Lagrange Points

circular orbit approximation

• Colinear L1, L2, L3

• Earth-Sun Lagrange points

• Moon-Earth Lagrange points

• Asteroid Lagrange points

Colinear L1, L2, L3

Test orbital radius

deviations perhaps due to eccentricity and tidal effects

\large \omega_e ~ = ~ 2 \pi / t_e ~ = ~ = 1.9909865e-7 radians/second

{ \large r_e } ~ = ~ { \LARGE \left( { \mu_s + \mu_e } \over { \omega_e }^2 \right) ^ {1 \over 3} } ~ = ~ 1.4959787e11 ~ m ~ \approx ~ { \LARGE \left( { \mu_s } \over { \omega_e }^2 \right) ^ {1 \over 3} } ~ = ~ 1.4959772e11 m (ignoring barycenter)

\large \omega_m ~ = ~ 2 \pi / t_m ~ = ~ = 2.6616995e-6 radians/second

{ \large r_m } ~ = ~ { \LARGE \left( { \mu_e + \mu_m } \over { \omega_m }^2 \right) ^ {1 \over 3} } ~ = ~ 3.8474861e8 ~ m ~ \approx ~ { \LARGE \left( { \mu_s } \over { \omega_e }^2 \right) ^ {1 \over 3} } ~ = ~ 3.831833e08 m (ignoring barycenter)

Earth-Sun Lagrange points

Circular orbits, ignoring barycenter

\large { k ~ ≡ \mu_e / \mu_s ~ ~ ~ ~ f ~ ≡ ~ r / r_e }

Earth/Sun L1

{ \large {\omega_e}^2 r_1 ~ = ~ } { \LARGE { { \mu_s r_1 } \over {r_e}^3 } } { \large ~ = ~ } { \LARGE { \mu_s \over {r_1}^2 } - { \mu_e \over ( r_e - r_1 )^2 } }

\large f ~ = ~ 1 - f { \Large \sqrt{ k \over { 1 - f^3 } } } . . . iterate until convergence . . . r_1 ~ = ~ f * r_e

Earth/Sun L2

{ \large {\omega_e}^2 r_1 ~ = ~ } { \LARGE { { \mu_s r_2 } \over {r_e}^3 } } { \large ~ = ~ } { \LARGE { \mu_s \over {r_2}^2 } + { \mu_e \over ( r_e - r_1 )^2 } }

\large f ~ = ~ 1 + f { \Large \sqrt{ k \over { f^3 - 1 } } } . . . iterate until convergence . . . r_2 ~ = ~ f * r_e

Earth/Sun L3

{ \large {\omega_e}^2 r_3 ~ = ~ } { \LARGE { { \mu_s r_3 } \over {r_e}^3 } } { \large ~ = ~ } { \LARGE { \mu_s \over {r_3}^2 } - { \mu_e \over ( r_e + r_1 )^2 } }

\large f ~ = ~ \left( 1 + k \left( { \LARGE f \over { f + 1 } } \right)^2 \right)^{ 1 \over 3 } . . . iterate until convergence . . . r_3 ~ = ~ f * r_e

• Earth-Sun Lagrange at 235 Re original

• magnetotail at 200 Re original

Moon-Earth Lagrange points

Circular orbits, ignoring barycenter

spreadsheet

Asteroid Lagrange points

How far away are the L1/L2 collinear Lagrange points from an asteroid? For a small secondary mass, the radius is approximately the Hill radius:

r \approx R { \Large \left( M_2 \over { 3 M_1 } \right)^{1 \over 3} }

For a spherical asteroid with density ρ and radius r₂, the mass is M_2 = { { 4 \pi } \over 3 } \rho {r_2}^3 . So, the Hill radius becomes:

r \approx R r_2 { \Large \left( { 4 \pi \rho } \over { 9 M_1 } \right)^{ 1 \over 3 } }

For R = 150e9 m × AU, M₁ = 2e30 kg, and ρ= 2000 kg/m³, this works out to:

r ≈ 170 × AU × r₂ (in the same units as r₂)

So, for a 100 meter radius asteroid 3 AU from the Sun, r ≈ 50 km. At that distance, the apparent angular diameter of the asteroid is 200/50000 radians or 400 microradians, or 0.23 degrees. The Sun's angular diameter is 0.09 degrees at that distance. Both angular sizes are proportional to the Sun's distance.

I can imagine a slowly rotating large asteroid with pivots at its "poles", strung with a pair of thin filaments to a "satellite" in the shadow of the asteroid, somewhat beyond its outer L₂ point. Imagine large PV wings cantilevered beyond the shadow. Wrap the "equator" of the asteroid in thin reflective foil to reduce its albedo, while exposing the poles to deep space. The asteroid will grow very cold, the emissivity will be small, and the satellite will be in a very cold shadow; a good place for an asteroid observation telescope.

The same geometries apply to an Atira/Apohele asteroid orbiting inside Earth's orbit; this would be a good place for a telescope designed to find more Atira/Apohele asteroids.