User:Mark Widmer/sandbox

Sandbox. Mark Widmer (talk) 01:17, 5 August 2021 (UTC)

Draft for additions to Hill_sphere New sections:

Hill sphere and L1 Lagrange point

-- added note in Formulas section

Hill sphere of the Sun

-- added to article

Hill sphere of objects that orbit Earth

The Moon -- added to article

Artificial satellites in low-Earth orbit -- added to article

Hill sphere of an object orbiting with another comparable-mass object

OR change to Hill sphere of two equal-mass object

The earlier formulas assume are approximately valid for a planetary mass that is much less than the star's mass. This no longer applies if the orbiting objects have comparable masses. This is the case for many binary star systems. For example, in the Alpha Centauri system, the stars Alpha Centauri A and B have masses that are 1.1 and 0.9 times that of the Sun, respectively.

For two equal-mass objects, let R be the distance between the objects. Each object is then in a circular orbit of radius R/2 about the center of mass, which is halfway between them.

Outline:

Follow the derivation for small mass ratio given at http://www.phy6.org/stargaze/Slagrang.htm

Mass ratio = x = m/M, with 0 < x <= 1

Equate Grav force (distance R) with centripetal force ( radius R /(1+x) )

GmM / R^2 = m v^2 / (R/(1+x)) = m v^2 (1+x) / R

Mult by R/m:

GM / R = (1+x) v^2

Solve for v^2 ?

Substitute for v = 2 pi (R/(1+x)) / T

GM / R = (1+x) 4pi^2 R^2 / (T^2 (1+x)^2)

GM / R^3 = 4pi^2 / T^2 (1+x)

${\frac {GM}{R^{3}}}={\frac {4\pi ^{2}}{(1+x)T^{2}}}$

An small-mass object at the L1 point, a distance r from object m, will have an orbit with radius ${\frac {R}{1+x}}-r$ and the same period T:

${\frac {GM}{(R-r)^{2}}}-{\frac {Gm}{r^{2}}}={\frac {v^{2}}{{\frac {R}{1+x}}-r}}$

$v={\frac {2\pi ({\frac {R}{1+x}}-r)}{T}}$,

so

${\frac {GM}{(R-r)^{2}}}-{\frac {Gm}{r^{2}}}={\frac {4\pi ^{2}({\frac {R}{1+x}}-r)^{2}}{T^{2}}}{\frac {1}{{\frac {R}{1+x}}-r}}={\frac {4\pi ^{2}({\frac {R}{1+x}}-r)}{T^{2}}}$

Since T is the same for the planet and an object at the L1 point,

${\frac {4\pi ^{2}}{T^{2}}}={\frac {GM(1+x)}{R^{3}}}={\frac {GM}{(R-r)^{2}({\frac {R}{1+x}}-r)}}-{\frac {Gm}{r^{2}({\frac {R}{1+x}}-r)}}$

Divide through by GM

${\frac {(1+x)}{R^{3}}}={\frac {1}{(R-r)^{2}({\frac {R}{1+x}}-r)}}-{\frac {x}{r^{2}({\frac {R}{1+x}}-r)}}$

${\frac {(1+x)}{R^{3}}}({\frac {R}{1+x}}-r)={\frac {1}{(R-r)^{2}}}-{\frac {x}{r^{2}}}$

${\frac {R-r(1+x)}{R^{3}}}={\frac {1}{(R-r)^{2}}}-{\frac {x}{r^{2})}}$

${\frac {R-r(1+x)}{R}}{\frac {(R-r)^{2}}{R^{2}}}=1-{\frac {x(R-r)^{2}}{r^{2}}}$

$(1-\rho (1+x))(1-\rho ^{2})=1-x({\frac {1}{\rho ^{2}}}-1)$

template math object:

${\frac {1}{R}}={\frac {1}{T}}$

${\frac {1}{r}}$