# Period of a circular orbit given the path [closed]

A satellite is in circular orbit $500$ miles above the surface of the earth. What is the period of the orbit? (You may take the radius of the earth to be $4000$ miles, or $6.436 \times 10^6$ meters.)

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## closed as off topic by Ross Millikan, Micah, Paul, Trevor Wilson, robjohn♦Mar 1 '13 at 5:49

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This belongs on physics, though joshphysics has provided a good approach. The mass of the earth and the gravitational constant are still required, neither of which belongs here. –  Ross Millikan Mar 1 '13 at 4:31
@RossMillikan: it was migrated, but closed as too localized on physics.SE. It can be reopened here, if desired. –  robjohn Mar 1 '13 at 9:52
This question concerned the arc length of a path function. i.e. $L(t) = \int_{t_0}^{t} \|\mathbf{c}'(s)\| ds$ –  Sean Haugh Mar 1 '13 at 22:27
@SeanHaugh: it is a circular orbit. –  robjohn Mar 2 '13 at 9:07
@RossMillikan: we don't need to know the mass of the Earth and $G$ if we know the gravitational acceleration at the surface of the Earth: $\sim9.8\frac{\text{m}}{\text{sec}^2}$. –  robjohn Mar 2 '13 at 9:10
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Useful facts:

1. The force on the satellite has magnitude $\frac{GMm}{r^2}$ where $G$ is Newton's gravitational constant, $M$ is the Earth's mass, $m$ is the satellite's mass, and $r$ is its distance the the Earth's center, and the force is directed radially inward.

2. The acceleration of the satellite has magnitude $\frac{v^2}{r}$ where $v$ is its speed, and the acceleration is directed radially inward.

3. $\mathbf F=m \mathbf a$

4. The speed $v$ of the satellite is related to its orbital period $T$ by $vT = 2\pi r$.

Kepler's Third Law says $$P^2=\frac{4\pi^2r^3}{GM}$$ Since $g\sim9.8\frac{\text{m}}{\text{sec}^2}$ and $g=\frac{GM}{r_e^2}$, this becomes $$P^2=\frac{4\pi^2r^3}{gr_e^2}$$