This is not entirely true. At least in principle, the Sun can be divided into two black holes. One consists the Sun’s north half, the other one of the Sun’s south half, each going in the opposite direction with the nearly light speed, perpendicular to the ecliptic plane.
Earth would feel a rapidly fading gravitational pull and the darkness.
The question is, can we do it even better? Can the Sun disappear in the opposite direction, away from us with the (nearly) speed of light, without spraying us with some deadly rays?
It would work, to a degree. But for the majority of the Sun’s mass to escape from us with the nearly light speed, it wouldn’t. The small pieces should be very close to the speed of light and thus very heavy for an observer here on Earth. That would mean an even bigger gravitational pull toward the ex-Sun, for a while.
The question is, can we do it even better? Can the Sun disappear in the opposite direction, away from us with the (nearly) speed of light, without spraying us with some deadly rays?
Sure, it just has to make sure it’s rays (or relativistically accelerated mass packets) miss us. Shooting two (or more) simultaneous streams of mass on appropriate vectors that add up to directly away from us should work fine… or it would in a true vacuum. Depending on the details of the implementation I’d be a little worried about indirect rays hitting earth due to collisions with dust.
Yes, it can. Can anybody figure it out how?
What is your solution? The obvious one (above) requires the Sun sacrificing rather a lot of mass in the process of running away. Does your solution avoid this?
I have a solution, which by its effect, is nearly indistinguishable from the “instantly disappearing Sun”.
You have to convert all the Sun’s mass into a Dyson sphere at the Mercury’s orbit. For the Sun’s now missing light, it can be a hot spot somewhere on the Dyson. Arbitrary perfect imitation of the Sun shinning on the sky for us.
Now, the sphere has started to inflate at the rate of a few km per second. Our orbit would be just the same until the moment the Dyson sphere reaches us. At the last few million kilometers the sphere’s inflation accelerates and we are enveloped inside otherwise empty sphere in a matter of seconds. We are in zero gravity of a hollow sphere, shortly after everything still seemed normal. Of course, there is a hole in the surface of the DS just big enough that no collision takes place.
An aside for those curious about the Gauss Law argument. The law in its integral form states that the flux of the gravitational field inward through any closed surface encompassing the Sun is proportional to the Sun’s mass.
As long as the mass distribution is spherically symmetric the gravity outside of the sun is the same as if the mass was all located at the center. It’s the same for electrostatic force since that goes like 1/r^2 too :D.
Oh, that’s what the gravity from a hollow sphere all adds/multiplies out to? Uniform zero (net) gravity inside, normal outside the sphere? Neat.
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center. (Suppose that there was a normal net effect everywhere at and just above the surface and zero net effect just below the surface: that would require that the gravitational field not be continuous, which is not the case)
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center.
Remember that this is a spherical shell, not a sphere. When calculating the force of gravity inside the Earth, for example, you ignore all mass at higher radius that the location you’re interested in, but not the mass at lower radius.
Why does this work? Imagine opposite cones that originate at the center of the spherical shell. The intersection of that cone with the shell will have surface area that increases with r^2, but the inverse square law decreases with r^2, and so the gravitational effect only depends on the angle of the cone, regardless of distance- but the two cones are pointed in opposite directions, and so cancel out. (This is obvious at the center, but works just as well elsewhere inside the spherical shell.)
that would require that the gravitational field not be continuous, which is not the case
For an infinitely thin shell, the drop in gravitation is infinitely steep. But shells with real thickness will have a gradual dropoff that corresponds to that thickness.
It’s not clear to me where that model came from. We shouldn’t expect the density of celestial bodies to be uniform unless they’re made of something incompressible, and it’s important to separate out the net force of gravity and the weight of mass above you. In steady state, all of the mass in the celestial body is being pulled towards the center by gravity, pushed upwards by the mass below it, and has to push upwards on the mass above it (with the net force being 0). I haven’t done the math to see what the pressure function would look like for a gaseous celestial body, and it seems like the full calculation will have lots of complications, but we can note that the mass below you has to push up harder than the mass above you is pushing down, suggesting the pressure is highest at the center.
delta-P is the first derivative of pressure; it would have to be zero at the center for there to be a pressure maximum at zero.
I would expect a gaseous body to have a roughly spherically symmetric mass distribution, which is all we need. Treat it as an infinite number of infinitely thin spheres each of uniform density, and we can do calculus on it.
We can also do this though experiment with a perfect liquid of uniform density; at least it will have a surface that we can stop at. Pressure is still highest at the center and reality is continuous, meaning dP/dR is zero at the center and approaches zero as R approaches zero.
Surface gravity of a sphere of constant density and radius R is proportional to R? Mass is proportional to volume (R^3) and surface gravity is proportional to mass/R^2, or R^3/R^2, or R.
Okay, I’ve got a new respect for the problems involved with using barometric pressure to measure altitude, and the advantages of using barometric pressure directly for navigational purposes at high altitudes.
At the time of writing, I thought I was paraphrasing Newton. Downbranch I realized that the behavior of being attracted to the center of a shell while inside the shell would, in a body composed entirely of fluid, yield a maximum first derivative of pressure at the center, and no maximum of pressure within the body.
This is not entirely true. At least in principle, the Sun can be divided into two black holes. One consists the Sun’s north half, the other one of the Sun’s south half, each going in the opposite direction with the nearly light speed, perpendicular to the ecliptic plane.
Earth would feel a rapidly fading gravitational pull and the darkness.
The question is, can we do it even better? Can the Sun disappear in the opposite direction, away from us with the (nearly) speed of light, without spraying us with some deadly rays?
Yes, it can. Can anybody figure it out how?
This still conserves mass, since black holes have mass. It also obeys the Gauss Law
Who’s downvoting this? It’s obviously correct, while the post it’s responding to is completely wrong on its first, most important statement.
Would this work?
It would work, to a degree. But for the majority of the Sun’s mass to escape from us with the nearly light speed, it wouldn’t. The small pieces should be very close to the speed of light and thus very heavy for an observer here on Earth. That would mean an even bigger gravitational pull toward the ex-Sun, for a while.
Sure, it just has to make sure it’s rays (or relativistically accelerated mass packets) miss us. Shooting two (or more) simultaneous streams of mass on appropriate vectors that add up to directly away from us should work fine… or it would in a true vacuum. Depending on the details of the implementation I’d be a little worried about indirect rays hitting earth due to collisions with dust.
What is your solution? The obvious one (above) requires the Sun sacrificing rather a lot of mass in the process of running away. Does your solution avoid this?
I have a solution, which by its effect, is nearly indistinguishable from the “instantly disappearing Sun”.
You have to convert all the Sun’s mass into a Dyson sphere at the Mercury’s orbit. For the Sun’s now missing light, it can be a hot spot somewhere on the Dyson. Arbitrary perfect imitation of the Sun shinning on the sky for us.
Now, the sphere has started to inflate at the rate of a few km per second. Our orbit would be just the same until the moment the Dyson sphere reaches us. At the last few million kilometers the sphere’s inflation accelerates and we are enveloped inside otherwise empty sphere in a matter of seconds. We are in zero gravity of a hollow sphere, shortly after everything still seemed normal. Of course, there is a hole in the surface of the DS just big enough that no collision takes place.
Oh, that’s what the gravity from a hollow sphere all adds/multiplies out to? Uniform zero (net) gravity inside, normal outside the sphere? Neat.
As long as the mass distribution is spherically symmetric the gravity outside of the sun is the same as if the mass was all located at the center. It’s the same for electrostatic force since that goes like 1/r^2 too :D.
adds
Yep! Newton had a proof in Principia, but here’s a more recent, geometric one.
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center. (Suppose that there was a normal net effect everywhere at and just above the surface and zero net effect just below the surface: that would require that the gravitational field not be continuous, which is not the case)
Remember that this is a spherical shell, not a sphere. When calculating the force of gravity inside the Earth, for example, you ignore all mass at higher radius that the location you’re interested in, but not the mass at lower radius.
Why does this work? Imagine opposite cones that originate at the center of the spherical shell. The intersection of that cone with the shell will have surface area that increases with r^2, but the inverse square law decreases with r^2, and so the gravitational effect only depends on the angle of the cone, regardless of distance- but the two cones are pointed in opposite directions, and so cancel out. (This is obvious at the center, but works just as well elsewhere inside the spherical shell.)
For an infinitely thin shell, the drop in gravitation is infinitely steep. But shells with real thickness will have a gradual dropoff that corresponds to that thickness.
… So pressure in a gaseous celestial body doesn’t increase linearly; delta-p falls off to zero in the center?
It’s not clear to me where that model came from. We shouldn’t expect the density of celestial bodies to be uniform unless they’re made of something incompressible, and it’s important to separate out the net force of gravity and the weight of mass above you. In steady state, all of the mass in the celestial body is being pulled towards the center by gravity, pushed upwards by the mass below it, and has to push upwards on the mass above it (with the net force being 0). I haven’t done the math to see what the pressure function would look like for a gaseous celestial body, and it seems like the full calculation will have lots of complications, but we can note that the mass below you has to push up harder than the mass above you is pushing down, suggesting the pressure is highest at the center.
delta-P is the first derivative of pressure; it would have to be zero at the center for there to be a pressure maximum at zero.
I would expect a gaseous body to have a roughly spherically symmetric mass distribution, which is all we need. Treat it as an infinite number of infinitely thin spheres each of uniform density, and we can do calculus on it.
We can also do this though experiment with a perfect liquid of uniform density; at least it will have a surface that we can stop at. Pressure is still highest at the center and reality is continuous, meaning dP/dR is zero at the center and approaches zero as R approaches zero.
Surface gravity of a sphere of constant density and radius R is proportional to R? Mass is proportional to volume (R^3) and surface gravity is proportional to mass/R^2, or R^3/R^2, or R.
Okay, I’ve got a new respect for the problems involved with using barometric pressure to measure altitude, and the advantages of using barometric pressure directly for navigational purposes at high altitudes.
Can you please use questions instead of confident contradiction when you’re up against Newton?
At the time of writing, I thought I was paraphrasing Newton. Downbranch I realized that the behavior of being attracted to the center of a shell while inside the shell would, in a body composed entirely of fluid, yield a maximum first derivative of pressure at the center, and no maximum of pressure within the body.
This will indeed work, but it has no relation to your original puzzle:
It is your frequent remark: “What has this to do with what has been previously said?”
;)
Why bother with black holes?
Be cause they are cool. They don’t shine very bright. As long as they encapsulate all the matter and don’t have any accretion discs.
Otherwise you have to cool down everything for many million Kelvins.
Better question: why not bother with black holes? I mean, since we’re talking about the implications doing crazy impossible stuff to reality anyway...
Mind you, I can’t find any reference or allusion to black holes in the grandparent so I’m not entirely sure why you are asking me.