# Electric orbits Electrons in a beam are moving in a circle in a magnetic field (cyclotron motion). Lighting is caused by excitation of atoms of gas in a bulb. Credit: Marcin Białek.

This laboratory is an activity for you to calculate an electric or magnetic orbit of an astronomical object. While it is part of the astronomy course principles of radiation astronomy, it is also independent.

Some suggested entities to consider are electric fields, magnetic fields, mass, charge, Euclidean space, Non-Euclidean space, or spacetime.

## Evaluation

evaluation activity

Okay, this is an astronomy orbits laboratory, specifically to try out electric/magnetic orbits and where possible compare them to those calculated using gravity.

Yes, this laboratory is structured.

I will provide an example of an electric/magnetic orbit. The rest is up to you.

Questions, if any, are best placed on the discussion page.

## Control groups

For an electric or magnetic orbit, what would make an acceptable control group? Think about a control group to compare your calculations to.

I will provide one for my example.

## Conic sections

In terms of Cartesian coordinates, the general equation in three dimensions (x,y,z) for an object moving through the solar system is

$Ax^{2}+By^{2}+Cz^{2}+Dxyz+Exy+Fxz+Gyz+Hx+Iy+Jz+K=0.$ In terms of right ascension (RA) and declination (Dec), the Cartesian coordinates become

$r={\sqrt {x^{2}+y^{2}+z^{2}}}$ $RA=\operatorname {arccos} \left({\frac {z}{r}}\right)$ $Dec=\operatorname {arccotan} \left({\frac {y}{x}}\right).$ To recover the Cartesian coordinates

$x=r\,\sin {RA}\,\sin {Dec}$ $y=r\,\sin {RA}\,\cos {Dec}$ $z=r\,\cos {RA}.$ ## Sources

The Lorentz force is the combination of electric and magnetic force on a charge due to both fields.

For a particle of charge q moving with a velocity v through an electric field Ef and a magnetic field B, there is a force. There is an opposite reactive force. In the case of the magnetic field:

$\mathbf {F_{L}} =q\left(\mathbf {E_{f}} +\mathbf {v} \times \mathbf {B} \right)$ The gravitational force is given by

$F_{g}=G{\frac {m_{1}m_{2}}{r^{2}}}\$ ,

where:

• Fg is the force between the masses,
• G is the gravitational constant,
• m1 is the first mass,
• m2 is the second mass, and
• r is the distance between the centers of the masses.

## Objects

Consider an object of mass (m = 0.1M) passing through the heliosphere of the Sun at a velocity of 20 km s-1 from a galactic location originally outside the heliosphere. There are no other large objects in orbit around the Sun at the time of this object's entry. Jupiter, Saturn, Uranus, and Neptune are not present.

The object has a surface negative charge of 0.2Q and an intrinsic magnetic field of 100 Gauss (G) directed vertically. A current of 5 e- per second is following along magnetic field lines toward the Sun from the object. A comparable number of protons are following magnetic field lines from the Sun to the object.

At t0 the object is at 1,000 AU. Its RA is 14h 30m Dec +46° 30'.

Let the Sun have a magnetic field of 1 G directed vertically. The charge on the Sun (Q) is -3 x 1027 e.s.u.

## Gravity

The apparent forces on the Sun and the object are

$F_{g}=G{\frac {0.1M_{\odot }^{2}}{r^{2}}}\$ ,

where:

• Fg is the force between the masses,
• G is the gravitational constant,
• 0.1M is the first mass,
• M is the second mass, and
• r is the distance between the centers of the masses.

Conversion of units:

1. 1M ≈ 2 x 1030 kg
2. 1 AU ≈ 1.5 x 108 km
3. G ≈ 6.67 x 10-17 Nˑ(km/kg)2

Replacing the symbols with their physical equivalents yields

$F_{g}={\frac {6.67\times 10^{-17}km^{2}N}{kg^{2}}}{\frac {0.1(2\times 10^{30}kg)^{2}}{[(1.5\times 10^{8}km)(1000)]^{2}}}\,,$ $F_{g}={\frac {(6.67)(0.1)(4)}{(1.5)^{2}}}{\frac {10^{-17}\times 10^{60}}{10^{16}\times 10^{6}}}{\frac {km^{2}Nkg^{2}}{kg^{2}km^{2}}}\,,$ $F_{g}=1.19\times 10^{21}N\,.$ But, each of the constants used are approximates so Fg ≈ 1.19 x 1021 N.

### Electrostatics

Coulomb's law states that the electrostatic force $F_{q}$ experienced by a charge, $0.2Q_{\odot }$ at position $r_{q}$ , in the vicinity of another charge, $Q_{\odot }$ at position $r_{Q}$ , in vacuum is equal to:

$F_{q}={0.2Q_{\odot }^{2} \over 4\pi \varepsilon _{0}}{1 \over {r^{2}}},$ where $r$ is the distance between the two charges and the constant $\varepsilon _{0}$ is called the electric constant or the permittivity of free space and in SI units of C2 m−2 N−1, where C is Coulombs.

ε0 ≈ 8.854 x 10-12 C2 m−2 N−1, or

ε0 ≈ 8.854 x 10-12 C2 10-6 km−2 N−1, then

ε0 ≈ 8.854 x 10-18 C2 km−2 N−1.

1 e.s.u. ≈ 3.34 x 10-10 C.

$F_{q}={\frac {0.2}{4\pi }}{\frac {[(-3\times 10^{27}e.s.u.)(3.34\times 10^{-10}C/e.s.u.)]^{2}}{(8.854\times 10^{-18}C^{2}km^{-2}N^{-1})[(1.5\times 10^{8})10^{3}km]^{2}}},$ $F_{q}={\frac {0.2(-3)^{2}(3.34)^{2}}{4\pi (8.854)(1.5)^{2}}}{\frac {(10^{27}\times 10^{-10})^{2}}{10^{-18}(10^{8}\times 10^{3})^{2}}}{\frac {[(e.s.u.)(C/e.s.u.)]^{2}}{C^{2}km^{-2}N^{-1}km^{2}}},$ $F_{q}=8.02\times 10^{28}N.$ Each of the constants used are approximates so Fq ≈ 8.02 x 1028 N.

### Magnetostatics

If the velocity noted on approach is the vector component pointed toward the Sun, then the angle between the velocity vector and the magnetic field of the Sun is about 180°.

$1G(Gauss)={\frac {1T(Tesla)}{10,000}}={\frac {1Ns}{10^{4}C10^{-3}km}}.$ $F_{qB}=0.2(-3\times 10^{27}e.s.u.)(3.34\times 10^{-10}C/e.s.u.)(20kms^{-1}){\frac {{-1}Ns}{10^{4}Ckm10^{-3}}},$ $F_{qB}=0.2(-3)(3.34)(20)(-1){\frac {10^{27}\times 10^{-10}}{10^{4}\times 10^{-3}}}{\frac {(e.s.u.)(C/e.s.u.)(kms^{-1})Ns}{Ckm}},$ $F_{qB}=4.01\times 10^{17}N.$ Approximately, FqB ≈ 4.01 x 1017 N.

The total Lorentz force is predominantly the electrostatic force.

## Statuses

The Larmor radius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

“[F]or a particle of energy E in EeV and charge Z in a magnetic field B in µG [the Larmor radius (RL)] is roughly”

$R_{L}=1kpc{\frac {E}{ZB}}$ where

• $R_{L}\$ is the Larmor radius,
• $E\$ is the energy of the particle in EeV
• $Z\$ is the charge of the particle, and
• $B\$ is the constant magnetic field.

Or,

$R_{Lm}={\frac {mv_{\perp }}{|q_{m}|B_{\odot }}}$ $R_{L\odot }={\frac {M_{\odot }(-v_{\perp })}{|q_{\odot }|B_{m}}}$ where

• $R_{L}\$ is the Lamor or gyroradius,
• $m\$ is the mass of the charged particle,
• $v_{\perp }$ is the velocity component perpendicular to the direction of the magnetic field,
• $q_{m}\$ is the charge of the particle, and
• $B_{\odot }\$ is the constant magnetic field for the Sun.

For the object, as deflected by the Sun,

$1G(Gauss)={\frac {1T(Tesla)}{10,000}}={\frac {1Ns}{10^{4}C10^{-3}km}}={\frac {1kg10^{-3}kms^{-2}s}{10^{4}C10^{-3}km}},$ $1G={\frac {1kg}{10^{4}Cs}}.$ $R_{Lm}={\frac {(0.1)(2\times 10^{30}kg)(20km/s)}{(0.2)(3\times 10^{27}e.s.u.)(3.34\times 10^{-10}C/e.s.u.)(1kg/(10^{4}Cs))}},$ $R_{Lm}={\frac {(0.1)(2)(20)}{(0.2)(3)(3.34)(1)}}{\frac {10^{30}\times 10^{4}}{10^{27}\times 10^{-10}}}{\frac {kgkmCs}{(e.s.u.)(C/e.s.u.)kgs}},$ $R_{Lm}=2\times 10^{17}km.$ Or,

$R_{Lm}=1.33\times 10^{9}AU.$ For the Sun, as deflected by the object,

$R_{L\odot }={\frac {(2\times 10^{30}kg)(20km/s)}{(3\times 10^{27}e.s.u.)(3.34\times 10^{-10}C/e.s.u.)(100kg/(10^{4}Cs))}},$ $R_{L\odot }={\frac {(2)(20)}{(3)(3.34)(100)}}{\frac {10^{30}\times 10^{4}}{10^{27}\times 10^{-10}}}{\frac {kgkmCs}{(e.s.u.)(C/e.s.u.)kgs}},$ $R_{L\odot }=3.99\times 10^{15}km.$ Or,

$R_{L\odot }=2.66\times 10^{7}AU.$ ### Escape velocity

For a spherically symmetric body, the escape velocity at a given distance is calculated by the formula:

$v_{e}={\sqrt {\frac {2GM}{r}}},$ where G is the universal gravitational constant (G = [6.67×10−17 km2 kg−2 N]), M the mass of the planet, star or other body, and r the distance from the center of gravity. The value GM is called the standard gravitational parameter, or μ, and is often known more accurately than either G or M separately.

The escape velocity for the Sun at 1,000 AU is

$v_{e}={\sqrt {\frac {2(6.67\times 10^{-20}km^{3})(2)(10^{30}kg)}{kgs^{2}(10^{3})(1.5\times 10^{8})km}}},$ $v_{e}={\sqrt {{\frac {2(6.67)(2)}{1.5}}{\frac {10^{-20}\times 10^{30}}{(10^{3})\times (10^{8})}}{\frac {(km^{3})kg}{kgs^{2}km}}}},$ $v_{e}={\sqrt {1.779km^{2}/s^{2}}}=1.33km/s.$ The escape velocity for the object at 1,000 AU is

$v_{e}={\sqrt {\frac {2(6.67\times 10^{-20}km^{3})(0.1)(2)(10^{30}kg)}{kgs^{2}(10^{3})(1.5\times 10^{8})km}}},$ $v_{e}={\sqrt {{\frac {2(6.67)(0.1)(2)}{1.5}}{\frac {10^{-20}\times 10^{30}}{(10^{3})\times (10^{8})}}{\frac {(km^{3})kg}{kgs^{2}km}}}},$ $v_{e}={\sqrt {0.1779km^{2}/s^{2}}}=0.422km/s.$ ### Pauli's exclusion principle

Another possible concern is Pauli's exclusion principle that incoming charged particles of the same sign must have opposite spins.

### Initial position

At initial detection

1. the Lamor radius is 1.33 x 10^9 AU,
2. the escape velocity of the Sun is 1.33 km/s at 1,000 AU,
3. the escape velocity of the object is 0.422 km/s at 1,000 AU.

## Electric orbits

In order for the object to have an orbit around the Sun, its approach velocity of 20 km/s must be reduced to below the Sun's escape velocity of 1.33 km/s.

The electrostatic force is the major available force of repulsion to slow down the object. As it continues to travel closer to the Sun, the forces increase. But, an estimate can be calculated of how much time is required to reduce the object's velocity to below the Sun's escape velocity.

$v_{1}=20km/s-(F_{q}/m)t<1.33km/s.$ $20km/s<1.33km/s+(F_{q}/m)t,$ $(F_{q}/m)t>20km/s-1.33km/s,$ $t>{\frac {20km/s-1.33km/s}{F_{q}/m}},$ $t>{\frac {20km/s-1.33km/s}{(8.02\times 10^{28}kg10^{-3}km/s^{2})/((0.1)(2\times 10^{30}kg))}},$ $t>{\frac {20-1.33}{8.02/((0.1)(2))}}{\frac {10^{30}}{10^{28}\times 10^{-3}}}{\frac {(km/s)kg}{kg(km/s^{2})}},$ $t>4.66\times 10^{4}s.$ Or,

$t>12.9h.$ ## Evaluation

To assess your orbit calculations, including your justification, analysis and discussion, I will provide such an assessment of my example for comparison.

Back of the Envelope Electric Orbit

by Marshallsumter (discusscontribs) 22:49, 21 January 2014 (UTC).

Abstract

Using elementary formula an object's approach to the Sun is described using gravity and the Lorentz force. As both the Sun and the object have charges, the possible Larmor radii are calculated. Each object has an escape velocity with respect to the other. With an initial velocity toward the Sun it is shown that the overwhelming force is the electrostatic repulsion between the Sun and the object. Should this force continue to operate as the object moves toward the Sun, after more than about 12.9 hours the object's velocity will be below the escape velocity of the Sun. Whether the object's orbit is stable may depend on any likely tangential velocity. If there is none the object will eventually be repelled back out of the heliosphere. As there are currents of electrons and positive charges flowing between the two objects, the electrostatic repulsion could be reduced until only gravity holds the object in an orbit around the Sun.

Introduction

The objective is to test whether the measured charge on the Sun and a similar astronomical object entering the heliosphere would be sufficient to slow the object down sufficiently to allow its incoming velocity to drop below the Sun's escape velocity. In the past only gravity has been considered, with the result that the Sun cannot capture another astronomical object into an orbit unless its incoming velocity is less than 0.5 km/s and another object such as Jupiter assists.

Experimentation

Using an object of mass 10 % of the Sun's and 20 % of the Sun's charge with an incoming velocity toward the Sun of 20 km/s several calculations were made of the instantaneous forces between the objects at the initial approach.

Larmor radii were calculated to show the likely deflection of each object by the other. The escape velocities were also calculated as a simple way of assessing whether the forces involved could brake the object sufficiently to allow capture.

Discussion

The possible effect of the Pauli exclusion principle was not assessed. The forces were not allowed to vary differentially as the object continued to approach which would likely reduce the time to capture. If no tangential velocity exists the object would eventually be repelled back out of the solar system. Charge dissipation was not considered even though initial currents were indicated. Increases in either currents may have mitigated the repulsion and either sped up or slowed down the capture.

Conclusion

Although all likely conditions were not considered, a back-of-the-envelope calculation suggests that an object entering the heliosphere with a charge comparable to that of the Sun, but at 40 times the currently accepted incursion and capture velocity, could be captured into some kind of orbit. The stability of the orbit with time was not assessed.

## Hypotheses

1. Electric currents between major astronomical objects can produce stable orbits.