A satellite is launched to orbit the Earth at an altitude of $1.55\times10^7$ m for use in the Global Positioning System (GPS). Take the mass of the Earth to be $5.97\times10^{24}$ kg and its radius
$6.38\times10^6$ m.
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What is the orbital period of this GPS satellite, in hours?
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With what speed does it orbit the Earth, in m/s?
I'm a math major that was assigned a couple of physics problems. I've been on this problem for 30 minutes and can't get anywhere close, if you could leave me step by step instructions I'd really appreciate it, thanks for your time.
Best Answer
For circular orbits, we get to use this short derivation:
The centripetal acceleration holding the satellite on its orbit is $ \ a_c = \frac{v^2}{R} \ . $ It is the force of Earth's gravity at that distance from the center of the planet that provides that acceleration, $ \ a_g = \frac{GM}{R^2} \ . $ Here, $ \ R \ $ is the radius of the orbit, $ \ M \ $ is the mass of Earth, and $ \ G \ $ is the Universal Gravitational Constant...constant...consta--
We may thus set these two accelerations equal and solve for the "circular orbital velocity", $ \ v_c = \sqrt{\frac{GM}{R}} \ . $ You will need $ \ G = 6.67 \cdot 10^{-11} \ $ in System Internationale (SI) units. Don't forget that the orbital radius $ \ R \ $ is Earth's radius plus the orbital altitude. The distances must be in meters, as given; the mass in kilograms is fine. This answers question 2 (in $ \ \frac{\text{m.}}{\text{sec.}^2} \ $ ) , and question 1 is the circumference of the orbit divided by the circular orbital speed.
I will note, incidentally, that the orbital altitude given in the problem does not agree with that given in the articles on GPS ($ \ 2.02 \cdot 10^7 \ $ m.), so you will not get the period given there of half a sidereal day, or 43,082 seconds. [The altitude used is peculiar to that "constellation" of satellites, so this is often referred to as a "GPS orbit". Geosynchronous satellites complete their orbits in a full sidereal day, so they are at the still higher altitude of $ \ 3.58 \cdot 10^7 \ $ m.]