# Lagrange Points

### circular orbit approximation

 Period s Radius m Gravitational Parameter m3/s2 Sun \mu_s = 1.3271244002e20 Earth t_e = 31558149.76 r_e = 1.49598023e11 \mu_e = 3.98600442e14 Moon t_m = 2360591.51 r_m = 3.84399e8 \mu_m = 4.904869e12

## Colinear L1, L2, L3

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

 L1 r_1 Lagrange point between Earth and Sun 0.990029594 r_e 1,491,550 km in front of earth L2 r_2 Lagrange point behind Earth from Sun 1.0100371234 r_e 1,501,530 km behind earth L3 r_3 Lagrange point behind Sun from Earth 1.0001286404 r_e 19,244 km behind earth orbit

\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

## Moon-Earth Lagrange points

#### Circular orbits, ignoring barycenter

 L1 r_1 Lagrange point between Moon and Earth 0.848883087 r_m 57,905 km in front of moon L2 r_2 Lagrange point behind Moon from Earth 1.1681346921 r_m 64,426 km behind moon L3 r_3 Lagrange point behind Earth from Moon 1.001025435 r_m 393 km behind moon orbit

## 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.

Lagrange (last edited 2019-02-21 18:40:30 by KeithLofstrom)