# I Are we about to lose the Moon?

1. Feb 9, 2017

### Dave Wrixon

This extract from my TOE suggests yes, if Newton Gravitational Laws apply. Unless you can spot my error:

Why would the Moon rotate about the Earth if it is more strongly attracted to the Sun? The quoted solution to this issue is the Hill/Roche Sphere Radius. This radius is calculated by setting the Centrifugal Force equal to the combined Gravitational attraction of Planet and Star when the satellite is at its greatest distance from the Star. However, this approach seems rather flawed when considering that the due to its greater attractive force the Sun would tend to strip the Moon from the Earth at it closest approach to Sun. Taking a similar approach to Hill/Roche Sphere the equilibrium condition for the planet is:

mRΩ^2 = GmM/R^2

Where the satellite or moon (mass μ) is orbiting the star (mass M) with the same angular velocity ω at the distance R + r as the planet (mass m) at the distance R (permanent full moon position).

Ω^2 = GM/R^3

Consideration of what happens on Moon’s closest approach to the Sun, taking into account that Sun Gravitational pull on the Earth and Moon combined is already cancelled by their orbital angular momentum around the Sun, would give us:

μrω2 = Gμm/r^2 - (GμM/(R - r)^2 - GμM/R^2)

ω^2r^3R^2(R - r)^2 = Gm(R - r)^2R^2 - (GMr^2R^2 - GMr^2(R - r)^2)

ω^2r^3R^2(R - r)^2 = GmR^2(R - r)^2 – GMr^2(2rR - r^2))

Neglecting small terms:

ω^2r^3R^3 = GmR^3 - 2GMr^3

r = (GmR^3/(ω^2R^3 + 2GM))^1/3

r = (6.674 × 10^-11 x 5.972 × 10^24 × (149.6 × 10^9)^3 /((2.662 × 10^-6)^2 × (149.6 × 10^9)^3 + 2 x 6.674 × 10^-11 x 1.989 × 10^30))^1/3

r = (1.334 × 10^48/(2.373 × 10^22 + 2.655 × 10^20))^1/3

r = (5.559 × 10^25)^1/3

r = 382 × 10^6 m

This compares with an actual Lunar Orbit radius of 384.4 x 10^6 m.