What is the escape velocity of a neutron particle (not neutron star) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Moderator Election Q&A - Question Collectionwhat is the difference between a blackhole and a point particleHas anyone else thought about gravity in this way?Why is there an escape velocity?How does escape velocity relate to energy and speed?What is the relation between orbital velocity and escape velocity in strongly relativistic situations?How can escape velocity be independent of the direction of projection of a body?How can escape velocity and gravitational potential be higher in weaker gravitational fields?Escape velocity at an angleNeutron star core understandingIs this definition of “escape velocity” correct?What happens if a neutron flies towards a nucleus?
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What is the escape velocity of a neutron particle (not neutron star)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Question Collectionwhat is the difference between a blackhole and a point particleHas anyone else thought about gravity in this way?Why is there an escape velocity?How does escape velocity relate to energy and speed?What is the relation between orbital velocity and escape velocity in strongly relativistic situations?How can escape velocity be independent of the direction of projection of a body?How can escape velocity and gravitational potential be higher in weaker gravitational fields?Escape velocity at an angleNeutron star core understandingIs this definition of “escape velocity” correct?What happens if a neutron flies towards a nucleus?
$begingroup$
I'm not sure if this question makes sense (if not maybe you can explain why)
But if the neutron has mass and have a size, then it should have a escape velocity in the "surface" right?
I know the gravity force generated by a neutron is really low, since it is also really small in size, what happens with that force when we are really close? is not almost infinite?
quantum-mechanics gravity quantum-gravity neutrons escape-velocity
asked 5 hours ago
EnriqueEnrique
1235
$endgroup$
add a comment |
$begingroup$
I'm not sure if this question makes sense (if not maybe you can explain why)
But if the neutron has mass and have a size, then it should have a escape velocity in the "surface" right?
I know the gravity force generated by a neutron is really low, since it is also really small in size, what happens with that force when we are really close? is not almost infinite?
quantum-mechanics gravity quantum-gravity neutrons escape-velocity
asked 5 hours ago
EnriqueEnrique
1235
$endgroup$
add a comment |
$begingroup$
I'm not sure if this question makes sense (if not maybe you can explain why)
But if the neutron has mass and have a size, then it should have a escape velocity in the "surface" right?
I know the gravity force generated by a neutron is really low, since it is also really small in size, what happens with that force when we are really close? is not almost infinite?
quantum-mechanics gravity quantum-gravity neutrons escape-velocity
asked 5 hours ago
EnriqueEnrique
1235
$endgroup$
I'm not sure if this question makes sense (if not maybe you can explain why)
But if the neutron has mass and have a size, then it should have a escape velocity in the "surface" right?
I know the gravity force generated by a neutron is really low, since it is also really small in size, what happens with that force when we are really close? is not almost infinite?
quantum-mechanics gravity quantum-gravity neutrons escape-velocity
quantum-mechanics gravity quantum-gravity neutrons escape-velocity
asked 5 hours ago
EnriqueEnrique
1235
asked 5 hours ago
EnriqueEnrique
1235
asked 5 hours ago
EnriqueEnrique
1235
asked 5 hours ago
EnriqueEnrique
1235
asked 5 hours ago
EnriqueEnrique
1235
1235
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I know the gravity force generated by a neutron is really low
To define the force, you need to define a second object, with some mass, that is being acted on. That can't be a second neutron, because then there would be an attraction due to the strong nuclear force that would be much greater than the attraction due to gravity. You would want to talk about a particle such as an electron or some other lepton that doesn't participate in the strong force.
what happens with that force when we are really close? is not almost infinite?
The neutron is similar to an object like the earth, in that its mass is distributed over some volume. Therefore you can't get to zero distance from all its mass. The radius of a neutron is roughly 0.8 fm ($10^-15$ m). (It's fuzzy, but the number is well defined to within about 20%, if you take some criterion like where the density falls off to half of the value at the center.) Using this radius, the escape velocity is $sqrt2Gm/r=1.7times10^-11$ m/s. The extreme smallness of this velocity confirms that gravity is too weak to matter at the atomic scale.
In reality, if you take a particle such as an electron and try to put it right at the surface of the nucleus, constraining its position to within less than ~1 fm, then by the Heisenberg uncertainty principle, it will be moving many orders of magnitude faster than escape velocity. To get this zero-point velocity to be as small as the escape velocity, you would need a very massive particle -- much more massive than any subatomic particle we know of.
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
$endgroup$
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
add a comment |
$begingroup$
I'm sure a full quantum mechanical explanation exists that takes complexities at these size scales into account. But using the classical formula for escape velocity ($v_esc=sqrt2GM/R$, derived by determining the total amount of work to move a massive point particle from the surface of a massive object to an infinite distance), an estimate of the escape speed using the neutron mass ($1.67times 10^-27$ kg), and radius ($sim 1.5$ fm) comes to about $1times 10^-11$ m/s.
answered 4 hours ago
Cam HoughCam Hough
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
2
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
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$begingroup$
I know the gravity force generated by a neutron is really low
To define the force, you need to define a second object, with some mass, that is being acted on. That can't be a second neutron, because then there would be an attraction due to the strong nuclear force that would be much greater than the attraction due to gravity. You would want to talk about a particle such as an electron or some other lepton that doesn't participate in the strong force.
what happens with that force when we are really close? is not almost infinite?
The neutron is similar to an object like the earth, in that its mass is distributed over some volume. Therefore you can't get to zero distance from all its mass. The radius of a neutron is roughly 0.8 fm ($10^-15$ m). (It's fuzzy, but the number is well defined to within about 20%, if you take some criterion like where the density falls off to half of the value at the center.) Using this radius, the escape velocity is $sqrt2Gm/r=1.7times10^-11$ m/s. The extreme smallness of this velocity confirms that gravity is too weak to matter at the atomic scale.
In reality, if you take a particle such as an electron and try to put it right at the surface of the nucleus, constraining its position to within less than ~1 fm, then by the Heisenberg uncertainty principle, it will be moving many orders of magnitude faster than escape velocity. To get this zero-point velocity to be as small as the escape velocity, you would need a very massive particle -- much more massive than any subatomic particle we know of.
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
$endgroup$
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
add a comment |
$begingroup$
I know the gravity force generated by a neutron is really low
To define the force, you need to define a second object, with some mass, that is being acted on. That can't be a second neutron, because then there would be an attraction due to the strong nuclear force that would be much greater than the attraction due to gravity. You would want to talk about a particle such as an electron or some other lepton that doesn't participate in the strong force.
what happens with that force when we are really close? is not almost infinite?
The neutron is similar to an object like the earth, in that its mass is distributed over some volume. Therefore you can't get to zero distance from all its mass. The radius of a neutron is roughly 0.8 fm ($10^-15$ m). (It's fuzzy, but the number is well defined to within about 20%, if you take some criterion like where the density falls off to half of the value at the center.) Using this radius, the escape velocity is $sqrt2Gm/r=1.7times10^-11$ m/s. The extreme smallness of this velocity confirms that gravity is too weak to matter at the atomic scale.
In reality, if you take a particle such as an electron and try to put it right at the surface of the nucleus, constraining its position to within less than ~1 fm, then by the Heisenberg uncertainty principle, it will be moving many orders of magnitude faster than escape velocity. To get this zero-point velocity to be as small as the escape velocity, you would need a very massive particle -- much more massive than any subatomic particle we know of.
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
$endgroup$
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
add a comment |
$begingroup$
I know the gravity force generated by a neutron is really low
To define the force, you need to define a second object, with some mass, that is being acted on. That can't be a second neutron, because then there would be an attraction due to the strong nuclear force that would be much greater than the attraction due to gravity. You would want to talk about a particle such as an electron or some other lepton that doesn't participate in the strong force.
what happens with that force when we are really close? is not almost infinite?
The neutron is similar to an object like the earth, in that its mass is distributed over some volume. Therefore you can't get to zero distance from all its mass. The radius of a neutron is roughly 0.8 fm ($10^-15$ m). (It's fuzzy, but the number is well defined to within about 20%, if you take some criterion like where the density falls off to half of the value at the center.) Using this radius, the escape velocity is $sqrt2Gm/r=1.7times10^-11$ m/s. The extreme smallness of this velocity confirms that gravity is too weak to matter at the atomic scale.
In reality, if you take a particle such as an electron and try to put it right at the surface of the nucleus, constraining its position to within less than ~1 fm, then by the Heisenberg uncertainty principle, it will be moving many orders of magnitude faster than escape velocity. To get this zero-point velocity to be as small as the escape velocity, you would need a very massive particle -- much more massive than any subatomic particle we know of.
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
$endgroup$
I know the gravity force generated by a neutron is really low
To define the force, you need to define a second object, with some mass, that is being acted on. That can't be a second neutron, because then there would be an attraction due to the strong nuclear force that would be much greater than the attraction due to gravity. You would want to talk about a particle such as an electron or some other lepton that doesn't participate in the strong force.
what happens with that force when we are really close? is not almost infinite?
The neutron is similar to an object like the earth, in that its mass is distributed over some volume. Therefore you can't get to zero distance from all its mass. The radius of a neutron is roughly 0.8 fm ($10^-15$ m). (It's fuzzy, but the number is well defined to within about 20%, if you take some criterion like where the density falls off to half of the value at the center.) Using this radius, the escape velocity is $sqrt2Gm/r=1.7times10^-11$ m/s. The extreme smallness of this velocity confirms that gravity is too weak to matter at the atomic scale.
In reality, if you take a particle such as an electron and try to put it right at the surface of the nucleus, constraining its position to within less than ~1 fm, then by the Heisenberg uncertainty principle, it will be moving many orders of magnitude faster than escape velocity. To get this zero-point velocity to be as small as the escape velocity, you would need a very massive particle -- much more massive than any subatomic particle we know of.
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
answered 4 hours ago
Ben CrowellBen Crowell
54.5k6165314
54.5k6165314
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
add a comment |
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
is really low, I canno understand how a black hole can exist then, I mean for a black hole we need a very dense object right? and what is more dense than a neutron?
$endgroup$
– Enrique
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
$begingroup$
@Enrique: To make a black hole, the relevant figure is not density, which would scale like $m/r^3$, but $m/r$. It actually wouldn't make sense if a neutron was an ultrarelativistic object. See physics.stackexchange.com/questions/12404/…
$endgroup$
– Ben Crowell
1 hour ago
add a comment |
$begingroup$
I'm sure a full quantum mechanical explanation exists that takes complexities at these size scales into account. But using the classical formula for escape velocity ($v_esc=sqrt2GM/R$, derived by determining the total amount of work to move a massive point particle from the surface of a massive object to an infinite distance), an estimate of the escape speed using the neutron mass ($1.67times 10^-27$ kg), and radius ($sim 1.5$ fm) comes to about $1times 10^-11$ m/s.
answered 4 hours ago
Cam HoughCam Hough
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
2
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
add a comment |
$begingroup$
I'm sure a full quantum mechanical explanation exists that takes complexities at these size scales into account. But using the classical formula for escape velocity ($v_esc=sqrt2GM/R$, derived by determining the total amount of work to move a massive point particle from the surface of a massive object to an infinite distance), an estimate of the escape speed using the neutron mass ($1.67times 10^-27$ kg), and radius ($sim 1.5$ fm) comes to about $1times 10^-11$ m/s.
answered 4 hours ago
Cam HoughCam Hough
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
2
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
add a comment |
$begingroup$
I'm sure a full quantum mechanical explanation exists that takes complexities at these size scales into account. But using the classical formula for escape velocity ($v_esc=sqrt2GM/R$, derived by determining the total amount of work to move a massive point particle from the surface of a massive object to an infinite distance), an estimate of the escape speed using the neutron mass ($1.67times 10^-27$ kg), and radius ($sim 1.5$ fm) comes to about $1times 10^-11$ m/s.
answered 4 hours ago
Cam HoughCam Hough
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm sure a full quantum mechanical explanation exists that takes complexities at these size scales into account. But using the classical formula for escape velocity ($v_esc=sqrt2GM/R$, derived by determining the total amount of work to move a massive point particle from the surface of a massive object to an infinite distance), an estimate of the escape speed using the neutron mass ($1.67times 10^-27$ kg), and radius ($sim 1.5$ fm) comes to about $1times 10^-11$ m/s.
answered 4 hours ago
Cam HoughCam Hough
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Cam HoughCam Hough
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Cam HoughCam Hough
211
answered 4 hours ago
Cam HoughCam Hough
211
211
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Cam Hough is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
2
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
add a comment |
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
2
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
4 hours ago
2
2
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
$begingroup$
My point is using the classical description for escape velocity puts us so far away from precision of that degree that there's no point worrying about it. Call it a mistake if you want.
$endgroup$
– Cam Hough
4 hours ago
add a comment |
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