How would gravity on a moon be affected based on orientation relative to its parent planet?
I am writing a sci-fi novel in which humanity has colonized a majority of the solar system, and a few minor colonies on nearby stars. Although calculating mass and gravity on individual planets and moons is mostly simple, I am struggling to find out if there is any major difference in gravity based on the moons orientation to the planet. For example, if I am standing on the surface of Titan and Saturn is directly above me, would Titan's gravity be somewhat countered by Saturn's?
Everything I keep finding is the basic equations to calculate gravity on an individual stellar body, but nothing that accounts very well for a counter source of gravity unless it starts getting into General Relativity and Einstein's Field Equation which is way over my head. I imagine in most cases that the difference would be negligible, but I want to account for reality.
Are there any equations or that might allow me to calculate the difference in gravity on a moon based on lunar orientation?
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1 answer
Let's make some assumptions:
- The moon is essentially spherical symmetric, with uniform density.
- There are no other gravitational effects from other celestial bodies.
Denote the position of the center of the planet by Cartesian coordinates $(x_p,0,0)$ and the position of the center of the moon by $(0,0,0)$, the origin. We can use spherical coordinates ($r,\theta,\phi$) to identify a point on the moon's surface. Since we've assumed that the moon is essentially a sphere, then any point on the moon's surface will always satisfy $r=R$, the radius of the moon.
The distance between a point on the moon (with Cartesian coordinates $(x,y,z)$) and the center of the planet is given by $$d=\sqrt{(x_p-x)^2+(0-y)^2+(0-z)^2}=\sqrt{(x_p-x)^2+y^2+z^2}$$ We can simply use spherical coordinates to expand this: $$x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta$$ $$d=\sqrt{(x_p-r\sin\theta\cos\phi)^2+r^2\sin^2\theta\cos^2\phi+r^2\cos^2\theta}$$ We then plug $d$ into the classical equation for gravity (and set $r=R$), and we find that the force on a body of mass $m$ due to the planet of mass $m_p$ is $$F=G\frac{m_pm}{(x_p-r\sin\theta\cos\phi)^2+r^2\sin^2\theta\cos^2\phi+r^2\cos^2\theta}$$ The difference in gravity is most drastic for two objects at antipodal points on the moon. Let's say that one is at $p_1=(R,0,0)$ and the other is at $p_2=(-R,0,0)$, where I'm using Cartesian coordinates again. We find that $$F(p_1)=G\frac{m_pm}{(x_p-R)^2},\quad F(p_2)=G\frac{m_pm}{(x_p+R)^2}$$ Since $R\ll x_p$ in most situations, we can use the binomial approximation. We first rewrite $$(x_p\pm R)^{-2}=x_p^{-2}\left(1\pm\frac{R}{x_p}\right)^{-2}\approx x_p^{-2}\left(1\mp2\frac{R}{x_p}\right)$$ Therefore, $$F(p_1)\approx Gm_pmx_p^{-2}\left(1+2\frac{R}{x_p}\right),\quad F(p_2)\approx Gm_pmx_p^{-2}\left(1-2\frac{R}{x_p}\right)$$ and the ratio is $$\frac{F(p_1)}{F(p_2)}\approx\frac{1+2\frac{R}{x_p}}{1-2\frac{R}{x_p}}$$ It should be apparent that this ratio is very close to $1$, but you can plug in some numbers and check, if you want. We've gone through a procedure quite similar to calculating the tidal acceleration.
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