Editing Openai/67b71e2c-cc10-8008-9939-dfe494d1384d (section)

== gpt-4o ==
=== User: during lunar nights, do we know the minimum surface voltage of the lunar surface? ===
during lunar nights, do we know the minimum surface voltage of the lunar surface?

=== Assistant: During the lunar night, the surface voltage of the Moon is influenced primarily by the interaction of the surface with … ===
During the lunar night, the surface voltage of the Moon is influenced primarily by the interaction of the surface with the solar wind plasma and the ambient space environment, rather than direct photoelectric charging from sunlight.

===== - The minimum surface voltage of the lunar night side is estimated to be around -100 V to -200 V, but it can vary depending on location and plasma conditions. =====
* This negative charging occurs because: - The lunar surface is in shadow and does not experience photoelectron emission (which occurs on the sunlit side). - Electrons from the solar wind (or the plasma sheet when the Moon is in Earth's magnetotail) deposit a net negative charge on the surface. - The lack of photoemission means that there is no balancing positive charge, leading to an accumulation of negative charge.

===== 1. Solar Wind Plasma Density & Energy - In the normal solar wind (when the Moon is not inside Earth's magnetotail), the night-side voltage is estimated to be around -100 V to -200 V. - If solar wind density is higher, the negative charge accumulation could increase. =====
# Earth’s Magnetotail Influence - During part of its orbit, the Moon passes through Earth's magnetotail plasma sheet, which can cause the surface potential to vary. - Some measurements and models suggest that voltages could drop to -500 V or even lower in these conditions.
# Local Surface Composition & Topography - Rough terrain or crater edges may have locally stronger electric fields. - The presence of insulating dust (regolith) affects how charge distributes.

===== - Day Side: Lunar surface voltage can be +5V to +10V, and in extreme cases, up to +100V, due to UV-driven photoemission. =====
* Night Side: There is no UV, so charging is controlled by the balance of electrons and ions in the plasma environment, leading to a negative surface potential.

===== - Minimum surface voltage during lunar night: Typically -100V to -200V, but can drop to -500V or lower in Earth’s magnetotail. =====
* Why? Lack of photoemission, electron accumulation from the solar wind or magnetotail plasma.
* Implications: Strong negative charging could influence dust movement and spacecraft charging when landed on the Moon’s dark side.

Would you like more details on how these values were measured or modeled?

=== User: how much self-repulsion force is expected among the Regolith particles during lunar nights with a surface charge of -10… ===
how much self-repulsion force is expected among the Regolith particles during lunar nights with a surface charge of -100 volts?

=== Assistant: To estimate the self-repulsion force among lunar regolith particles due to electrostatic charging at -100 V during the … ===
To estimate the self-repulsion force among lunar regolith particles due to electrostatic charging at -100 V during the lunar night, we need to consider the Coulomb force acting between similarly charged dust particles.

===== The charge (QQQ) on a dust particle can be estimated using capacitance theory, assuming the regolith grains act like small conducting spheres: =====

Q=CVQ = C VQ=CV
where:
* C=4πϵ0rC = 4 \pi \epsilon_0 rC=4πϵ0r is the capacitance of a spherical dust grain,
* V=−100VV = -100VV=−100V is the surface potential,
* ϵ0=8.854×10−12\epsilon_0 = 8.854 \times 10^{-12}ϵ0=8.854×10−12 F/m is the vacuum permittivity,
* rrr is the radius of the dust grain.

For typical lunar dust grains, let’s consider a radius of r=1r = 1r=1 μm (10−610^{-6}10−6 m):

C=4π(8.854×10−12)(10−6)C = 4 \pi (8.854 \times 10^{-12}) (10^{-6})C=4π(8.854×10−12)(10−6)
C≈1.11×10−16 FC \approx 1.11 \times 10^{-16} \text{ F}C≈1.11×10−16 F
Thus, the charge on a 1-μm grain is:

Q=(1.11×10−16 F)(−100 V)Q = (1.11 \times 10^{-16} \text{ F}) (-100 \text{ V})Q=(1.11×10−16 F)(−100 V)
Q≈−1.11×10−14 CQ \approx -1.11 \times 10^{-14} \text{ C}Q≈−1.11×10−14 C

===== The Coulomb force (FFF) between two like-charged particles separated by a distance ddd is given by: =====

F=14πϵ0Q2d2F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}F=4πϵ01d2Q2
Assuming the two particles are 10 μm apart (d=10−5d = 10^{-5}d=10−5 m):

F=(9×109)(10−10)×(1.11×10−14)2F = \frac{(9 \times 10^9)}{(10^{-10})} \times (1.11 \times 10^{-14})^2F=(10−10)(9×109)×(1.11×10−14)2
F=(9×109)×(1.23×10−28)F = (9 \times 10^9) \times (1.23 \times 10^{-28})F=(9×109)×(1.23×10−28)
F≈1.1×10−18 NF \approx 1.1 \times 10^{-18} \text{ N}F≈1.1×10−18 N

===== - Compared to Lunar Gravity: The gravitational force acting on a single 1-μm particle of density ρ≈3000\rho \approx 3000ρ≈3000 kg/m³ is: Fg=mg=43πr3ρgF_g = mg = \frac{4}{3} \pi r^3 \rho gFg=mg=34πr3ρg =43π(10−6)3(3000)(1.62)= \frac{4}{3} \pi (10^{-6})^3 (3000) (1.62)=34π(10−6)3(3000)(1.62) ≈2.0×10−14 N\approx 2.0 \times 10^{-14} \text{ N}≈2.0×10−14 N The electrostatic repulsion force (~10−18^{-18}−18 N) is much weaker than gravity for 1-μm grains, meaning it won’t easily lift regolith off the surface. =====
* For Smaller Grains (0.1 μm): Since charge scales with radius, but weight scales with r3r^3r3, smaller grains experience relatively stronger repulsion. For 0.1 μm grains, the electrostatic force can approach or exceed their weight, possibly leading to dust lofting.

===== - The electrostatic self-repulsion force between 1-μm lunar dust particles at -100V is about 10−1810^{-18}10−18 N. =====
* This force is much weaker than lunar gravity for 1-μm grains, so large grains won’t levitate.
* However, submicron grains (0.1 μm and smaller) can experience forces comparable to or greater than their weight, supporting the idea of electrostatic dust lofting on the Moon.

Would you like calculations for different dust sizes or conditions (e.g., -500V charging in Earth's magnetotail)?

=== User: please redo these calculations but for a particle size of 40 microns and a spacing of 1 micron ===
please redo these calculations but for a particle size of 40 microns and a spacing of 1 micron

=== Assistant:  ===
===== 1. Electrostatic Repulsion Force: Felectric≈1.78×10−3 NF_{\text{electric}} \approx 1.78 \times 10^{-3} \text{ N}Felectric≈1.78×10−3 N =====
# Gravitational Force on the Particle: Fgravity≈1.30×10−9 NF_{\text{gravity}} \approx 1.30 \times 10^{-9} \text{ N}Fgravity≈1.30×10−9 N

===== - The electrostatic repulsion force is vastly stronger than the gravitational force by a factor of ~10⁶ (one million times). =====
* This means that two 40-micron regolith grains at a 1-micron distance would strongly repel each other, easily overcoming lunar gravity.
* Such strong repulsion could lead to significant dust particle lofting and redistribution, particularly in highly charged environments.

Would you like further refinements, such as calculations under different charge conditions (e.g., -500V)?

=== User: please show your work again, but this time use a spacing of 55 microns instead of 1 micron. ===
please show your work again, but this time use a spacing of 55 microns instead of 1 micron.  

Please show your work.

=== Assistant:  ===
===== #### =====

The capacitance of a sphere is given by:

C=4πϵ0rC = 4 \pi \epsilon_0 rC=4πϵ0r
where:
* ϵ0=8.854×10−12\epsilon_0 = 8.854 \times 10^{-12}ϵ0=8.854×10−12 F/m (vacuum permittivity),
* r=40r = 40r=40 microns = 40×10−640 \times 10^{-6}40×10−6 m.

C=4π(8.854×10−12)(40×10−6)C = 4 \pi (8.854 \times 10^{-12}) (40 \times 10^{-6})C=4π(8.854×10−12)(40×10−6)
C≈4.45×10−15 FC \approx 4.45 \times 10^{-15} \text{ F}C≈4.45×10−15 F
====== Charge is calculated as: ======

Q=CVQ = C VQ=CV
where V=−100VV = -100VV=−100V:

Q=(4.45×10−15)×(−100)Q = (4.45 \times 10^{-15}) \times (-100)Q=(4.45×10−15)×(−100)
Q≈−4.45×10−13 CQ \approx -4.45 \times 10^{-13} \text{ C}Q≈−4.45×10−13 C
====== The electrostatic force between two particles separated by distance ddd is given by Coulomb’s law: ======

F=14πϵ0Q2d2F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}F=4πϵ01d2Q2
Substituting:
* Q=−4.45×10−13Q = -4.45 \times 10^{-13}Q=−4.45×10−13 C,
* d=55d = 55d=55 microns = 55×10−655 \times 10^{-6}55×10−6 m,

F=(9×109)(55×10−6)2×(4.45×10−13)2F = \frac{(9 \times 10^9)}{(55 \times 10^{-6})^2} \times (4.45 \times 10^{-13})^2F=(55×10−6)2(9×109)×(4.45×10−13)2
F=(9×109)×(1.98×10−253.025×10−9)F = (9 \times 10^9) \times \left(\frac{1.98 \times 10^{-25}}{3.025 \times 10^{-9}}\right)F=(9×109)×(3.025×10−91.98×10−25)
F≈5.88×10−7 NF \approx 5.88 \times 10^{-7} \text{ N}F≈5.88×10−7 N
====== The gravitational force acting on the particle is: ======

Fg=mg=43πr3ρgF_g = mg = \frac{4}{3} \pi r^3 \rho gFg=mg=34πr3ρg
where:
* ρ=3000\rho = 3000ρ=3000 kg/m³ (density of lunar regolith),
* g=1.62g = 1.62g=1.62 m/s² (lunar gravity).

Fg=43π(40×10−6)3(3000)(1.62)F_g = \frac{4}{3} \pi (40 \times 10^{-6})^3 (3000) (1.62)Fg=34π(40×10−6)3(3000)(1.62)
Fg≈1.30×10−9 NF_g \approx 1.30 \times 10^{-9} \text{ N}Fg≈1.30×10−9 N

===== - Electrostatic Repulsion Force: Felectric≈5.88×10−7 NF_{\text{electric}} \approx 5.88 \times 10^{-7} \text{ N}Felectric≈5.88×10−7 N =====
* Gravitational Force: Fgravity≈1.30×10−9 NF_{\text{gravity}} \approx 1.30 \times 10^{-9} \text{ N}Fgravity≈1.30×10−9 N

===== - The electrostatic force is about 450 times stronger than lunar gravity at this spacing of 55 microns. =====
* This means that electrostatic repulsion can easily counteract gravity, potentially leading to particle levitation or dust motion under these conditions.

Would you like me to extend this to different surface voltages or environments (e.g., Earth's magnetotail at -500V)?

=== User: now please do this again, show your work, but do it for particle size 80 microns, with 100 micron particles spacing. ===
now please do this again, show your work, but do it for particle size 80 microns, with 100 micron particles spacing.

=== Assistant:  ===
===== #### =====

The capacitance of a sphere is given by:

C=4πϵ0rC = 4 \pi \epsilon_0 rC=4πϵ0r
where:
* ϵ0=8.854×10−12\epsilon_0 = 8.854 \times 10^{-12}ϵ0=8.854×10−12 F/m (vacuum permittivity),
* r=80r = 80r=80 microns = 80×10−680 \times 10^{-6}80×10−6 m.

C=4π(8.854×10−12)(80×10−6)C = 4 \pi (8.854 \times 10^{-12}) (80 \times 10^{-6})C=4π(8.854×10−12)(80×10−6)
C≈8.91×10−15 FC \approx 8.91 \times 10^{-15} \text{ F}C≈8.91×10−15 F
====== Charge is calculated as: ======

Q=CVQ = C VQ=CV
where V=−100VV = -100VV=−100V:

Q=(8.91×10−15)×(−100)Q = (8.91 \times 10^{-15}) \times (-100)Q=(8.91×10−15)×(−100)
Q≈−8.91×10−13 CQ \approx -8.91 \times 10^{-13} \text{ C}Q≈−8.91×10−13 C
====== The electrostatic force between two particles separated by distance ddd is given by Coulomb’s law: ======

F=14πϵ0Q2d2F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}F=4πϵ01d2Q2
Substituting:
* Q=−8.91×10−13Q = -8.91 \times 10^{-13}Q=−8.91×10−13 C,
* d=100d = 100d=100 microns = 100×10−6100 \times 10^{-6}100×10−6 m,

F=(9×109)(100×10−6)2×(8.91×10−13)2F = \frac{(9 \times 10^9)}{(100 \times 10^{-6})^2} \times (8.91 \times 10^{-13})^2F=(100×10−6)2(9×109)×(8.91×10−13)2
F=(9×109)×(7.94×10−2510−8)F = (9 \times 10^9) \times \left(\frac{7.94 \times 10^{-25}}{10^{-8}}\right)F=(9×109)×(10−87.94×10−25)
F≈7.12×10−7 NF \approx 7.12 \times 10^{-7} \text{ N}F≈7.12×10−7 N
====== The gravitational force acting on the particle is: ======

Fg=mg=43πr3ρgF_g = mg = \frac{4}{3} \pi r^3 \rho gFg=mg=34πr3ρg
where:
* ρ=3000\rho = 3000ρ=3000 kg/m³ (density of lunar regolith),
* g=1.62g = 1.62g=1.62 m/s² (lunar gravity).

Fg=43π(80×10−6)3(3000)(1.62)F_g = \frac{4}{3} \pi (80 \times 10^{-6})^3 (3000) (1.62)Fg=34π(80×10−6)3(3000)(1.62)
Fg≈1.04×10−8 NF_g \approx 1.04 \times 10^{-8} \text{ N}Fg≈1.04×10−8 N

===== - Electrostatic Repulsion Force: Felectric≈7.12×10−7 NF_{\text{electric}} \approx 7.12 \times 10^{-7} \text{ N}Felectric≈7.12×10−7 N =====
* Gravitational Force: Fgravity≈1.04×10−8 NF_{\text{gravity}} \approx 1.04 \times 10^{-8} \text{ N}Fgravity≈1.04×10−8 N

===== - The electrostatic force is about 68 times stronger than lunar gravity at this spacing of 100 microns. =====
* This means that electrostatic repulsion can still strongly influence dust motion, even for larger grains.
* Though less extreme than smaller grains at closer distances, dust could still be lofted or transported due to charging effects.

Would you like to extend this to different charge conditions, such as -500V in Earth's magnetotail?

=== User: now show your work for this again, but for a particle spacing of 800 microns ===
now show your work for this again, but for a particle spacing of 800 microns

=== Assistant:  ===
===== #### =====

The capacitance of a sphere is given by:

C=4πϵ0rC = 4 \pi \epsilon_0 rC=4πϵ0r
where:
* ϵ0=8.854×10−12\epsilon_0 = 8.854 \times 10^{-12}ϵ0=8.854×10−12 F/m (vacuum permittivity),
* r=80r = 80r=80 microns = 80×10−680 \times 10^{-6}80×10−6 m.

C=4π(8.854×10−12)(80×10−6)C = 4 \pi (8.854 \times 10^{-12}) (80 \times 10^{-6})C=4π(8.854×10−12)(80×10−6)
C≈8.91×10−15 FC \approx 8.91 \times 10^{-15} \text{ F}C≈8.91×10−15 F
====== Charge is calculated as: ======

Q=CVQ = C VQ=CV
where V=−100VV = -100VV=−100V:

Q=(8.91×10−15)×(−100)Q = (8.91 \times 10^{-15}) \times (-100)Q=(8.91×10−15)×(−100)
Q≈−8.91×10−13 CQ \approx -8.91 \times 10^{-13} \text{ C}Q≈−8.91×10−13 C
====== The electrostatic force between two particles separated by distance ddd is given by Coulomb’s law: ======

F=14πϵ0Q2d2F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}F=4πϵ01d2Q2
Substituting:
* Q=−8.91×10−13Q = -8.91 \times 10^{-13}Q=−8.91×10−13 C,
* d=800d = 800d=800 microns = 800×10−6800 \times 10^{-6}800×10−6 m,

F=(9×109)(800×10−6)2×(8.91×10−13)2F = \frac{(9 \times 10^9)}{(800 \times 10^{-6})^2} \times (8.91 \times 10^{-13})^2F=(800×10−6)2(9×109)×(8.91×10−13)2
F=(9×109)×(7.94×10−256.4×10−7)F = (9 \times 10^9) \times \left(\frac{7.94 \times 10^{-25}}{6.4 \times 10^{-7}}\right)F=(9×109)×(6.4×10−77.94×10−25)
F≈1.11×10−8 NF \approx 1.11 \times 10^{-8} \text{ N}F≈1.11×10−8 N
====== The gravitational force acting on the particle is: ======

Fg=mg=43πr3ρgF_g = mg = \frac{4}{3} \pi r^3 \rho gFg=mg=34πr3ρg
where:
* ρ=3000\rho = 3000ρ=3000 kg/m³ (density of lunar regolith),
* g=1.62g = 1.62g=1.62 m/s² (lunar gravity).

Fg=43π(80×10−6)3(3000)(1.62)F_g = \frac{4}{3} \pi (80 \times 10^{-6})^3 (3000) (1.62)Fg=34π(80×10−6)3(3000)(1.62)
Fg≈1.04×10−8 NF_g \approx 1.04 \times 10^{-8} \text{ N}Fg≈1.04×10−8 N

===== - Electrostatic Repulsion Force: Felectric≈1.11×10−8 NF_{\text{electric}} \approx 1.11 \times 10^{-8} \text{ N}Felectric≈1.11×10−8 N =====
* Gravitational Force: Fgravity≈1.04×10−8 NF_{\text{gravity}} \approx 1.04 \times 10^{-8} \text{ N}Fgravity≈1.04×10−8 N

===== - The electrostatic force is now comparable to lunar gravity (about 1.07 times stronger). =====
* This suggests that at 800-micron spacing, electrostatic repulsion and gravity are nearly balanced, meaning some particles might hover or move slowly rather than being lofted away.
* If there are additional forces like surface adhesion, they may dominate at this spacing.

Would you like to explore a different particle charge condition, such as -500V, or include additional forces like adhesion?

=== User: please do this again, and show your work, but this time do it for particles of different sizes, with first particles si… ===
please do this again, and show your work, but this time do it for particles of different sizes, with first particles size of 10 microns and the second particle size at 80 microns, and with a spacing of 100 micron spacing

=== Assistant: Step-by-Step Calculations for 10-Micron and 80-Micron Particles with 100-Micron Spacing ===
===== Step-by-Step Calculations for 10-Micron and 80-Micron Particles with 100-Micron Spacing =====

====== Step 1: Capacitance of Both Particles ======

The capacitance of a spherical particle is given by:

C=4πϵ0rC = 4 \pi \epsilon_0 rC=4πϵ0r
where:
* ϵ0=8.854×10−12\epsilon_0 = 8.854 \times 10^{-12}ϵ0=8.854×10−12 F/m (vacuum permittivity).

For the 10-micron particle (r=10×10−6r = 10 \times 10^{-6}r=10×10−6 m):

C10=4π(8.854×10−12)(10×10−6)C_{10} = 4 \pi (8.854 \times 10^{-12}) (10 \times 10^{-6})C10=4π(8.854×10−12)(10×10−6)
C10≈1.11×10−15 FC_{10} \approx 1.11 \times 10^{-15} \text{ F}C10≈1.11×10−15 F
For the 80-micron particle (r=80×10−6r = 80 \times 10^{-6}r=80×10−6 m):

C80=4π(8.854×10−12)(80×10−6)C_{80} = 4 \pi (8.854 \times 10^{-12}) (80 \times 10^{-6})C80=4π(8.854×10−12)(80×10−6)
C80≈8.91×10−15 FC_{80} \approx 8.91 \times 10^{-15} \text{ F}C80≈8.91×10−15 F

====== Step 2: Charge on the Particles ======

Charge is calculated as:

Q=CVQ = C VQ=CV
For the 10-micron particle (V=−100VV = -100VV=−100V):

Q10=(1.11×10−15)×(−100)Q_{10} = (1.11 \times 10^{-15}) \times (-100)Q10=(1.11×10−15)×(−100)
Q10≈−1.11×10−13 CQ_{10} \approx -1.11 \times 10^{-13} \text{ C}Q10≈−1.11×10−13 C
For the 80-micron particle:

Q80=(8.91×10−15)×(−100)Q_{80} = (8.91 \times 10^{-15}) \times (-100)Q80=(8.91×10−15)×(−100)
Q80≈−8.91×10−13 CQ_{80} \approx -8.91 \times 10^{-13} \text{ C}Q80≈−8.91×10−13 C

====== Step 3: Electrostatic Repulsion Force ======

The electrostatic force between two differently sized charged particles is given by Coulomb’s law:

F=14πϵ0Q10Q80d2F = \frac{1}{4\pi\epsilon_0} \frac{Q_{10} Q_{80}}{d^2}F=4πϵ01d2Q10Q80
Substituting:
* Q10=−1.11×10−13Q_{10} = -1.11 \times 10^{-13}Q10=−1.11×10−13 C,
* Q80=−8.91×10−13Q_{80} = -8.91 \times 10^{-13}Q80=−8.91×10−13 C,
* d=100d = 100d=100 microns = 100×10−6100 \times 10^{-6}100×10−6 m,

F=(9×109)(100×10−6)2×(−1.11×10−13)×(−8.91×10−13)F = \frac{(9 \times 10^9)}{(100 \times 10^{-6})^2} \times (-1.11 \times 10^{-13}) \times (-8.91 \times 10^{-13})F=(100×10−6)2(9×109)×(−1.11×10−13)×(−8.91×10−13)
F=(9×109)×(9.89×10−2610−8)F = (9 \times 10^9) \times \left(\frac{9.89 \times 10^{-26}}{10^{-8}}\right)F=(9×109)×(10−89.89×10−26)
F≈8.90×10−8 NF \approx 8.90 \times 10^{-8} \text{ N}F≈8.90×10−8 N

====== Step 4: Gravitational Force on Each Particle ======

The gravitational force acting on each particle is:

Fg=mg=43πr3ρgF_g = mg = \frac{4}{3} \pi r^3 \rho gFg=mg=34πr3ρg
where:
* ρ=3000\rho = 3000ρ=3000 kg/m³ (density of lunar regolith),
* g=1.62g = 1.62g=1.62 m/s² (lunar gravity).

For the 10-micron particle:

Fg=43π(10×10−6)3(3000)(1.62)F_g = \frac{4}{3} \pi (10 \times 10^{-6})^3 (3000) (1.62)Fg=34π(10×10−6)3(3000)(1.62)
Fg≈2.04×10−11 NF_g \approx 2.04 \times 10^{-11} \text{ N}Fg≈2.04×10−11 N
For the 80-micron particle:

Fg=43π(80×10−6)3(3000)(1.62)F_g = \frac{4}{3} \pi (80 \times 10^{-6})^3 (3000) (1.62)Fg=34π(80×10−6)3(3000)(1.62)
Fg≈1.04×10−8 NF_g \approx 1.04 \times 10^{-8} \text{ N}Fg≈1.04×10−8 N

===== Final Results =====
* Electrostatic Repulsion Force: Felectric≈8.90×10−8 NF_{\text{electric}} \approx 8.90 \times 10^{-8} \text{ N}Felectric≈8.90×10−8 N
* Gravitational Force on 10-Micron Particle: Fgravity, 10\mum≈2.04×10−11 NF_{\text{gravity, 10\mu m}} \approx 2.04 \times 10^{-11} \text{ N}Fgravity, 10\mum≈2.04×10−11 N
* Gravitational Force on 80-Micron Particle: Fgravity, 80\mum≈1.04×10−8 NF_{\text{gravity, 80\mu m}} \approx 1.04 \times 10^{-8} \text{ N}Fgravity, 80\mum≈1.04×10−8 N

===== Comparison and Interpretation =====
* The electrostatic force is about 10610^6106 times stronger than gravity for the 10-micron particle.
* The electrostatic force is about 8.5 times stronger than gravity for the 80-micron particle.
* This suggests that even large regolith grains can be repelled from each other due to electrostatic forces, leading to potential levitation or transport.
* The smaller particle is far more affected than the larger one, meaning fine dust can be significantly influenced by electrostatic interactions.

Would you like to explore different charge environments (e.g., -500V in Earth's magnetotail) or include additional effects like adhesion?