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Is there any significant difference between Gravitational and Coulomb Force (/Δ PE)? 
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#1
Aug1011, 05:19 AM

P: 27

The definition of Δ (difference,drop of) Electrostatic Potential Energy (D EPE) says it
'is the work done on a unit charge (e) to bring it from r to r1 (against the force) with constant (without acceleration) velocity and extremely low absolute value of velocity (quasistatic, vanishingly small, etc..) afaik, v must be steady not to create magnetic disturbances or the like. It escapes me what would happen if value of v is , say, 1 cm/s. (constant) (We can shoot an electron in a G Field at 1 Km/ s without problems, and get exact D GPE) Could you explain this to me, possibly without formulas? I need only to grasp the general idea! Thanks (* to simplify discussion let's assume the following ideal conditions: A positive charge in vacuum distant from Galaxies equivalent of GM ( 0.0000184 C ?)at the origin r(o) and one electron (e) at r = r(earth) = 6.4x 10^6 m moving to r1 at v= 1 cm (m, km?)/ s) 


#2
Aug1011, 09:44 PM

P: 148

If on the other hand the question is (b) if they're still the same when motion is involved, the answer would be ERROR: electrostatics isn't valid if your charges / fields are moving / changing; though of course the smaller the speed / charge / fields, the smaller the difference between the actual physics and the electrostatic approximation gets  but in Newtonian gravity, as far as I know, there is no such limitation (particularly since Newton formulated it knowing that he was working with fastmoving "particles"). 


#3
Aug1111, 12:08 AM

P: 27

Both questions are related because both in Gfield (A) and Efield (B) one electron e (which is at same time both unitmass / unitcharge)) is 1) at rest at r(1) and then 2) moves falls towards M /(+charge) at origin r(o) until it bangs into something at r (e) Can you separate the two sides of the issue? I assume that when it reaches r (=r(e)) it has the same energy Ek in G or Efield. I proposed an example because that is the best way to cancel any doubt. if you set r = 6.4 x10^8 cm (= r(earth)) and r1 = 10^ 9 or any value you wish, you can figure out Ek (D PE) in both cases. Then you take the same e as test charge and move it against the charge from r to r(1) and see if you get different values at different speeds. just an additional question: as in the example (B) charge = 1 can we say that D EPE = Ek = Voltage (PE/ 1) ? 


#4
Aug1111, 10:47 AM

P: 70

Is there any significant difference between Gravitational and Coulomb Force (/Δ PE)?



#5
Aug1111, 10:49 AM

P: 148

In electrostatics or Newtonian gravity, the change in a particle's potential energy is equal to the change in its kinetic energy. So if a particle with q = 1C starts at rest (kinetic energy = 0J) any position with V = 10V, then its potential energy = 10J. If it moves to any other position where V = 0V  one meter away, or 10.551*10^6 lightyears away, if we're only dealing with electrostatics, its potential energy will be 0J, so its kinetic energy will be 10J. The same argument works exactly the same if you replace volts (electrostatic potential energy per coulomb) with gravitational potential energy and coulombs (C) with kilograms (kg). 


#6
Aug1211, 01:00 AM

P: 27

see next post 


#7
Aug1211, 02:49 AM

P: 27

unit is a very tricky word! I'll call it our unit* it is charge 1.6022x 10^ 19 C and at the same time 9.1x 10^ 31 kg OK? now if r = 6.4 x 10^6 (acc=9.8, PE 6.25 x10^7); and r1 = 10^ 7 (acc=4, PE= 4x10^7) when unit* falls from r1 to r gains D PE* (6.25  4 = 2.25) v at r = 6.7 km/s if we shoot it up from r at 6.7 km/s it will reach r1 and loose exactly PE°, now you say what happens if we shoot unit* in a similar Efield and why the result is different Thanks 


#8
Aug1211, 03:28 AM

P: 70




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