# 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?

• You are looking at the wrong forces. Disregarding tidal forces, the gravitational acceleration due to the primary cancels out because both the observer and the satellite are falling simultaneously towards the primary with the same acceleration. What the observer will observe is the tidal force due to the primary. For example, the tidal acceleration due to the Moon on Earth's surface is about 1.1E−7 g (maximum) along the Earth-Moon axis. – AlexP Sep 3 '17 at 20:51

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.

• This was surprisingly more precise an answer than I was expecting. – TitaniumTurtle Sep 5 '17 at 14:01

It seems that doesn't matter.

Earth gravity varies a bit with geographical location, but you don't have to re-calibrate high-resolution laboratory scales every half hour because the relative position of your lab to the moon has changed.

I'll have to think a bit why this is so, because we do have tides, which certainly have something to do with the moons gravity.

• Disregarding tidal forces, you don't need to recalibrate the scales because both the scales and the Earth are falling simultaneously towards the Moon with the same acceleration. The tidal acceleration due to the Moon on Earth's surface is about 1.1E−7 g (maximum). – AlexP Sep 3 '17 at 20:42
• @AlexP OK, I don't have a scale that gives me seven significant digits. What you say is if I had, I'd have to take the current relative position of the moon into account? – Karl Sep 3 '17 at 20:57
• If it's a spring scale, yes. If it's a balance scale, no. – AlexP Sep 3 '17 at 20:58
• @AlexP That's clear. OK, physics make sense again. Tnx. – Karl Sep 3 '17 at 21:07