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Neeuteron Star Falling Into Black Hole?



 
 
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Old May 3rd 19, 07:20 PM posted to sci.astro
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Default Neeuteron Star Falling Into Black Hole?

https://www.space.com/1865-black-hol...s-suggest.html

Here is something interesting about a neutron star falling in to a black hole.

the gravitational force on the near side of an object exceeds that of the far side of an object, because gravity decreases with distance. this is called the TIDAL FORCE. For people on the surface of Earth, the tidal forces between head and feet are negligible. However, Earth exerts significant tidal forces on the Moon, despite the Moon being further away, due to the Moon's size.

before we begin, let us calculate the force (in solar mass miles per second squared) exerted by a 1.5 solar mass, 12-mile radius neutron star upon a 0.001 solar mass "slice" on the surface.

In cubic miles per solar mass per second squared, the GRAVITATIONAL CONSTANT is

3.18479 E 10

The gravitational force of a neutron star upon its surface is

3.18749 E 10*1.5*0.001/12^2=33,030.2083 solar mass miles per second squared.

What happens if the neutron star's near side if forty-five miles from the center of a black hole with eight solar masses?

The black hole's pull on a 0.001 solar mass slice on the near side is

3.18749 E 10*8*0.001/45^2=125,925.5309 solar mass miles per second squared.

The far side is twenty-four miles from the near side, so the force exerted on the far side is

3.18749 E 10*8*0.001/69^2=53,560.0084 solar mass miles per second squared.

The tidal force is the difference. 125,925.5309-53560.0084= 72,365.5225 solar mass miles per second squared.

The tidal force exceeds the gravitational force of the neutron star itself, so the neutron star had already been torn apart.

an interesting thing happens. Neutron stars have this tremendous outbound pressure, called NEUTRON DEGENERACY PRESSURE. Only gravity from the neutron star's high mass and small size keeps the thing together. If a 1.5 solar mass neutron star is torn apart by tidal forces, then the pieces will not have enough mass to counteract the neutron degeneracy pressure, and they will explode. Some of the material will fall into the black hole, and some will escape to infinity. It is possible that the explosion of a neutron star fragment could create heavy elements, like the neutron star collision did.

for a 1.5 solar mass neutron star falling into an 8 solar mass black hole, when do tidal forces tear the neutron star apart? It has to be when the tidal force equals the neutron star's gravity. Using n as the near distance, we must solve these equations for n.

3.18749 E 10*8*0.001/n^2-3.18749E10*8*0.001/(n+24)^2=254,999,200/n^2-254,999,200/(n+24)^2=254,999,200(1/n^2-1/(n+24)^2=33,030.2083 solar mass miles per second squared.

1/n^2-1/(n+24)^2=1/n^2-1/(n^2+48n+576)=0.000129531 per mile squared per second squared.


1-n^2/(n^2+48n+576)=0.000129531n^2

n^2+48n+576-n^2=48n+576=0.000129531n^2(n^2+48n+576)=0.00012953 1n^4+0.006217488n^3+0.074609856n^2

0.000129531n^4+0.006217488n^3+0.074609856n^2-48n-576=0

We only need to solve the quartic. We get n=61.59325 miles (61 mi, 1044 yd, just under 100 km) So it is around sixty miles from an 8 solar mass black hole that a 1.5 solar mass neutron star with a 12-mile radius would start to break apart before its pieces explode.

But what about a supermassive black hole? Let us take the example of a 4 million solar mass black hole. The radius is

2*3.18749 E 10*4000000/186282^2=7,348,471.833 miles

The tidal forces across this same neutron star, at the event horizon, are


3.18749 E 10*4000000*0.001/7,348,471.833^2-3.18749 E 10*4000000*0.001/(7,348,471.833+24)^2=1.54226E-5 solar mass miles per second squared. Thuis is much less than the force betweeen the neutron star and a 0.001 solar mass sliver at the surface.

So the neutron star will enter the black hole in one piece. It will eventually be torn apart as it plunges towards the center, but no one outside the hole will observe it.


Michael
 




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