What is the escape velocity of a neutron particle (not neutron star)what is the difference between a blackhole and a point particleWhy 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 and gravitational potential be higher in weaker gravitational fields?Escape velocity at an angleNeutron star core understandingWhat happens if a neutron flies towards a nucleus?Escape Velocity - Won't the orbital path just become larger with higher initial velocity?
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What is the escape velocity of a neutron particle (not neutron star)
what is the difference between a blackhole and a point particleWhy 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 and gravitational potential be higher in weaker gravitational fields?Escape velocity at an angleNeutron star core understandingWhat happens if a neutron flies towards a nucleus?Escape Velocity - Won't the orbital path just become larger with higher initial velocity?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
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 Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
$endgroup$
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 Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
$endgroup$
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 Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
$endgroup$
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 Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
asked Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
asked Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
asked Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
asked Apr 17 at 14:33
EnriqueEnrique
1435 bronze badges
1435 bronze badges
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 Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
$endgroup$
$begingroup$
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
Apr 17 at 17:51
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
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 Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
$endgroup$
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
Apr 17 at 15:07
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
Apr 17 at 15:08
4
$begingroup$
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
Apr 17 at 15:09
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
$endgroup$
$begingroup$
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
Apr 17 at 17:51
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
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 Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
$endgroup$
$begingroup$
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
Apr 17 at 17:51
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
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 Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
$endgroup$
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 Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
answered Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
answered Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
answered Apr 17 at 15:15
Ben CrowellBen Crowell
61.7k6 gold badges179 silver badges347 bronze badges
61.7k6 gold badges179 silver badges347 bronze badges
$begingroup$
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
Apr 17 at 17:51
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
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
Apr 17 at 17:51
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
$begingroup$
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
Apr 17 at 17:51
$begingroup$
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
Apr 17 at 17:51
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
@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
Apr 17 at 18:03
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
Wouldn't a proper escape velocity take into consideration all 4 fundamental forces?
$endgroup$
– corsiKa
Apr 18 at 1:10
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
$begingroup$
@corsiKa: If you use an electron, then it's certainly true that the gravitational force would be much weaker than the force exerted by the neutron's dipole field. You could use a neutrino. I doubt that the weak force is significant here.
$endgroup$
– Ben Crowell
Apr 18 at 22:00
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 Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
$endgroup$
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
Apr 17 at 15:07
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
Apr 17 at 15:08
4
$begingroup$
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
Apr 17 at 15:09
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 Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
$endgroup$
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
Apr 17 at 15:07
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
Apr 17 at 15:08
4
$begingroup$
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
Apr 17 at 15:09
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 Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
$endgroup$
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 Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
answered Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
answered Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
answered Apr 17 at 15:04
Cam HoughCam Hough
212 bronze badges
212 bronze badges
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
Apr 17 at 15:07
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
Apr 17 at 15:08
4
$begingroup$
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
Apr 17 at 15:09
add a comment
|
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
$endgroup$
– Cam Hough
Apr 17 at 15:07
$begingroup$
It's much better defined than a factor of 2. You simply made a mistake.
$endgroup$
– Ben Crowell
Apr 17 at 15:08
4
$begingroup$
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
Apr 17 at 15:09
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
1.5 fm sounds more like the diameter, not the radius.
$endgroup$
– Ben Crowell
Apr 17 at 15:05
$begingroup$
It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
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– Cam Hough
Apr 17 at 15:07
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It's not well-defined, just an approximation to get an estimate using classical concepts. (en.wikipedia.org/wiki/Atomic_nucleus)
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– Cam Hough
Apr 17 at 15:07
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It's much better defined than a factor of 2. You simply made a mistake.
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– Ben Crowell
Apr 17 at 15:08
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It's much better defined than a factor of 2. You simply made a mistake.
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– Ben Crowell
Apr 17 at 15:08
4
4
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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.
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– Cam Hough
Apr 17 at 15:09
$begingroup$
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.
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– Cam Hough
Apr 17 at 15:09
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