Expression for components of gravitational force in spherical polar coordinates

Question

How do one find the gravitational force components of the earth of a satellite moving around the planet in spherical polar coordinates. Since gravitational always towards the centre does it meant that its components along the theta (co latitude) and phi (azimulth) are zero?

Answer 1

$\def \b {\mathbf}$ The components of gravitation force in inertial system are:

$$ \mathbf F_g=-\frac{m\,M\,G}{ |\b R|^{2}}\,\frac{\b R}{|\b R|}=\begin{bmatrix} F_x \\ F_y \\ F_z \\ \end{bmatrix}\tag 1$$

where $~\b R~$ is the position vector to the mass in polar coordinates

$$\b R= r\,\left[ \begin {array}{c} \cos \left( \theta \right) \sin \left( \phi \right) \\ \sin \left( \theta \right) \sin \left( \phi \right) \\ \cos \left( \phi \right) \end {array} \right] $$

thus: $$\b F_g=-\frac{m\,M\,G}{ r^2}\,\hat{\b{e}}_r$$

from here the component towards $~\hat{\b{e}}_\phi~$

$$\b F_g\,\cdot \hat{\b{e}}_\phi=\frac{m\,M\,G}{ r^2}\,\hat{\b{e}}_r\cdot \hat{\b{e}}_\phi=0$$

analog the component towards $~\hat{\b{e}}_\theta~$

$$\b F_g\,\cdot \hat{\b{e}}_\theta=\frac{m\,M\,G}{ r^2}\,\hat{\b{e}}_r\cdot \hat{\b{e}}_\theta=0$$

Answer 2

Lunthang Peter asked: "Since gravitational always towards the centre does it meant that its components along the theta (co latitude) and phi (azimulth) are zero?"

No, that would contradict the conservation of angular momentum since the angular velocity needs to increase when the radius becomes smaller. With the overdot representing differentiation by proper time and aligning the angle to the plane the acceleration is

$$\rm \ddot{r} = -\frac{G M}{r^2}+r \dot{\phi}^2 \color{#b0b0b0}{ -\frac{3 G M \dot{\phi}^2}{c^2}}$$

$$ \rm \ddot\phi = -\frac{2 \dot r \dot\phi}{r} $$ For Newtonian physics only use the black parts, and for Schwarzschild also the gray one.