How weak is the electrostatic gravitational force between two protons?
372031
Protons are positively charged particles.\n\nBetween protons there are two types of forces.\n\nOne is the electro-static repulsion between two protons.\n\nThe other is the gravitational force of attraction between the masses of protons.\n\nWhen two protons are separated by any distance,\n\nThe electro-static repulsive force is 1.29 x 10^28 times greater than the gravitational attraction between them.
678477
9.8 x .020 will give you the gravitational force between the balloon and the earth.  The electrostatic force between the balloon and the ceiling could be different but must be at least 9.8 x .020 (otherwise the balloon would fall).  If the electrostatic force was larger than the gravitational force, there would be a normal force from the ceiling pushing down on the balloon (otherwise the balloon would try to calapse into the ceiling).
503357
Maybe it's not really a mass generated force at all. It could just be that it is masses effect on space-time. Actually, I guess it could be that it's not the earth for example that causes gravity but space-times effort to put matter in clumps and roll it into a ball. The bigger the mass the more space-time there is to push against it. Einstein's example was to imagine a bowling ball sitting on a large rubber sheet. The bowling ball would deform the sheet and the curvature of the rubber represented gravity in space-time. The closer you get to the ball the more curvature you have. That would explain why gravity is weaker with added distance.
14731
Photons, gluons, gravitons and Z are neutral and elementary so they are their own anti-particles.\n\nThe W+ and W-, the weak force carriers, are charged and are anti-particles.  W+ and W- can annihilate each other.
800535
jinde, hi!\n\nwhat an amazing question!  i remember reading about this somewhere, so it took me so long to research it...\n\nsee, when we talk about weight, it varies from spot to spot in the universe.  you may weigh 60kgs on Earth.  but on the moon, you'd be 10kgs.  on Jupiter, you'd be 600000kgs!!!\n\nwhat we need to discuss is the mass of the object, because that would be the only thing that makes sense on the Galactic scale.  are you ready for the answer?  it's gonna be mind-boggling!\n\nThe Earth's mass is approximately 6,000,000,000,000,000,000,000,000 (6E+24) kilograms!!\n\ni know what you're next question will be?\n\n"How did they measure that?"  i have the answer for you!\n\nthe measurement of the planet's weight is derived from the gravitational attraction that the Earth has for objects near it.\n\nit turns out that any two masses have a gravitational attraction for one another. if you put two iron weights near each other, they will attract one another gravitationally. the attraction is extremely slight, but if your instruments are sensitive enough you can measure the gravitational attraction that they have on one another. from that measurement, you could determine the mass of the two objects. the same is true for two marbles too! but the attraction is even slighter because the amount of gravitational force depends on mass of the objects...\n\nNewton showed that, for spherical objects, you can make the simplifying assumption that all of the object's mass is concentrated at the center of the sphere... and the following equation expresses the gravitational attraction that two spherical objects have on one another :\nF = G * M1 * M2 / R2\n\n    * R is the distance separating the two objects.\n    * G is a constant that is 6.67259x10-11m3/s2 kg.\n    * M1 and M2 are the two masses that are attracting each other.\n    * F is the force of attraction between them. \n\nassume that Earth is one of the masses (M1) and a 1-kg sphere is the other (M2).  the force between them is 9.8 kg*m/s2 -- we can calculate this force by dropping the 1-kg sphere and measuring the acceleration that the Earth's gravitational field applies to it (9.8 m/s2)...\n\nthe radius of the Earth is 6,400,000 meters (6,999,125 yards)... if you plug all of these values in and solve for M1, you find that the mass of the Earth is 6,000,000,000,000,000,000,000,000 kilograms (6E+24 kilograms / 1.3E+25 pounds)...\n\nhope this made sense... and if it did, i hope it was useful to you...\n\ndo write me if you need more clarification...\n\nenjoy!
603730
The following refers to Coulomb force between two charges in a single uniform medium, not the two-medium situation described in  the question.\nAnswer 1 is helpful but the lack of exponent signs makes it misleading. The force is inversely proportional to the square of the distance, so the simplified formula is \nF = k * q1 * q2 / r^2 \nk is a constant = 8.99*10^9 N m^2 / C^2 (C is charge in Coulombs). Thus if q1 = q2 = 1 Coulomb, and q1 is separated from q2 by r = 1 meter, the force is 8.99*10^9 N or 2.01 billion pounds! \nThe above simplified formula is a contraction of the full formula (see ref.),\nF = q1 * q2 / (4 * pi * epsilon * r^2)\nin which k is replaced by 1/(4 * pi * epsilon), where epsilon is the product of epsilon0, the electrical permittivity of the vacuum (= 8.85 * 10^-12 c^2/(N m^2)), and kappa, the dielectric constant which = 1 for vacuum and is assumed in the simplified formula.\nFor two segments of distance as you specify, first solve for F1 and F2, each as a function of a particular distance (r or d - r) and dielectric constant. Then my inclination would be to try the sum-of-inverse square root solution which works for a similar situation, gravitational force F at distance R, computed from F1 and F2 at partial distances R1 and R2, where R = R1 + R2, \n1/SQRT(F) = 1/SQRT(F1) + 1/SQRT(F2)\nSorry I can't be more definite than that.
833894
On a subatomic level, forces involve the exchange of mediating particles.  In the case of the electromagnetic and gravitational forces, those particles do not carry electric charge, or any other quantum numbers, besides spin.  Therefore, releasing them and absorbing them does not alter the identity of a particle, although its energy and momentum are changed.  \n\nThe weak force is more complicated, however.  There are two types of mediating particles. One type (the W particles) carries electric charge.  In order to release or absorb a W particle, a subatomic particle must change its electric charge (and therefore its identity) in order for charge to be conserved.\n\nThis is essentially why beta decay happens.  A neutron produces a negatively charged W particle, and changes into a proton.  The W particle itself decays into an electron and a neutrino.  \n\nThe second decay, (of the mediating particle) is not possible with gravitational or electromagnetic forces, because their mediating particles are massless, and travel at the speed of light. (The rules of special relativity forbid light speed objects from decaying.)
432447
Yes.  Charge is just whatever the particle has that makes it interact with a given field.  So, just like electric charge makes the proton interact with the electromagnetic field, gravity charge is whatever would make the proton interact with the gravitational field-- that is, its mass.  Gravity charge is the same thing as mass.
16211
This is actually a subtle question. Why is there only one "charge" for gravity? That fundamentally depends on the details of General Relativy and the nature of gravity waves and the graviton. The result, of course, is that there is only one "charge." There is no antimatter.\n\nIt's a little hard to figure out exactly what you mean in the rest of the question. No offense, but punctuation would help. But, suffice it to say that there is only one "charge" of gravity, so the details are moot.
58372
the formula you need is Fg=Gm1m2/r^2.  so just plug in. G is the universal gravitational constant. so it's:\n(6.673 x 10^-11)(70)(70)/(1)^2 = 3.327 x 10^-7 N\n\nas you can see, this is negligable because people are not big enough to create a substancial gravitational force.
787952
The force of attraction is given by Newton's law, F = G* m1*m2/r^2, where G = universal gravitational constant, 6.67 * 10^-11 N*m^2/kg^2.  You have the two masses, and the radius of earth is the distance r.  Your weight is the same as the gravitational force.
205616
The Begy atomic model. It eliminates the nedd for the quite unbelievable strong and weak nuclear forces. It also defines gravity as a force acting upon the universe rather than from within it. This reduces the forces in this universe to just one the electromagnetic force. Is this a unified field theory?\n\nWeb search, "Begy atomic model"
196200
the fact is that in the nucleus there is no actual contact between the protons and the neutrons..they are held only by the electrostatic forces.so asolutely no friction
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
208985
The force of gravity has to do with an attraction of objects due to mass. The nuclear force has to do with attraction or repulsion of nucleous at very small distances. Electric Force is attraction or repulsion due to the charge of the object, and is very important in the actual structure of an atom. Inertia has to do with the continued movement of an object (an object in motion continues to move unless another force acts upon that object.)\n\nYou might think that gravity is a strong force, because you see it's effects all the time - however, the strongest forces will either be the nuclear force or electric force. If gravity is stronger than electric force, then you get black holes.
846277
It decreases to 1/4 of the original force.  This is because the force of gravity is inversely proportional to the distance between the two bodies.  Thus, increasing distance by a factor of n reduces force to 1/n^2.
904747
Its origins and nature are still open to postulation. But it is an attractive force only between any two masses and Newton concluded in his uneiversal laws of gravitation that the attractive force (F) was proportional to the masses interacting(M1 and M2) and inversely proportional to the square of their distance apart (r)\n\nie F α M1*M2/d²\nwhence \n\nF = G*M1*M2/r² \nwhere G is the universal gravitation constant\nG ≈ 6.6742 x 10^(-11) Nm²/kg²