Equality of Gravitational and Electrostatic force

In summary, the equality of gravitational and electrostatic forces refers to the concept that, under certain conditions, the magnitude of the gravitational force between two masses can be compared to the electrostatic force between two charged particles. This relationship is often illustrated through the equations governing each force: Newton's law of universal gravitation and Coulomb's law. Despite their different natures—gravity being an attractive force that acts on masses, while electrostatics can be both attractive and repulsive depending on the charges—both forces diminish with distance and have similar forms. The comparison highlights the relative strengths of these fundamental forces, emphasizing that electrostatic forces are significantly stronger than gravitational forces at the atomic and molecular levels.
  • #1
PhysicsEnjoyer31415
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So i was experimenting around with formulae and i thought to myself "could a planet rip a electron off a single electron system?" So i used this formula and i want to know if this is correct? . I know that gravity is magnitudes weaker than electrostatic force but if it were forced anyway would this be correct?
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  • #2
I feel two points
1 How we can arrange the experiment so that only electrons no ion cores are under influence of gravity ? Free fall of atoms is the most familiar situation. I do not know how we can fix the position of the cores with no influence to electrons.
2 If we can fix the cores, forgetting about EM force, I wonder which is bigger gravity from the Earth or gravity from the ion core to the electrons ?
 
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  • #3
For a hydrogen atom with a radius of 37 picometers, placed near the surface of the Earth, I get the following results for the forces on the electron:

Tidal Force: 1.185x10-16 newtons
Electrostatic Force: 1.68x10-7 newtons

Even if my electrostatic force is an order of magnitude off due to being a classical calculation and not a quantum one, we're looking at a difference in force in the realm of nine orders of magnitude. A planet is absolutely not ripping an electron off of an atom unless the electron is only bound to the atom in the loosest possible sense of the word 'bound'.
 
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  • #4
Drakkith said:
Tidal Force:
It's worth noting that the tidal force calculation addresses @anuttarasammyak's objection about the nucleus apparently being unaffected by gravity in the OP's calculation. This tidal force approach is the correct way to go about this, not the OP's calculation.
anuttarasammyak said:
I wonder which is bigger gravity from the Earth or gravity from the ion core to the electrons ?
Earth. Atomic masses are of order ##10^{-27}\mathrm{kg}##, electrons about three orders of magnitude lower, and atomic radii ##10^{-10}\mathrm{m}##. That gives you a gravitational force around ##10^{-48}\mathrm{N}##. Assuming Newtonian gravity is remotely valid for atoms.
 
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  • #5
Drakkith said:
For a hydrogen atom with a radius of 37 picometers, placed near the surface of the Earth, I get the following results for the forces on the electron:

Tidal Force: 1.185x10-16 newtons
Electrostatic Force: 1.68x10-7 newtons

Even if my electrostatic force is an order of magnitude off due to being a classical calculation and not a quantum one, we're looking at a difference in force in the realm of nine orders of magnitude. A planet is absolutely not ripping an electron off of an atom unless the electron is only bound to the atom in the loosest possible sense of the word 'bound'.
I think you are right and this is only a theoretical possibility , i thought it could become true only if we have mass in 50s of magnitude , neglect gravitational force of proton ,and have radius not more than magnitudes of 6 or 7. Also i was using the bohrs model where radius of hydrogen atom wpuld be 0.529 angstrong
 
  • #6
PhysicsEnjoyer31415 said:
I think you are right and this is only a theoretical possibility , i thought it could become true only if we have mass in 50s of magnitude , neglect gravitational force of proton ,and have radius not more than magnitudes of 6 or 7. Also i was using the bohrs model where radius of hydrogen atom wpuld be 0.529 angstrong
In order to get large tidal force, putting an atom very closed to a tiny densed mass instead of planet would be interesting.
 
  • #7
anuttarasammyak said:
In order to get large tidal force, putting an atom very closed to a tiny densed mass instead of planet would be interesting.
Would definitely love to see that
 
  • #8
anuttarasammyak said:
In order to get large tidal force, putting an atom very closed to a tiny densed mass instead of planet would be interesting.
I think it still might not be enough to pull a electron even if we hold down the proton somehow. That would require a huge density which i am not sure is possible or not , maybe a neutron star or something but it would surely be interesting
 
  • #9
PhysicsEnjoyer31415 said:
I think you are right and this is only a theoretical possibility.
Actually, he demonstrated that it's a theoretical impossibility.
 
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  • #10
PhysicsEnjoyer31415 said:
I think it still might not be enough to pull a electron even if we hold down the proton somehow. That would require a huge density which i am not sure is possible or not , maybe a neutron star or something but it would surely be interesting
There's a little thing called Quantum Mechanics that may be relevant here. You don't get far with a Newtonian model of the atom.
 
  • #11
PeroK said:
There's a little thing called Quantum Mechanics that may be relevant here. You don't get far with a Newtonian model of the atom.
Yes
 

FAQ: Equality of Gravitational and Electrostatic force

What is the principle of equality of gravitational and electrostatic forces?

The principle of equality of gravitational and electrostatic forces states that under certain conditions, the magnitudes of the gravitational force between two masses can be equal to the electrostatic force between two charged particles. This occurs when comparing the forces at specific distances and with appropriate charge and mass values.

How can we mathematically express the equality of gravitational and electrostatic forces?

The gravitational force (F_g) between two masses (m1 and m2) is given by Newton's law of gravitation: F_g = G(m1 * m2) / r^2, where G is the gravitational constant and r is the distance between the centers of the masses. The electrostatic force (F_e) between two charges (q1 and q2) is given by Coulomb's law: F_e = k(q1 * q2) / r^2, where k is Coulomb's constant. Setting these two equations equal to each other allows us to find conditions under which they are equal.

What are the implications of the equality of gravitational and electrostatic forces?

The implications of this equality suggest that at very small scales, such as those involving elementary particles, the forces acting between charged particles can be comparable to gravitational forces. This has implications for understanding fundamental interactions in physics and can influence theories in particle physics and cosmology.

Can the equality of gravitational and electrostatic forces be observed experimentally?

While direct observation of the equality of these forces is challenging due to the relative weakness of gravity compared to electrostatic forces, experiments can be designed to measure the forces acting on charged particles and masses at small scales. Such experiments often involve precision measurement techniques to detect the subtle balance of forces.

What factors can affect the equality of gravitational and electrostatic forces?

Several factors can affect the equality of these forces, including the distance between the objects, the magnitudes of the masses and charges involved, and the medium in which the forces are acting. Additionally, environmental factors, such as the presence of other charges or masses, can also influence the net forces experienced by the objects in question.

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