Is There Any Significant Difference Between Gravitational and Coulomb Forces?

[formal]

The definition of Δ (difference,drop of) Electrostatic Potential Energy (D E-PE) 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 G-PE)

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)

 

[jjustinn]

The definition of Δ (difference,drop of) Electrostatic Potential Energy (D E-PE) says it...
Could you explain this to me, possibly without formulas? I need only to grasp the general idea!

I'm not sure I understand the question, or how the question follows from the subject, but I'll give it a shot anyway: if the question is (a) whether electrostatics (via the coulomb potential) and Newtonian gravity are equivalent, then the answer is Yes, with the caveat that the "charge" in the case of gravity is not only always attractive, but is also the same quantity that determines how much energy it takes to accelerate the body (internal = gravitational mass).

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 fast-moving "particles").

 

[formal]

1) if the question is (a) whether electrostatics (via the coulomb potential) and Newtonian gravity are equivalent, then the answer is Yes,
2) If on the other hand the question is (b) if they're still the same when motion is involved, the answer would be ERROR: .

Hi jjustinn,

Both questions are related because
both in G-field (A) and E-field (B) one electron -e (which is at same time both unit-mass / unit-charge)) 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 E-field.

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 E-PE = Ek = Voltage (PE/ 1) ?

 

[thebiggerbang]

Both questions are related because
both in G-field (A) and E-field (B) one electron -e (which is at same time both unit-mass / unit-charge)) 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)
?

An electron is not assumed to have a unit mass and I don not think that it 'bangs' into something at r. Our concern is to just get it to r and not to see that it gets there and stops or so!

 

[jjustinn]

Hi jjustinn,

Both questions are related because
both in G-field (A) and E-field (B) one electron -e (which is at same time both unit-mass / unit-charge)) 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 E-field.

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 E-PE = Ek = Voltage (PE/ 1) ?

I think part of my problem is I don't follow your notation...

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 light-years 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).

 

[formal]

An electron is not assumed to have a unit mass!

Unit mass in the example, not Unit in dimensional analysis
see next post

 

[formal]

I think part of my problem is I don't follow your notation.

I'm awfully sorry, jjustinn,
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 E-field
and why the result is different
Thanks

 

[thebiggerbang]

I'm awfully sorry, jjustinn,
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)
when it moves at any speed from r to r1 loses always same PE*
if this is at last clear, move on to voltage, please!

AFAIK,in the definition the unit charge is a 1C charge and not a charge of e (that is the fundamental unit of charge). But while experimenting, they using a vanishingly small charge to avoid disturbing the existing EF.

 

[Evo]

formal is a previously banned crackpot, sorry that you had to deal with his nonsense.