# Why the gravity of a black hole is so strong by the Bad Astronomer

In this bad astronomer video, Phil Plait says two things:

1) He first demonstrates the effect of gravity using paper, and he keeps a fixed mass and distance. However as he makes the paper smaller and smaller keeping the mass the same, he says that the particle can get more closer than before, or gravity increases. However, towards the end of the video, he says that the star before collapsing has a mass of 20-40 times the mass of the sun, but once it does, it reduces to 3-4 times. But in this example he is changing the mass for two different situations. How does this fit in with the crushed paper analogy?

2) Again, towards the end of the video, he says that if the sun were to become a black hole, the Earth would continue orbiting as usual because it experiences the same gravitational force as before. So if there is no change in the gravitational pull what makes the sun a black hole? Just the fact that it is the core of a collapsed star?

And using the above example of the sun becoming a black hole, when does a black hole actually start "sucking" things into it?

• "when does a black hole actually start "sucking" things into it?" It doesn't. It's gravitational effect is the same as that of the star it was. – hdhondt Apr 20 '19 at 5:08

The point Phil Plait is making is that gravity depends only on mass and distance. For example the gravitational acceleration the Earth experiences due to the Sun's gravity is:

$$a = \frac{GM}{r^2}$$

where $$M$$ is the mass of the Sun and $$r$$ is the distance from the Earth to the Sun ($$G$$ is a constant called the gravitational constant). The size of the Sun does not appear in this equation, so if we able to magically shrink the size of the Sun until it became a black hole the Earth wouldn't feel any difference in its gravity.

But in real life we can't magically shrink stars. A black hole forms when a star much more massive than the Sun goes supernova, and the explosion blasts away most of the star leaving only a much smaller mass behind. That's why Phil Plait says that a star 20-40 times as massive as the Sun leaves behind a black hole only 3-4 times more massive than the Sun. Most of the star's mass got blasted away in the supernova that formed the black hole.

• Yes but he's changing the mass. When in the paper ball situation, he doesn't. I don't get this part – noorav Apr 20 '19 at 5:13
• @noorav In real life you couldn't do the experiment Phil Plait describes, because the only way to make a black hole is from a supernova and that changes the mass. He is describing some purely theoretical experiment of shrinking down a star to make a black hole without the accompanying supernova. – John Rennie Apr 20 '19 at 5:26
• @JohnRennie The size of the Sun does not appear in this equation, so if we able to magically shrink the size of the Sun until it became a black hole the Earth wouldn't feel any difference in its gravity. This statement must be qualified. First, it's far from trivial. In GR it follows from Birkhoff's theorem - and if it's a theorem then it has a proof, requiring Einstein's equations etc. – Elio Fabri Apr 20 '19 at 9:34
• @JohnRennie Second, stated as you did it's not exactly true, but only asymptotically, i.e. at distances ${}\gg GM/c^2$. What is exactly true, for a spherically symmetrical sun, is that planets motions wouldn't depend on sun's size but only on its mass. But motion in GR isn't the same as it's in Newtonian gravitation (see e.g. perihelion precession). – Elio Fabri Apr 20 '19 at 9:35
• @ElioFabri yes, I agree, but the OP is asking about an elementary level video. I don't think it's appropriate to go into a lot of detail. If there's a follow up question wanting a deeper analysis that would be the place for GR to rear its head. – John Rennie Apr 20 '19 at 10:09

Ok, stay with me on this. There are two vectors acting on the earth. The first deals with the inertial mass and the other with the gravitational mass. These two vectors are set at 90 deg. to each other (the angle of least action). The velocity of the first vector (the orbital vector) keeps the earth in stable orbit and can not vary because a faster velocity will push the earth out of orbit and into outer space, a slower velocity will cause the earth to spiral in and crash into the sun. Think of the earth as traversing the surface of a mathematical sphere. The earth can make no progress toward or away from the sun because the curvature of the sun exactly matches the action produced by the gravitational (radial) vector deflecting the earth off a straight line path. The mass of the sun and earth have no extension in space. They are just mathematical points with X and Y magnitudes of mass. The earth stays in stable orbit per Newton's third law based on adding the earth and suns mass, the radial distance between the earth and the sun, Newton's gravitational constant, and the orbital velocity which is not part of Newtons law. If there were no orbital velocity (vector velocity set to zero) the earth would drop directly towards the center of the sun. On the way, however, it would encounter another sphere representing the limit of the suns mass density. Now imagine a series of nested spheres from here on down getting smaller and smaller as we approach the suns center. Pick any sphere and it will have a gravitational force associated with it. The smaller the sphere the greater the force (swapping sphere surface area for force). Now the questions are, is there any sphere who's gravitational force is great enough to accelerate the earth up to light speed? Has the sun's mass density been squeezed sufficiently enough tho allow the earth to drop to this surface? In a black hole both the mass and mass density will be concentrated at a single mathematical point called a singularity per Einstein but not per quantum theory. Is the suns mass sufficient to generate this force at this radius? If no then a Swartzchild radius (the mathematical spherical radius where the gravitational force counterbalances the force of light trying to get out) and black hole will not form. The sun does not have sufficient mass to create a Swartzchild radius at any distance (radial). The minimum mass I believe is 1.3 suns mass. The mathematical sphere with Swartzchild radius is real in the sense that it prevents light from behind it from reaching our eye (except for lensing effects). In essence we are looking at a region of nothingness.