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I've seen in a documentary that when a star collapses and becomes a black hole, it starts to eat the planets around.

But it has the same mass, so how does its gravitational field strength increase?

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Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes. – Kyle Kanos Aug 14 '14 at 1:22
As an aside, power is almost certainly not what you mean. Power is the rate of energy transfer with respect to time. I expect what you mean is gravitational field strength. – Tom W Aug 14 '14 at 10:55
Yes, that's what i meant. Thank you. – xxxo Aug 14 '14 at 11:07
If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Aug 14 '14 at 12:50
up vote 41 down vote accepted

Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes.

The Schwarzschild radius is the radius at which, if an object's mass where compressed to a sphere of that size, the escape velocity at the surface would be the speed of light $c$; this is given by $$ r_s=\frac{2Gm}{c^2} $$ For a 3-solar mass black hole, this amounts to about 10 km. If we measure the gravitational acceleration from this point, $$ g_{BH}=\frac{Gm_{BH}}{r_s^2}\simeq10^{13}\,{\rm m/s^2} $$ and compare this to the acceleration due to the precursor 20 solar mass star with radius of $r_\star=5R_\odot\simeq7\times10^8$ m, we have $$ g_{M_\star}=\frac{Gm_\star}{r_\star^2}\simeq10^3\,{\rm m/s^2} $$ Note that this is the acceleration due to gravity at the surface of the object, and not at some distance away. If we measure the gravitational acceleration of the smaller black hole at the distance of the original star's radius, you'll find it is a lot smaller (by a factor of about 7).

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A star can be so big that its collapse to a black hole permits no supernova, hence no mass is lost in this way even locally. Some mass/energy is lost by a gravitational wave which may 10-20% of the available mass/energy. This may allow some planets escape from the system. The local amplitude of the gravitational wave might do far more damage to planets than the supernova. More stuff for science fiction. The distant observer, us with our instruments and LIGO, are in for a much more interesting time. Science fiction is boring compared to science fact.

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If you measure the large-distance strength of the gravitational acceleration $g\approx \frac{GM}{r^2}$ of a star / black hole with the assumption that your distance $r$ is much further out than the various mass parts, shock wave, and ejected material; then $g\approx \frac{GM}{r^2}$ is (within a percent or so) the same before and after the supernova. This is so (i) because we can (to that precision) ignore the fraction of energy in the supernova coming from ultra-relativistic particles consisting mainly of neutrinos, and (ii) because of averaging effects similar to Newton's shell theorem and Birkhoff's theorem.

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Hah. The other answerers thought they were smarty-pants by pointing out that "when a star become a black hole, much is blown off!" BUT. Qmechanic gets the last laugh. of course, all the mass is still there in the "same spot". Awesome answer. – Joe Blow Aug 15 '14 at 11:23
@JoeBlow If the $r$ is the same before and after, then it would be about the same. But traditionally $r$ would be larger only after the nova and then it might include some planets or other bodies making it more massive. – user6972 Aug 15 '14 at 18:27
Note that $r$ in the answer (v2) refers to the distance to an observer, not the radius of the star / black hole. – Qmechanic Aug 15 '14 at 18:52
@JoeBlow: The ejecta moves at about 1/3 the speed of light. After a day, that's already $\sim10^{15}$ cm which is about 100 AU, well past your "same spot." For reference, former planet Pluto is between 30 and 50 AU from our sun. – Kyle Kanos Aug 15 '14 at 19:29
When @Qmechanic says "large-distance", he means really, really large. Once you're inside a shell, its mass no longer contributes to the gravitational acceleration. – Snowbody Aug 15 '14 at 20:55

When you watch a pop-sci TV show, you need to take everything you see with a very healthy grain of salt. This is particularly the case if the show's host isn't a scientist, but even when a scientist is the host, you need to be suspicious.

Stellar black holes do not turn into monsters that reach out and pluck objects from the heavens. From far away, a black hole behaves no differently gravitationally than an ordinary object of equal mass. It's only when an object gets very close that black holes behave differently. Note that this "very close" means what would be well into the interior of the ordinary object.

If anything, stellar black holes are little kitty cats rather than big monsters compared to the stars that generated them. The supernovae that generate stellar black holes blow away a very large portion their mass during the supernova event, both as energy and ejected matter. The resulting black hole has a much smaller mass than did the parent star.

If the parent star is a member of a close binary star, the black hole might still be able to draw mass from the other star. But reaching out and inhaling planets? That's just bad pop-sci.

Except for the outer atmosphere of a red giant that is a close binary pair of a stellar black hole, it would be quite amazing for a stellar black hole to gobble up anything. It would take a lot of energy to intentionally send something very close to a black hole. By way of analogy, mankind has sent four satellites out of the solar system (with a fifth on the way) but we have only sent two missions to Mercury. The reason is that takes a lot of energy (a whole lot of energy!) to get to Mercury. Escaping the solar system is a piece of cake compared to getting to Mercury. It would take even more energy to get very close to the surface of the Sun. If our Sun was instead a one solar mass black hole, it would take a whole, whole lot more energy to send something within a few Schwarzschild radii of the black hole.

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Regarding Mercury, the tricky thing about getting there is the delta-V between Earth's orbit and Mercury's. A stationary object outside the solar system would not need to expend any energy to get to (and fly-by or crash into) Mercury, as it would just be falling down the Sun's gravity well. Killing the speed gained during this process enough to enter orbit around Mercury would of course mean large energy expenditure. – engineerC Aug 14 '14 at 3:31
@LorenPechtel -- With sci-fi I know that it's not real. The problem is pop-sci shows that try to portray what science is all about. Unfortunately, most of them mystify rather than clarify. I do know some science, and of the science I do know, I know the stuff on those pop-sci shows is often pure garbage. That leads me to wonder, how can I trust those pop-sci shows on the subjects where I am just an interested layman? Popularization of science can't use equations, I understand that. But that doesn't mean they have to fall into quantum woo / chemical woo / biological woo / whatever woo. Yech! – David Hammen Aug 14 '14 at 3:33
"The reason is that takes a lot of energy (a whole lot of energy!) to get to Mercury. Escaping the solar system is a piece of cake compared to getting to Mercury. It would take even more energy to get very close to the surface of the Sun." I can understand that escaping from orbit around Mercury or the Sun, or indeed not crashing into them at all would be very difficult; but talking about just getting till there, wouldn't going towards a massive object be easier than going away from it due to gravity? – YatharthROCK Aug 14 '14 at 6:19
@YatharthROCK: If you were a stationary object at rest relative to the Sun, yes, you could just wait and fall. But if you're orbiting, the situation is very different. To stay in orbit, you need a lot of sideways velocity - and to fall into the Sun, you need to ditch a large fraction of that velocity. Indeed, the velocity change is larger than what's needed to escape from the Solar System! – JohannesD Aug 14 '14 at 14:58
@YatharthROCK: The solar orbit, of course! The Earth is orbiting the Sun, moving at about 30 km per second, and of course everything else in the Solar System is orbiting the Sun as well. To fall into the Sun from a 1 AU radius heliocentric orbit, you need to get rid of almost all that speed; on the other hand, to escape the Solar System from the same orbit you only need circa 12 km/s of extra speed. – JohannesD Aug 14 '14 at 22:03

It actually goes the other way around: when a star collapses to form a black hole, its planets (if it has any) will become unbound and fly away to infinity.

Simple reason: when the star explodes to form a compact object (neutron star or black hole), it releases most of its mass in the form of a SuperNova explosion, so that the central object around which the planet is orbiting has a much smaller mass than the original star. The least decrease is roughly from an $8 M_\odot$ star to a $1.4 M_\odot$ neutron star, giving a minimum reduction of about a factor of 6.

Now let us consider what happens to the planet. Before the explosion, its kinetic energy $K$ is half the modulus of its potential energy $W$: $$ K = -\frac{1}{2} W $$ so that its total energy $E = T+W = -T/2 < 0$, and the planet is bound to the star.

But after the explosion, while the planet speed is left unchanged, its potential energy $W = -GM_\star M_{planet}/r$ is reduced because $M_\star$ has decreased by at least a factor of $6$: the new potential energy $-W_{final} < -W_{initial}/6$. Hence the new energy

$$ E = T + W_{final} = -\frac{1}{2}W_{initial} + W_{final} > -\frac{1}{3} W_{initial} > 0\;: $$

the final , total energy is positive, the planet is unbound from the star, it will just fly away from it.

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This applies once the ejecta pass the planet's orbit, right? – Charles Aug 14 '14 at 13:40
Actually, when a star collapses to form a black hole, its planets are typically fried. The supernova that precedes the formation of the black hole takes care of the pesky planet problem. – David Hammen Aug 14 '14 at 15:05
@DavidHammen Not necessarily, depends on planet mass and distance from the star. For instance, Jupiter has a binding energy $G M_J^2/R_J \approx 4\times 10^{43}\; erg$. However, of the total energy released by a typical SN ($10^{51}\;erg$, neutrinos do not count), it intercepts a fraction $\delta\Omega/4\pi = R_J^2/4D_J^2 \approx 3\times 10^{-9}$, which gives $3\times 10^{42}$ erg, which is only $0.1$ of the binding energy. Thus a Jupiter at its current location would not be evaporated. – MariusMatutiae Aug 14 '14 at 16:37
@MariusMatutiae although it would be fried :) – Tim B Aug 15 '14 at 9:57
@TimB Yes, by all means, no hope to get out alive... too many X-ray and UV photons. – MariusMatutiae Aug 15 '14 at 10:28

protected by Qmechanic Feb 15 at 8:13

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