Who knows how to calculate Sun's quadrupole moment on Mercury?

[guskz]

I believe the Roche limit is when the tidal force = the gravity force keep a planet together (R = distance between two celestial body, r= radius of smaller celestial body).

F_tide_far = GMm/(R+r)^2
F_tide_near = GMm/(R-r)^2
F_gravity_smaller_body = Gm/r^2

F_tide_near-F_tide_far = F_gravity_smaller_body

GMm (1/(R-r)^2 - 1/(R+r)^2) > = Gm/r^2

Wiki reduces it to: a_tidal = 2r GM/R^3

______________________________

Above the Tidal force = the front force - the rear force of the smaller planet

For the quadrupole moment (which they say also affects precession) I believe it is:

The front, rear, left, right forces of the larger planet while also taking into account the spin speed and direction of the larger planet? If so can anyone do the math using similar math reasoning as the tidal above?

[guskz]

On Thursday, February 26, 2015 at 12:15:10 AM UTC-5, guskz wrote:
> I believe the Roche limit is when the tidal force = the gravity force keep a planet together (R = distance between two celestial body, r= radius of smaller celestial body).
>
> F_tide_far = GMm/(R+r)^2
> F_tide_near = GMm/(R-r)^2
> F_gravity_smaller_body = Gm/r^2
>
> F_tide_near-F_tide_far = F_gravity_smaller_body
>
> GMm (1/(R-r)^2 - 1/(R+r)^2) > = Gm/r^2
>
> Wiki reduces it to: a_tidal = 2r GM/R^3
>

What's funny is my equation above G would be removed completely from the equation???

[Sam Wormley]

Sam writes:

Take a look at Wiki's Roche Limit Article:
> https://en.wikipedia.org/wiki/Roche_limit
> https://en.wikipedia.org/wiki/Roche_limit#Derivation_of_the_formula
> https://en.wikipedia.org/wiki/Roche_limit#A_more_accurate_formula
> https://en.wikipedia.org/wiki/Roche_limit#Physical_meaning_of_Roche_limit.2C_Hill_sphere_and_radius_of_the_planet

And finally look at the calculated examples:
> https://en.wikipedia.org/wiki/Roche_limit#Roche_limits_for_selected_examples