3
$\begingroup$

How strong would the combined forces of electromagnetism on the earth and planets need to be, to mimic, and therefore, replace gravity?

$\endgroup$
11
$\begingroup$

There is no strength that the EM field could be to mimic gravity.

EM forces are proportional to charge and charge can be positive or negative. Gravitational forces are proportional to mass and mass can only be positive.

The lowest order of EM radiation is dipole. The lowest order of gravitational radiation is quadrupole.

The force between two charges of the same sign is repulsive. The force between two masses of the same sign is attractive.

In a fixed gravitational field all test objects accelerate at the same rate regardless of their mass. In a fixed EM field different test objects with different charge accelerate at different rates.

Gravitation respects the equivalence principle. EM does not respect the equivalence principle.

All of this is to say that it is not a matter of strength. They have different characteristics, regardless of strength.

$\endgroup$
5
$\begingroup$

The gravitational force between the Earth and Sun is easily calculated to be about $3.5\times10^{22}$ newtons.

If the Earth and the Sun had opposite charges of magnitude $3.0\times10^{17}$ coulombs, their electrostatic attraction could "replace" this gravitational attraction. (Unequal charges would also work as long as their product produced the same result.)

This would require about $1.9\times10^{36}$ excess electrons on the Earth and the same number of excess protons on the Sun, or vice versa.

The Earth is estimated to have between $10^{49}$ and $10^{50}$ atoms, so there would need to be one extra electron or proton on the Earth for about every ten trillion atoms.

The voltage at the surface of the Earth would be $4.2\times10^{20}$ volts and the electric field strength would be $6.6\times10^{13}$ volts per meter. This would be a problem for atomic structure as this kind of field strength would ionize atoms.

These calculations are for amusement. They are not an endorsement of the silly Electric Universe "theory".

$\endgroup$
1
$\begingroup$

It's not possible for an electromagnetic field to mimic gravity. This is because gravity is strictly attractive, while electric forces can be both attractive and repulsive.

You could, as G Smith's answer calculates, have an attractive force between the Sun and the Earth. That's as far as you can get however. If the Sun is negatively charged and the Earth is positively charged, then a positively-charged Mars would be attracted by the Sun, but would be repulsed by the Earth. Similarly if the Sun were positively charged and the Earth negatively charged, a positively charged Mars would be repulsed by the Sun and therefore cannot stay in orbit around it.

To first order, you could have the Sun be negatively charged and all the planets positively charged. However, the repulsive forces between planets aren't negligible in this case, and the Solar System will blow itself apart. Therefore, it's not possible for an electromagnetic field to mimic gravity.

$\endgroup$
0
$\begingroup$

Under certain field configurations the magnetic force on a magnetic dipole moment $m_z$ can be $$F=m_z\frac{\partial B_z}{\partial z}$$ Where $B_z$ is the $z$-component of the magnetic field. The magnetic field of a magnetic dipole can be as strong as $$B_z=\frac{\mu_0}{4\pi}\cdot\frac{2m_z}{r^3}$$ Where $\mu_0$ is the permeability of free space and $r$ is the distance from the dipole. Then $$\frac{\partial B_z}{\partial z}=-\frac{3B_z}r$$ So the dipole-dipole force between $2$ planets of the same dipole moment $m_z$ could be up to $$F=\frac{\mu_0}{4\pi}\cdot\frac{6m_z^2}{r^4}$$ To be the same as the gravitational between $2$ planets of mass $m$ $$F=\frac{Gm^2}{r^2}$$ Where $G$ is the gravitational constant would require $$m_z=\sqrt{\frac{4\pi Gm^2r^2}{6\mu_0}}=9.423\times10^{26}\text{A}\cdot\text{m}^2$$ Assuming planets of similar mass to earth's. The magnetic field at the surface would be $$B_z=\frac{\mu_0}{4\pi}\cdot\frac{2m_z}{R^3}=0.7303\,\text{T}$$ Where $R$ is the radius of the planet, assumed that of earth. A spherical neodymium magnet has roughly this field at its poles. If you wanted to float like a frog, your magnetic moment would be $$m_z=\frac{\chi_mm}{\mu_0\rho}B_z$$ If your magnetic susceptibility $\chi_m$ and density $\rho$ are roughly that of water, to produce the same force on your mass $m$ as the gravitational force $mg$ where $g$ is the gravitational acceleration at the surface of the earth, we would need $$-\frac{3\chi_mm}{\mu_0\rho R}B_z^2=mg$$ so $$B_z=\sqrt{-\frac{\mu_0\rho gR}{3\chi_m}}=5.380\times10^4\,\text{T}$$ Comparing the actual field at the earth's surface of perhaps $6.5\times10^{-5}\,\text{T}$ it seems pretty wimpy but it is strong enough to offer significant protection of our planet and atmosphere from extraterrestrial radiation, which other forces don't do a comparable job of.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.