How would gravity on a moon be affected based on orientation relative to its parent planet?

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 ($r,\theta ,\varphi$) 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: $$x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi ,z=r\cos \theta$$ $$d=\sqrt{(x_p-r\sin \theta \cos \varphi {)}^2+r^2{\sin }^2\theta {\cos }^2\varphi +r^2{\cos }^2\theta }$$ 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 $$F=G\frac{m_{p}m}{(x_p-r\sin \theta \cos \varphi {)}^2+r^2{\sin }^2\theta {\cos }^2\varphi +r^2{\cos }^2\theta }$$ 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 $$F(p_1)=G\frac{m_{p}m}{(x_p-R{)}^2},F(p_2)=G\frac{m_{p}m}{(x_p+R{)}^2}$$ Since $R\ll x_p$ in most situations, we can use the binomial approximation. We first rewrite $$(x_p\pm R{)}^{-2}=x_p^{-2}{(1\pm \frac{R}{x_p})}^{-2}\approx x_p^{-2}(1\mp 2\frac{R}{x_p})$$ Therefore, $$F(p_1)\approx G{m}_{p}m{x}_p^{-2}(1+2\frac{R}{x_p}),F(p_2)\approx G{m}_{p}m{x}_p^{-2}(1-2\frac{R}{x_p})$$ 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.

posted over 7 years ago

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