I was reading this page on The supposed "Official String Theory" website that posed this question.

Question: "Under what conditions would the escape velocity from the surface of some planet or star be equal to the speed of light?"

Answer: "For a planet the mass of the Earth, this distance is only about a centimeter. So if the Earth were less than a centimeter in diameter, the escape velocity at the surface would be greater than the speed of light."

The answer it gave is what brought me to this question. Am I misunderstanding this answer, or is it really saying that if we shrunk the Earth to a size slightly smaller than a centimete and kept its mass in tact, you would have to travel greater than the speed of light to escape? In other words that centimeter sized earth's gravity would be as powerful as a black hole?

This leads me to a couple of questions....

1) Does Size affect gravity or is it only the mass that affects gravity?
2) If so, how?
3) Could you not leave the ground in a machine going 300mph, or so, and fly stright up into space and eventually beyond the earths gravity? Or do you really have to be going 11km/sec to get into space. Or does escape velocity only refer to the speed you have to be going to keep a stable orbit around the earth?

Thanks!

AFAWK, gravitational pull is directly proportional to the mass of an object, and inversely proportional to the distance from the center of mass of the object (if you are outside the object).

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Escape velocity is the velocity required to escape the gravitational attraction of a body without further acceleration, ie coasting. (and of course, ignoring factors such as friction from the atmosphere or the interplanetary medium).

Since gravitational attraction is inversely proportional to the distance from the center of mass of an object, at 300mph you will eventually reach a distance for which the escape velocity is 300mph.

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It would be a black hole.
That is pretty much the definition of a BH.

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So you are saying yes then? I know what inversely proportional means, but I am not sure I really understand your answer heh.

But I think you are saying that the closer you are to the center of mass (Assuming you are outside the object) the more the gravitational pull is.

It seems that size has waaaaay more effect on gravity than actual mass does. We experience 1g with the earth the size it now, but if the earth were say half its size (with same amount of mass) we would experience not twice as much gravity, but (without doing the actual math) something like 2000 times as much gravity, for example? Or if it were twice as large (with same amount of mass) we would experience like 1/2000th of a g? The numbers are arbitrary, but is that right?

Yah, umm thats what I said. As powerfull as, same as , etc...

I see, so if you had enough fuel, you could esacpe earths gravity going 1mph..that is rediculous of course, but that is what I was wondering.

Thanks!

Yes, absolutely. As long as you had enough fuel, you could escape the earth's gravity at 1 mm/hour. It would take a VERY long time, though. The point being, the force grows weaker as distance increases, so you would eventually get to a point where 1 mph would be faster than the escape velocity.

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It's not that size has more of an effect than mass, it's the relationship between the two. I get the impression that you understand this, at least intuitively, and are just having trouble phrasing it, but, just to be clear: if you have a marble, and you have the mass of the earth compressed to the size of a marble, the earth is going to have a WAY greater force of gravitation.

The radius any given mass has to be compressed to to become a black hole is called the Schwarzschild radius, and the smaller the mass, the smaller the radius. This link has a good explanation of this effect, so rather than babble on I will send you to the experts (scroll down, black holes are the last heading).

This explanation has been interesting as we do not often ask the why to questions that seem so self explanatory. Planet Earth escape velocity from surface to space 11 k/m per sec.
On reading this thread I now sagest that the method to launch space craft is daft... (This must be wrong ). Load up the shuttle strap on a couple of solid boosters attach it all to the back of a B52. (somewhat modified...) and take off. No, I must be missing something... It cant be this simple.

I'm not sure what you mean. First of all, a B52 couldn't be modified to do that. Each solid rocket booster weighs 600 tons, and a B52's maximum payload is only 30 tons or so.

But even then, a B52 can't fly 11 km/second upward! It can't fly anywhere near that speed even horizontally. It's 39,000 km/hr.

Plus, the figure of 11 km/second would only work if there were no atmosphere. You have to accelerate as you get into the thinner atmosphere. Which a B52 can't do.

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He was kiddin' Jens.

Ah, that yellow thingie at the end of the sentence. My eyes must be giving out.

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I'm not sure where you got the "2000" figure ("arbitrary"?), but a half radius with the same mass would make the gravitational pull four times as much--it is inversely proportional to the square of the distance. That is the 1/R² in the GMm/R² formula for gravitational force.

Thanks hh, yah like I said I didn't do any math and didn't know the math to do. Thanks for clearing that up. So by that formula, if we halfed the earths size we would have 4g, if we halfed that again we would have 16g, etc..?

Exactly how many g's does it take to keep light from escaping? Or is that like asking how many licks does it take to get to the center of a tootsie pop?

You've just shown that can calculate it yourself, using the inverse quare law and the sizes of the real Earth and the hypothetical one shrunk to a black hole (1 cm). It's the ratio of their radii, squared.

Yah I was hoping there was a well known answer or someone else would do the math because I am lazy

How does that convert to g force? In other words, the worlds mass packed into 1cm sphere would = how many g's? Or in even other words, how many times would you have to half the earths size to get to 1cm? Simple enough I suppose. I guess I will do the math myself. Sorry for being lazy.

and, sorry Jens...

Arcane. If light at 300,000 km / sec. is held back by the force required to do that, exceeding 300,000 km / sec. how would you express that ? The term G is one gravity as we know it here. we can test this with just a set of scales. Any change to any of the mass densities and its all going to be wrong. How many G's of gravity to hold light. Lots. Many. Can we express this as foot pounds of force...No, not really.
The gravitational forces at or near the event horizon of any black hole are immense. The mathematics do not interest me other than to say the force is so strong that it distorts the very fabric of time and space. yep thats strong.

Ok, i think i did this wrong, but the number of G's that it takes to keep light from escaping is 4503599627370496?

I figured you would have to half the earth 27 times to get under 1cm, not sure that's exactly right though. Being off by 1 can change that number significantly.

That number would be universal no? Even though you are dealing with diffrent amounts of mass and different sizes depending onthe black hole, the bare minimum amount of G force should be universal....or not?

The term km/sec is only as we know it here, but we universaly use it to describe speed, why would G be any different?

Not

Escape velocity is calculated by summing the deceleration due to gravity from the surface to infinity. But something escaping from a one centimeter object experiences that tremendous gravity only for the first cm, and then it's a fourth of that. Whereas, if you had a Jupiter-wide object that had that gravity at its surface (OK, that's not going to happen, but you get the picture), you'd expereince that tremendous gravity all the way out to a distance the radius of the planet.

If the object is big enough, it can have a density less than water, and have an escape velocity greater than the speed of light.

I understand all of that. The point was, the bare minimum amount of G's that any object has to have in order for light to not be able to escape is the same, regardless of the size/matter ratio... If it could get past the event horizon then that amount of speed gets lower and lower in order to be able to escape.

You point about how long it takes for that amount of escape velocity to last is agreeable.

Am I still clueless? It doesn't seem like I am, but I could very well be

I guess what I am getting at is, it is interesting why at 4503599627370497 G all our laws of physics still apply, but at 4503599627370496 G they suddenly break down. Of course, those numbers are not precise, but that is the general idea.

We need to find out what happens in that 1 G difference that pushes our physics over the edge.

Actually it doesn't work that way at all. "G", the gravitational acceleration, is a coordinate dependent thing anyway. In Newton, 'g' has some notion of absolute meaning. In General Relativity, it does not. It is far, far from any absolute meaning, indeed.

And another, "indeed", just what you mean by 'g' becomes ambiguous. If you mean the proper acceleration required to remain stationary, which would be the instantaneous acceleration measured by a local stationary observer if he dropped something, then that goes to infinity at the event horizon anyway, no matter what the size of the black hole. So, 'g' tends to infinity at the horizon always. But this 'g' is not the same thing as the coordinate acceleration a far away observer would see using his own coordinates.

What is blowing up at the singularity of a black hole is an invariant, the *curvature* of space-time. (And there's all sorts of ways the curvature, which is a tensor, can blow up. The singularity of Schwarzschild is but one way that thing blows up).

The curvature of space-time, amongst many other many splendored things, correlates with the tidal forces felt by free fallers. I say correlates because components of that sucker, the tidal tensor, can blow up for free fallers even though the curvature tensor itself is well behaved. An example is a circular orbit as it approaches the the light speed, r = 1.5R limit. The radial and tangential tides along the velocity vector blow up there (you can see this in terms of something trying to remain stationary with you. At r = 1.5R, anyting trying to hold stationary below you would have to move faster than light. So obviously something has to break there. )

All that gets a wee bit complicated, especially when you throw in the magnetogravitic (tidal frame dragging) behavor. A wee bit complex indeed. Sort of makes me feel very small sometimes. Anyway, we tend to equate curvature with inertial tides, and this shows that relation is bit less exact than you might tend to think. But, a strong field is strong curvature, and that means free fallers experience stronger tiders, and no curvature means no free-fall tides.

Now I fear I've done gone and made things way too complicated again, but my point is this Newtonian 'g' business doesn't mean much outside the weak field, low velocity limit where GR approaches Newton.

-Richard

So an Earth mass has about (density varriations produce a small error)the same gravity 16,000 kilometers from the center, weather the radius is 15,000 kilometers or 15 kilometers. The surface gravity however is much greater for the high density Earth mass. When we go below the surface, the gravity increase is no longer the inverse square of the distance from the center. We can infact get a gravity decrease if the density does not increase much with depth. According to another thread, gravity of Earth is close to constant until you get about 2000 kilometers from the center, then the gravity drops rapidly to zero at the center. Neil

Publius, do you not agree that if the earth were 1.0001 cm in size that all laws of physics would work the same as they do on the earth we stand on now?

Do you not agree that if we made the earth 0.9999cm in size our laws of physics would break down and we would have a singularity?

Of course those numbers are not exact, but they are close and you get the idea.

At one specific point everything is fine, then you take a little away from that and all of a sudden everything breaks down.

What exactly happens at that point when everything breaks down. It would be nice to be able to observe how something reacts when it is teetering on the edge of being a black hole and then to see it as it actually transform into a black hole.

If you have a black hole, then General Relativity theory predicts a singularity within it. The problem is, we are sure that GR is incomplete, and in future theories, it is quite likely the singularities will go away.

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I was trying to explain, that that was not true.

For instance, the sun has 332,000 times as much mass as earth, but it is 109 times the radius. So the force of gravity at the surface of the sun is 332,000/109², or just 27 times stronger than earth gravity. If the sun were a black hole, its radius would be about 3 km, and its surface gravity would be (695,000/3)² more. In other words, at the surface of that black hole, it's only 1445520000000 g, which is about a thousandth of your 4503599627370496 (BTW, I tried to recalculate that, and I get 637014000², or 4x10¹⁷, which seems to be a hundred times as big as your value. In that case, the sun black hole would have a hundred thousandth the gravity.)

I will have to ponder on that, because I get what you are saying, but it just doesn't make any sense.

Thanks though.

Hey!

It makes sense. Unless I've made a mistake in the calculation, the numbers are right, but the point is that your basic contention, that there is some fundamental value of intense gravity which marks the threshold, is wrong.

These calculations were done a long time ago, long before relativity (even before Albert was a quantum), and were pondered over even then. The simplest way is to look at conservation of energy, kinetic plus potential: ¹/₂mv² - GMm/r

If an object leaves the surface at escape velocity v and ends up "at infinity" with zero velocity, both kinetic and potential energy (by the definition of potential energy) will be zero. By conservation, their sum must have been zero at the surface. That is, r = 2GM/v². Now, let v be the speed of light, 299792458 m/s. For earth mass 5.976x10²⁴ kg (radius = 6,378,140 m) and G=6.67x10^-11 m³kg^-1s^-2, we get an r of .0089 m, or about a centimeter, as was said before.

So, the calculation is simple, but it is an approximation--and would involve general relativity, really.

PS: So, to complete the calculations, the acceleration of gravity is GM/r², so at the surface of the earth, that is 9.8 m/s². At that one centimeter level, it's 5.07 x 10¹⁸ m/s², which is 5.17x10¹⁷g, again about a hundred times what you calculated.