Charge and gravitation

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Offline jeffreyH

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Charge and gravitation
« on: 14/06/2015 14:31:41 »
In another thread the equation

[tex]+ \frac {GM}{rc} + C[/tex]

was the final result of integrating

[tex]\frac {-GM}{r^2}[/tex]

In that thread I assumed the constant of integration, C, to disappear. What if it doesn't?

In another forum I found the following equation

[tex]v=\sqrt{\frac{2G}{r} \left( M - \frac{Q^2}{2r} \right)}[/tex]

Here we have an increase in charge showing a decrease in escape velocity. Could C represent this?
« Last Edit: 14/06/2015 14:34:09 by jeffreyH »
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Offline chiralSPO

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Re: Charge and gravitation
« Reply #1 on: 14/06/2015 18:50:03 »
Even if the algebraic solution to the indefinite integral includes an unspecified constant, "C," any time we evaluate the integral for any definite answer, the C will (must) always cancel out.

The added term in the third equation presented is a function of r, so it cannot be your constant.

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Offline jeffreyH

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Re: Charge and gravitation
« Reply #2 on: 14/06/2015 20:07:39 »
Even if the algebraic solution to the indefinite integral includes an unspecified constant, "C," any time we evaluate the integral for any definite answer, the C will (must) always cancel out.

The added term in the third equation presented is a function of r, so it cannot be your constant.

As you say C will disappear in a definite integral. I am playing around with some ideas but they cannot ultimately be in this form, as you indicated. If it was that easy someone would have done such an analysis. I will be posting more to this thread and would appreciate your input to keep me on track.
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Offline jeffreyH

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Re: Charge and gravitation
« Reply #3 on: 14/06/2015 23:01:40 »
In another thread I derived the following equation via the mass to charge ratio.

[tex]m = \frac{kQ}{Er^2} \frac{G}{gr^2}[/tex]

However this equation is actually giving an invalid value for mass. Simply because mass was used to derive it. Therefore this is not deriving mass but something entirely different. The equation in the OP showing a decrease in escape velocity due to charge can not be of this form but must have charge as one component. Charge is viewed separately from mass and its energy is not part of mass equations. The field is exterior to the mass for the most part. Whatever part of the field is internal cannot be observed. It may be that an equation of the form [tex]x = \frac{Gm}{gr^2} - \frac{kQ}{Er^2}[/tex] is better suited where x is to be determined.

EDIT: Note that the gravitational factor equals unity. If the electric field portion also equals unity then x = 0
« Last Edit: 14/06/2015 23:13:13 by jeffreyH »
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Offline jeffreyH

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Re: Charge and gravitation
« Reply #4 on: 16/06/2015 00:24:11 »
An equation of the form

[tex]\frac{Gm}{gr^2} - \frac{kQ}{Er^2} = 0[/tex]

illustrates that an individual particle, or a mass with net zero charge is balanced. That is, the electric field cannot be modified to affect gravity. Only an imbalance in charge distribution should be able to do this. However, there will still be a net positive gravitational effect due to this imbalance. It is unclear how this would actually affect gravitation. It would mean that say two particles of negative charge could affect the gravitational energy of one particle of positive charge. So that instead of three particles contributing fully to the gravitational field only two would behave this way with the third then having a reduced contribution.

It is unclear whether or not this effect would be significant or whether it is actually true. It would need some observational evidence to back up the hypothesis. It may  only be noticeable when studying superconductors. Also the range may be very limited in scope. It would very likely be unfeasible to be able to generate a strong enough effect to lower escape velocity to a useful level in normal conditions.
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Offline jeffreyH

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Re: Charge and gravitation
« Reply #5 on: 16/06/2015 01:27:36 »
An important consideration concerning the above equation is how the relationship operates for a neutron star. Can the equation still equal zero?
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Offline jeffreyH

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Re: Charge and gravitation
« Reply #6 on: 16/06/2015 01:41:03 »
If we consider that a system with net zero charge due to cancellation is equivalent to the neutron then no effect will be seen. So therefore no effect upon gravitation. A net imbalance in charge is another matter. Although an ionized system will still fall at the same rate under the influence of a mass with net zero charge this may not be the same for the nature of the attraction towards the ionized source.
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Offline PmbPhy

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Re: Charge and gravitation
« Reply #7 on: 16/06/2015 18:13:42 »
Even if the algebraic solution to the indefinite integral includes an unspecified constant, "C," any time we evaluate the integral for any definite answer, the C will (must) always cancel out.

The added term in the third equation presented is a function of r, so it cannot be your constant.
When its the indefinite integral that one is looking for as when one wants to find the potential function, its the anti-derivative, aka integral, of the force. In this case the constant is chosen to be zero at a place which is convenient. In this case it's chosen to be zero so that the potential function at infinity is zero.