Escape Velocity

[Seeka]

I was at my Astronomy clubs monthly meeting Monday night & we had a really entertaining, talented Professor give us a talk on black holes.
He put a whole load of math up on the board and started talking about escape velocity & kinetic energy with regard to an object escaping from a black hole, i think he was trying to convey that nothing can escape and went off on a tangeant and i could not grasp what he was talking about, there were tooo many questions at the end of the session to fit in my 3rd query so i tried wiki and i was still unable to take it in, can anybody enlighten me in easy terms?

Much Appreciated

Steff

[gzhpcu]

When a massive star collapses under gravity, it is theoretically a mathematical point with virtually zero volume, where it is said to have infinite density (this however represents a breakdown of GR, so we are not sure what it looks like). This is referred to as a singularity. When this happens, escape would require a velocity greater than the speed of light. No object can reach the speed of light. The distance from the black hole at which the escape velocity is just equal to the speed of light is called the event horizon.

Hope this helps...

[Seeka]

Yes it does Gzhpcu, thanks!

[StupendousMan]

about escape velocity can be found at

http://spiff.rit.edu/classes/phys230...res/bh/bh.html

Perhaps it might help.

[Seeka]

Excellent link, will study it alot thank you

[alainprice]

Escape velocity is the speed you need to throw something straight up for it to never come back down. Anything less than that, and it will come back, eventually.

Ever since people realized the speed of light, some have wondered what would happen if an object had gravity so strong that the escape velocity is more than the speed of light. This object is now called a black hole. The 'surface', in this case an imaginary shell(like a beach ball), where the escape velocity is exactly the speed of light is called the event horizon.

Don't quote me on the event horizon though, since the analogy of throwing something up doesn't apply perfectly to light.

[Seeka]

Ok so where does kinetic energy fit in?

Also, if two galaxies collide and saying they both have a black hole, do they migrate to the centre and join as one supermassive black hole?
Is it possible they both get ejected from the galaxy and render the galaxy as an orphan galaxy?

[WayneFrancis]

Ok so where does kinetic energy fit in?

Also, if two galaxies collide and saying they both have a black hole, do they migrate to the centre and join as one supermassive black hole?
Is it possible they both get ejected from the galaxy and render the galaxy as an orphan galaxy?

With galactic mergers the black holes would merge in most cases I would say. There might be a VERY odd situation where 2 galaxies collide and the black holes don't merge but this would be a very weird case where the 2 galaxies just happen to have proper motions that intersect, not sure if this is even possible. Then if the 2 have enough speed they might not stay merged but just pass through each other. This would be a very interesting event.

In any case many star systems are expected to get kicked out of galaxies that are undergoing a merger. The norm for black holes would most likely be for them to merge.

[korjik]

Ok so where does kinetic energy fit in?

Also, if two galaxies collide and saying they both have a black hole, do they migrate to the centre and join as one supermassive black hole?
Is it possible they both get ejected from the galaxy and render the galaxy as an orphan galaxy?

Properly done, you find escape velocity by finding the kinetic energy required to go from the surface of your object out to infinity, and to have the kinetic energy, and therefore velocity, go to zero at infinity.

The situation with two black holes is complex. It all depends on the total energy of the system and the kinetic energies of the black holes. If they are moving fast enough, they may shoot out of their galaxies, but it is likely that the stars of the galaxies will follow the 'holes. This leaves you with gasless galaxies.

You can also have the two holes end up in orbit with one another. In this case, they will eventually merge.

[Seeka]

Great! Thank you for your replies

[mugaliens]

It's:

v_e=sqrt(2GM/r)

[Spaceman Spiff]

The situation with two black holes is complex. It all depends on the total energy of the system and the kinetic energies of the black holes. If they are moving fast enough,they may shoot out of their galaxies, but it is likely that the stars of the galaxies will follow the 'holes. This leaves you with gasless galaxies.

That can't be right. These black holes (you did not mention their mass) will drag the stars of the entire galaxy with them? Even a pair of supermassive black holes which might undergo interactions in a pair of merging galaxies won't be dragging a significant fraction of stars should one or both be ejected. And why gasless? How does the gas escape the host galaxy due to the ejection of the supermassive black hole? I can believe that they are able to maintain their gaseous accretion disks, but that content is tiny in comparison to mass of gas in entire galaxies.

Consider -- the typical supermassive black hole sitting at the center of a typical massive galaxy has a radius of influence of just a few 10s of light years. This characteristic radius looks like this:

r_bh = GM_bh/(sigma)^2

where M_bh is the mass of the black hole and sigma is the velocity dispersion of the stars due to the gravitational potential of the galaxy. So this is the characteristic distance beyond which the galaxy's gravitational influence dominates over that of the central supermassive black hole.

[Seeka]

Well i'll admit the part about a gasless galaxy did confuse me but thought i was just not 'getting it' so i didn't want to challenge it.

[Seeka]

It's:

v_e=sqrt(2GM/r)

Mugaliens this means absolutely nothin to me!! but thanks for posting it

[korjik]

That can't be right. These black holes (you did not mention their mass) will drag the stars of the entire galaxy with them? Even a pair of supermassive black holes which might undergo interactions in a pair of merging galaxies won't be dragging a significant fraction of stars should one or both be ejected. And why gasless? How does the gas escape the host galaxy due to the ejection of the supermassive black hole? I can believe that they are able to maintain their gaseous accretion disks, but that content is tiny in comparison to mass of gas in entire galaxies.

Consider -- the typical supermassive black hole sitting at the center of a typical massive galaxy has a radius of influence of just a few 10s of light years. This characteristic radius looks like this:

r_bh = GM_bh/(sigma)^2

where M_bh is the mass of the black hole and sigma is the velocity dispersion of the stars due to the gravitational potential of the galaxy. So this is the characteristic distance beyond which the galaxy's gravitational influence dominates over that of the central supermassive black hole.

I didnt say that the holes dragged the stars, but that the stars take the same path.

Compared to the total mass acting on the hole or the star, the star or hole is a dense low mass object. In any local region, the dynamics acting on one star will be similar to that on its neighbors. What happens to one will be similar to what happens to all. So, if the hole ends up on a parabolic orbit away from the galaxy collision, most of the stars will too, and you will end up with most of the stars in approximately the same orientations after the collision is over.

When you get down to free gas in two colliding galaxies, the gas is low density low mass objects. The dynamic friction between the two galaxies acts on the free gas much more strongly, and can pull it out of the galaxies.

This does assume that the interaction between the two galaxies is not so great to completely disrupt one or both.

[korjik]

Mugaliens this means absolutely nothin to me!! but thanks for posting it

escape velocity is the square root of the product of twice the mass of the object you are escaping from (M) and the gravitational constant (G) divided by the distance to the center of the object you are escaping from (r).

[Chunky]

the speed at which you reach escape velocity, is that the "default" speed of space....


is this hijacking a thread?
if it is then this isnt what i did last time.

[mugaliens]

Mugaliens this means absolutely nothin to me!! but thanks for posting it

LOL!

Ok, let's go into some detail...

v_e=sqrt(2GM/r)

It says the escape velocity is equal to the square root of 2 times the gravitational constant, G, times the mass of the body, M, divided by the distance from the center of the body.

G is always 6.67428 x 10^-11 m³/kg-s

Thus, for the Earth:

M = 5.9736 x 10²⁴ kg

r = 6,371 km = 6,371,000 m (this is the mean radius of the Earth's surface from it's center)

Thus, plugging and chugging...

v_e=sqrt(2GM/r)=sqrt(2*6.67428 x 10^-11 m³/kg-s*5.9736 x 10²⁴ kg/6,371,000 m)

v_e=11,187 m/s
v_e=36,704 ft/s
v_e=25,026 mph

This is the escape velocity at the surface of the Earth. It doesn't take into account the limitating effects of air friction.

Put simply, if there were no atmosphere on the Earth, and if you were the Flash, and you got up to 25,026 mph, you would be weightless, and would continue falling around the Earth. Actually, since this is .36 mph faster than the actual escape velocity, you would spiral outward until you left the Earth's gravitational influence altogether.

Because the operators are under the square root, it also tells you that if you were to go four times the distance from Earth's center, your escape velocity requirement would only be halved. Alternatively, for a planet four times as massive as Earth, but with the same dimensions, your escape velocity requirement would only be doubled.

[Jeff Root]

Put simply, if there were no atmosphere on the Earth, and if you were
the Flash, and you got up to 25,026 mph, you would be weightless, and
would continue falling around the Earth. Actually, since this is .36 mph
faster than the actual escape velocity, you would spiral outward until
you left the Earth's gravitational influence altogether.

"Spiral" probably isn't the most accurate term. In a universe where
the Earth is all alone in space, not orbiting the Sun, the trajectory
would be a hyperbola. Just barely different from a parabola. If you
jumped straight up at the same speed, instead of getting a running
start along the ground, your trajectory would be a straight line.

The weirdest thing about escape speed is that it doesn't matter
which direction you are headed, the speed is the same. It took
some time for me to comprehend how that can be. It is also the
reason that the term really should be "escape speed" rather than
"escape velocity".

-- Jeff, in Minneapolis

[Seeka]

Ok thanks for breaking that down for me i can understand the equation better now. I think that equation was written up on the board at my astronomy club meeting, the profressor sang it off like it was nothin to him
Will that equation always be the same no matter what you're applying it to?

Thanks again!

[Seeka]

The weirdest thing about escape speed is that it doesn't matter
which direction you are headed, the speed is the same. It took
some time for me to comprehend how that can be. It is also the
reason that the term really should be "escape speed" rather than
"escape velocity".

-- Jeff, in Minneapolis

The Prof said that also Jeff, it was the wrong term to use.

[mugaliens]

"Spiral" probably isn't the most accurate term. In a universe where
the Earth is all alone in space, not orbiting the Sun, the trajectory
would be a hyperbola.

Ever have one of those days when you wish you could (legally) go back and edit your own post to correct an obviously out to lunch oversite???

True! All velocities less than escape velocity will result in an ellipse. Even the trajectory of a baseball on an airless world, assumed to be a parabola, is really just the pointy end of an ellipse.

Thus, I was wrong! Flash would orbit the Earth at orbital velocity, which, for a circle, is given by v_o=sqrt(GM/r).

Thus, it's easy to see that v_e=sqrt(2)*v_o



So, unless Flash can swim through air, or add about 41% more velocity in that last step, he'll never leave orbit.

The weirdest thing about escape speed is that it doesn't matter
which direction you are headed, the speed is the same.

Well, so long as it's up (above the horizon), not down. When it's down, the speed rapidly diminishes to zero.

Ok thanks for breaking that down for me i can understand the equation better now. I think that equation was written up on the board at my astronomy club meeting, the profressor sang it off like it was nothin to him

Will that equation always be the same no matter what you're applying it to?

Thanks again!

Yes, it will. The neat thing is when you're in a circular orbit, and you want to compute how much additional velocity you need to add in order to reach escape velocity, you can use v_o to compute your current velocity, and v_e to compute your escape velocity. Then, simply subtract the two. Assuming your burn is relatively short, use F=ma and it's derivative, v_f²=v_i²+2*a*t, to compute the required duration of your burn.

[Seeka]

Ever have one of those days when you wish you could (legally) go back and edit your own post to correct an obviously out to lunch oversite???

True! All velocities less than escape velocity will result in an ellipse. Even the trajectory of a baseball on an airless world, assumed to be a parabola, is really just the pointy end of an ellipse.

Thus, I was wrong! Flash would orbit the Earth at orbital velocity, which, for a circle, is given by v_o=sqrt(GM/r).

Thus, it's easy to see that v_e=sqrt(2)*v_o



So, unless Flash can swim through air, or add about 41% more velocity in that last step, he'll never leave orbit.



Well, so long as it's up (above the horizon), not down. When it's down, the speed rapidly diminishes to zero.



Yes, it will. The neat thing is when you're in a circular orbit, and you want to compute how much additional velocity you need to add in order to reach escape velocity, you can use v_o to compute your current velocity, and v_e to compute your escape velocity. Then, simply subtract the two. Assuming your burn is relatively short, use F=ma and it's derivative, v_f²=v_i²+2*a*t, to compute the required duration of your burn.

Excellent stuff. Mugaliens can i ask what you do for a living?
Regards
Steff.

[mugaliens]

Excellent stuff. Mugaliens can i ask what you do for a living?
Regards
Steff.

Thanks, Steffanie! Among other things, I teach, though not in what you might call a traditionally academic setting.