For example, a geostationary satellite travels slower than the ISS. What is going on?
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$\begingroup$ Have you considered looking at the force balance that keeps satellites in orbit? $\endgroup$– AtmosphericPrisonEscapeCommented Apr 6, 2017 at 0:43
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$\begingroup$ @AtmosphericPrisonEscape That is a bit misleading, there is no force balance as the only force acting on the orbiting object is gravity. $\endgroup$– SE - stop firing the good guysCommented Apr 6, 2017 at 7:35
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$\begingroup$ @Hohmannfan: That doesn't matter. It doesn't seem to be a university-level question, and the most intuitive 'layman' way of understanding Keplerian orbits is through inertial forces. $\endgroup$– AtmosphericPrisonEscapeCommented Apr 6, 2017 at 14:48
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Kepler's third law in this situation tells us that $$ \frac{R^3}{P^2} = \frac{GM}{4\pi^2}, $$ where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P \propto R^{3/2}$.
To calculate the orbital speed, we divide the orbital circumference by the orbital period $$ v = \frac{2\pi R}{P}. $$
Then substituting for $P$ from Kepler's third law, we get $$ v = \sqrt{\frac{GM}{R}}$$
Thus, the larger the orbital radius, the slower the orbital speed.
Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force: $$ m \frac{v^2}{R} = G\frac{Mm}{R^2},$$ where $m$ is the mass of the satellite, which leads directly to the same result.
