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**Posts:**97**Joined:**Thu Jul 17, 2014 9:41 pm**Location:**Sydney - Australia

After reading about Vera Rubin's discovery that the outer stars in a galaxy have an anomalous velocity (anomalous rotation of galaxies) I realised that bodies such as planets and stars, simply move along equipotential paths equal to their surface potential.

GP theory explains relative velocity between bodies in a far simpler way than previous methods, it states that the relative velocity is a function of relative potential, it really is that simple.

\[\Delta v = c(\frac{\Delta \phi}{\Phi})\]

Further, the potential of a solid body is equal to it's surface potential which is a function of the total energy enclosed within it's surface, and can be simply expressed in Volts by expressing the total energy in eV and dividing by the number of nucleons.

Likewise if we imagine such an equipotential surface anywhere in a galaxy, we can be confident that the total energy inside that surface is equal to the surface potential times the number of nucleons it contains.

If we look at our Sun orbiting the galactic centre, we can estimate the Sun's surface potential to be around 930 Million Volts, which means it is travelling along a geodesic of 930 MV potential.

How do we know the Sun's surface potential?

From GP we already know the Earth's surface potential to be close enough to 930 MV so to work out the Sun's surface potential we only need to know the relative velocity. The earth orbits the Sun at around 30,000 m/s.

\[\Delta v = c*(\frac{\Delta \phi}{\Phi})\]

\[30000 \frac{m}{s} = c*(\frac{x}{938 MV})\]

\[x=-93,865 Volts\]

So the Sun's surface potential should be around 94,000 Volts below ours.

Steven

GP theory explains relative velocity between bodies in a far simpler way than previous methods, it states that the relative velocity is a function of relative potential, it really is that simple.

\[\Delta v = c(\frac{\Delta \phi}{\Phi})\]

Further, the potential of a solid body is equal to it's surface potential which is a function of the total energy enclosed within it's surface, and can be simply expressed in Volts by expressing the total energy in eV and dividing by the number of nucleons.

Likewise if we imagine such an equipotential surface anywhere in a galaxy, we can be confident that the total energy inside that surface is equal to the surface potential times the number of nucleons it contains.

If we look at our Sun orbiting the galactic centre, we can estimate the Sun's surface potential to be around 930 Million Volts, which means it is travelling along a geodesic of 930 MV potential.

How do we know the Sun's surface potential?

From GP we already know the Earth's surface potential to be close enough to 930 MV so to work out the Sun's surface potential we only need to know the relative velocity. The earth orbits the Sun at around 30,000 m/s.

\[\Delta v = c*(\frac{\Delta \phi}{\Phi})\]

\[30000 \frac{m}{s} = c*(\frac{x}{938 MV})\]

\[x=-93,865 Volts\]

So the Sun's surface potential should be around 94,000 Volts below ours.

Steven

Steven Sesselmann

Only a person mad enough to think he can change the world, can actually do it...

Only a person mad enough to think he can change the world, can actually do it...