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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}...


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I think the maximum velocity change from a flyby would help quantify this. $$\Delta v \leq \sqrt{\frac{GM}{r_P}}$$ That is, with perfect relative velocity and angle, the change in velocity from such a flyby perturbation is limited by the mass ($M$) of the asteroid, and the distance at closest encounter ($r_P$), which in the best case is to barely miss ...


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