round and plough into the ground some where near Highbury.

Ferg :)

ferg.c.

2004-Dec-02, 12:11 AM

Sorry Steve,

You're not wrong just not specific enough! Of course the surface of the Earth is some point in side the Earths atmosphere! It's a bit of lateral thinking though.

Ferg ;)

spacepunk

2004-Dec-02, 02:36 AM

Did some more checking on articles about black holes. There has been a lot of research done, so one does have to specify what kind/mass of black hole you are discussing. In response to my earlier question Hawking's black holes would be at the lower mass limit . Here's a quote from

http://www.newscientist.com/news/news.jsp?...p?id=ns99996151 (http://www.newscientist.com/news/news.jsp?id=ns99996151)

Hawking's black holes, unlike classic black holes, do not have a well-defined event horizon that hides everything within them from the outside world.

In essence, his new black holes now never quite become the kind that gobble up everything. Instead, they keep emitting radiation for a long time, and eventually open up to reveal the information within. "It's possible that what he presented in the seminar is a solution," says Gibbons. "But I think you have to say the jury is still out."

Matthew

2004-Dec-02, 08:29 AM

Instead, they keep emitting radiation for a long time, and eventually open up to reveal the information within.

Ah, how quickly we change. Once we would have never said black holes reveal their information.

Matthew

2004-Dec-02, 08:31 AM

Instead, they keep emitting radiation for a long time, and eventually open up to reveal the information within.

Ah, how quickly we change. Once we would have never said black holes reveal their information.

guest_star

2004-Dec-02, 10:07 AM

To make long story short, he (Hawking) abandoned his concept od singularity inside the black holes (which in turn raises the question about existence of Bing Bang singularity, I always thaught of Bing Bang as Big Collision rather).

We can't see singularity as it is information, well?

One question applied to "common" Black Holes.

Gravity is prevailing all other forces in a black hole.

What force prevents the material of black hole at radius let's say 1 mil. km., from

colapsing further into let's say Planck radius?

imported_WINSTON

2004-Dec-03, 06:39 AM

SORRY:The cannon question, I should have said shoot the cannonball from a non-spinning body with no atmosphere that is not near to or in orbit with any other body. I just wanted to point out that escape velocity does not inhibit all motion that is slower.

I see that the event horizon is not at the surface, but above it. Still, even with escape velocity >c, light should still travel a ways [even beyond the event horizon]before returning to the black hole.

I do believe in the classical dark star theory, not the black hole where matter "leaves" our universe and gravity is "remembered".

The center of a BH would have the least gravity force, as all forces have cancelled out.

The greatest area of gravitational force of a body could be imagined by removing a wedge [cut an acre of surface of the star all the way to a point in the center, forming a piece shaped like a pyramid]. Imagine the center of gravity of that wedge[about 2/3 of the way towards the surface]. Now assemble this and all of the other wedges together to draw a smaller sphere beneath the surface. This is where the greatest gravitational forces are.

zephyr46

2004-Dec-08, 04:30 AM

I have mentioned this link in other threads,

Jillian's Guide to Black Holes (http://www.gothosenterprises.com/black_holes/index.html)

Hawking radiation (http://www.gothosenterprises.com/black_holes/outside_black_holes.html)

The other interesting discussion I have seen is the fermion ball scenario (http://www.mpifr-bonn.mpg.de/gcnews/gcnews/Vol.13/bilic@physci.uct.ac.za_starmot.abs.shtml) and Quark stars (http://antwrp.gsfc.nasa.gov/apod/ap020414.html)

Ola D.

2004-Dec-09, 12:38 PM

Hey ferg,

was your calculations wrong or right?

I was thinking of another method but I am not sure if it actually would work. Please correct me if i am mistaken.

escape velocity=(sqr)2gr

new escape velocity (Vnew) = 90%xV(earth)

use v^2 = u^2 + 2gs

where u=Vnew , v=0 (since it'll be reaching its maximum height),

and g=9.8ms^-2

You'll find s which would be the maximum height reached by the object.

ferg.c.

2004-Dec-09, 12:59 PM

Hi Ola,

s = u((v-u)/gs^2) + .5((v-u)/ gs^2)^2 * gs^-2

Any up for differentiating this babe?

The added speed from the spin of the earth is only about 10m/s and the change of angle is so small as to be negligable.

Ferg B)

kashi

2004-Dec-09, 10:25 PM

With respect to which variable?

scott712

2004-Dec-10, 03:58 AM

It seems to me that by definition the radius of a Black Hole is that distance at which the escape velocity is the speed of light. On the other hand, perhaps there is a distinction to be made between the material radius of the black hole and the event horizon.

ferg.c.

2004-Dec-11, 04:07 PM

Hi Kashi

find ds/sv

ferg

TheThorn

2004-Dec-12, 09:03 PM

That is the definition of the event horizon. The material diameter of a black hole is 0, as all the mass is theoretically in a singularity. But then again, since we can't see inside the event horizon, even in theory, we'll never be able to see a singularity to observe it's 0 diameter.

alainprice

2004-Dec-12, 10:06 PM

I believe it safe to assume that the mass of a black hole is not a real singularity, but instead a virtual one. It is likely extremely, but not infinite.

bazbsg

2004-Dec-13, 12:52 AM

Originally posted by TheThorn@Dec 12 2004, 09:03 PM

That is the definition of the event horizon. The material diameter of a black hole is 0, as all the mass is theoretically in a singularity. But then again, since we can't see inside the event horizon, even in theory, we'll never be able to see a singularity to observe it's 0 diameter.

I'm glad someone added this and corrected the misconceptions posted earlier. The event horizon is not a physical object or shell. It is merely the radius at which the escape velocity equals the speed of light. Just outside the event horizon, it is possible to orbit the black hole and not fall in, but the orbital velocity will be the speed of light. At smaller radii, the escape velocity would be greater than c, were that possible. So once inside the event horizon, any matter cannot maintain its position or even orbit the singularity; it must eventually hit and join the singularity.

The inside of a black hole is still a black hole; blacker than black. Even when you cross the event horizon, you don't get the priviledge of seeing the singularity. Light from it cannot escape to rise up and hit your eyes. The direction of everything lower than the event horizon is towards the singularity.

(Note, after I've written this I must add that I am considering only non-rotating, non-charged black holes because frankly, I'm not sure if everything I've said applies to the general case. I think a rotating black hole's singularity is in the shape of a ring, not a point, so it might be possible to see the thing from the center of the ring, where gravity does cancel out???)

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