Post by dividavi on Jan 6, 2022 9:45:40 GMT
I watched a video about Neutronium, the material of Neutron Stars in which the narrator imagined what would be the characteristics of a small chunk of this matter. The narrator said that a 1 cubic centimeter chunk of Neutronium would have a noticeable gravitational field and if you hel it on your hand you could never let it go. According to my calculations it would be lots worse than he described.
Neutronium has been characterized as an element, similar to Iron, Uranium, Silver, Lithium and others. These elements are composed of positively charged protons and uncharged neutrons packed in close proximity in the nucleus with negatively charged electrons orbiting at some distance from the nucleus. Neutronium is quite different: There are only neutrons present and these are all packed together in one hideous mass. These neutrons are held together by the immense gravity of the Neutron Star. In a sense, a Neutron Star may be considered as a giant atom with all its component neutrons packed together.
So, you ask, what would be the gravity of a 1cc sphere of Neutronium? We know from Google than the Planet Earth has a radius (surface to center) of 6366 kilometers. The average density of our planet is 5.51 grams/cc, or 5.51 times greater than water. We know that the gravitational attraction at the surface is 9.8 meters/second squared. Finally, Google tells us that the density of Neutronium is, as expected, much greater than the density of Planet Earth. Google says the density of Neutronium is 4*10^17 kilograms/cubic meter. So now it's just a matter of plugging in the numbers.
First thing is to compare the densities of Earth material to Neutronium. As I said, Earth material is 5.5 gm/cc while Neutronium is 4*10^17 kilograms/cubic meter. We know that 1 kg = 1000 gm, that 1 m = 100 cm, and that 1 m^3 =10^6 cm^3 (or cc). PLugging in, we come up with the density of Neutronium is 4*10^14 gm/cc, or 72.59 trillion times denser that Earth.
Now, we know, that the gravitational attraction of Earth is 9.8 m/sec^2, but what would be the G force if the Earth had a radius of 0.62035 cm (to be 1cc in volume) instead 6366 km? We know that if a planet with the same density as Earth had a radius twice that of Earth its gravity would be two times greater than Earth or 2*9.8=19.6 m/sec^2. Another planet three time the radius of Earth (with a density of 5.5 gm/cc) would have a gravity three times that of Earth. Some planet one third the size of Earth with density not varying would have a field one third that of Earth and so on. This is because the mass of the planet increases with the cube of radius while the distance squared to the center varying with the square of radius. Since G varies inversely squared with distance to the center, the gravity overall = K*(r^3/r^2)=K*r.
The gravitational field of a 1 cc sphere of Earth density = 9.8*(0.62035cm/6366km)m/sec^2 = 9.8 * (0.62035cm/6.366 *10^8cm) m/sec^2 =9.55*10^(-9)m/sec^2. But we are concerned with Neutron Star Density which is 72.59 trillion times denser that Earth. So multiplying the value of 9.55*10^(-9)m/sec^2 by 72.59 trillion gives our answer of 693252m/sec^2 for the surface gravity of a 1 cc sphere of Neutronium.
So if you were touching the sphere with you finger the tip would experience a pull of 70740 times Earth gravity. Given the inverse square rule the tug on your finger 1centimeter from the tip would be much less, only 10368 times Earth gravity. That portion of your finger 2 cm from the tip would experience a pull of 3964 times Earth gravity. Your finger would be turned to bloody pulp along with your hand and probably your entire body. The closer you get to the Neutronium sphere the greater the gravity. I'm talking centimeters, not miles. Remember, Earth gravity =9.8 m/sec^2.
Here's what I calculated:
centimeter
Distance
From
sphere Times Earth
surface m/sec^2 Gravity
0 693252.5495 70740.05607
1 101612.6566 10368.63843
2 38855.00466 3964.796394
3 20354.69042 2077.009227
4 12497.2913 1275.233806
5 8445.763906 861.812643
10 2365.309831 241.358146
20 627.442092 64.024703
30 284.541371 29.034833
50 104.115512 10.624031
100 26.350816 2.688858
150 11.759759 1.199975
Neutronium has been characterized as an element, similar to Iron, Uranium, Silver, Lithium and others. These elements are composed of positively charged protons and uncharged neutrons packed in close proximity in the nucleus with negatively charged electrons orbiting at some distance from the nucleus. Neutronium is quite different: There are only neutrons present and these are all packed together in one hideous mass. These neutrons are held together by the immense gravity of the Neutron Star. In a sense, a Neutron Star may be considered as a giant atom with all its component neutrons packed together.
So, you ask, what would be the gravity of a 1cc sphere of Neutronium? We know from Google than the Planet Earth has a radius (surface to center) of 6366 kilometers. The average density of our planet is 5.51 grams/cc, or 5.51 times greater than water. We know that the gravitational attraction at the surface is 9.8 meters/second squared. Finally, Google tells us that the density of Neutronium is, as expected, much greater than the density of Planet Earth. Google says the density of Neutronium is 4*10^17 kilograms/cubic meter. So now it's just a matter of plugging in the numbers.
First thing is to compare the densities of Earth material to Neutronium. As I said, Earth material is 5.5 gm/cc while Neutronium is 4*10^17 kilograms/cubic meter. We know that 1 kg = 1000 gm, that 1 m = 100 cm, and that 1 m^3 =10^6 cm^3 (or cc). PLugging in, we come up with the density of Neutronium is 4*10^14 gm/cc, or 72.59 trillion times denser that Earth.
Now, we know, that the gravitational attraction of Earth is 9.8 m/sec^2, but what would be the G force if the Earth had a radius of 0.62035 cm (to be 1cc in volume) instead 6366 km? We know that if a planet with the same density as Earth had a radius twice that of Earth its gravity would be two times greater than Earth or 2*9.8=19.6 m/sec^2. Another planet three time the radius of Earth (with a density of 5.5 gm/cc) would have a gravity three times that of Earth. Some planet one third the size of Earth with density not varying would have a field one third that of Earth and so on. This is because the mass of the planet increases with the cube of radius while the distance squared to the center varying with the square of radius. Since G varies inversely squared with distance to the center, the gravity overall = K*(r^3/r^2)=K*r.
The gravitational field of a 1 cc sphere of Earth density = 9.8*(0.62035cm/6366km)m/sec^2 = 9.8 * (0.62035cm/6.366 *10^8cm) m/sec^2 =9.55*10^(-9)m/sec^2. But we are concerned with Neutron Star Density which is 72.59 trillion times denser that Earth. So multiplying the value of 9.55*10^(-9)m/sec^2 by 72.59 trillion gives our answer of 693252m/sec^2 for the surface gravity of a 1 cc sphere of Neutronium.
So if you were touching the sphere with you finger the tip would experience a pull of 70740 times Earth gravity. Given the inverse square rule the tug on your finger 1centimeter from the tip would be much less, only 10368 times Earth gravity. That portion of your finger 2 cm from the tip would experience a pull of 3964 times Earth gravity. Your finger would be turned to bloody pulp along with your hand and probably your entire body. The closer you get to the Neutronium sphere the greater the gravity. I'm talking centimeters, not miles. Remember, Earth gravity =9.8 m/sec^2.
Here's what I calculated:
centimeter
Distance
From
sphere Times Earth
surface m/sec^2 Gravity
0 693252.5495 70740.05607
1 101612.6566 10368.63843
2 38855.00466 3964.796394
3 20354.69042 2077.009227
4 12497.2913 1275.233806
5 8445.763906 861.812643
10 2365.309831 241.358146
20 627.442092 64.024703
30 284.541371 29.034833
50 104.115512 10.624031
100 26.350816 2.688858
150 11.759759 1.199975
