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Using Perifocal polar coordinates, where the x-axis points from the central body to the periapsis, and the polar equation for conic sections: $$r=\frac{a(1-e^2)}{1+e \cos(f)}$$ Provided parameters $\mu$ Standard Gravitational Parameter of the central body $r_p$ Periapsis Distance (Point P in the diagram) $v_p$ Speed at Periapsis $r$ Radial distance at ...


3

A diagram showing the cross section (longitudinal section) of the earth at the same longitude as the geostationary satellite is shown below. A location from the the geo stationary satellite aligns with the horizon is marked. Its latitude is also marked in the picture. At higher latitudes, the satellite is below the horizon. $$ cos(\mathrm{lat}) = \frac{6400}...


2

TL;DR It works for a very large amount of orbit changes First, let's take your equation and rearrange it to get rid of inconvenient values like force, time and mass. $$\frac{\Delta v_{\text{orb}}}{\Delta t} = \frac{F_{\text{retro}}}{m}$$ $$\Delta v_{\text{orb}} = \frac{F_{\text{retro}} \Delta t}{m} = - \Delta v_{\text{maneuver}}$$ $\frac{F\cdot t}{m}$ is ...


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