# Will Magnetic Field Changes Cause Orbit Decay?

no.

Thinsats are in a near-circular orbit - eccentricity is less than 6%. Most of the magnetic field change over an orbit happens because the dipole is offset from the rotation axis. The effective McIlwain L shell varies from 1.8 to 2.2 for the 14357 second orbit. The B field is:

B = B_0 ( R_E / R )^3 \sqrt{ 1 + 3 \sin( latitude ) }

B_0 = 30 μTesla. For the equatorial orbit, this simplifies to:

B = B_0 L^{-3}

... so the B field changes from 5.13 μT to 2.82 μT over the course of the orbit, and the \Delta B is 2.31 μT

The field change isn't sinusoidal, but let's pretend it is. The maximum field change rate is

{dB/dt}_{max} = \pi \Delta B / 14357 s = 500 pT/s

Magnetic induction : The line integral of the E field around a disk is the derivative of the flux through the disk:

2 \pi r E = \pi r^2 dB/dt

E = ( r/2 ) dB / dt

The power dissipated in an area element r ~ dr ~ d \theta is E^2 r ~ dr ~d \theta / \rho where \rho is the sheet resistance of the thinsat. Which will be difficult to compute, since the substrate is broken up with lots of slots, cavities for chips, etc., increasing the resistivity and reducing the power. But lets pick the worst case which is also easy to compute.

Assume an all aluminum ( density = 2.7 g/cm^{3} ) flat disk weighing 5 grams and 240 cm^{2}, 77 μm thick and 8.7 cm radius. The resistivity of aluminum is 2.33-8 Ω-m, the sheet resistance is ρ = 300 μ Ω per square .

The power dissipated in a ring radius r and width dr is

dP = ( 2 \pi r dr ) ( E^2 / \rho ) = 2 \pi r E^2 dr / \rho = 2 \pi r \left( ( r/2 ) dB/dt \right)^2 dr / \rho

dP = ( \pi/2 ) r^3 ( dB/dt )^2 / \rho dr

P = ( \pi r^4/8 ) ( dB/dt )^2 / \rho

P_{max} = \pi r^4 ( {dB/dt}_{max} )^2 /( 8 \rho )

P_{avg} = \pi r^4 ( {dB/dt}_{max} )^2 /( 16 \rho )

P_{avg} = π (0.087 m)^{4} (500 nT)^{2} / ( 16 * 300 μ Ω ) = 3E-21 Watts

Pretty small ...

This subtracts from orbital energy, and subtracts from altitude proportional to 2 / g ~ M = 2 /( 2.45m/s^2 * 0.005 g ) = 163 m/J at M288. The orbital decay rate is 5e-19 m/s, 15 picometers per year.

But that is for a disk flat to the magnetic equator. Thinsats are perpendicular to the equatorial plane, which is tilted about 11 degrees to the magnetic equator. SO, this already miniscule number is multiplied by (sin 21° )^{2} and the magnetic orbital decay is less than 1.5 pm/Y . Way too small to worry about.