If a communications satellite needs to be 36,000 km above theequator, how fast m

Question

If a communications satellite needs to be 36,000 km above theequator, how fast must it travel to remain at thatheight? a. Assuming the satellite travels in a perfect circle, what isthe radius of the satellite's travel. b. At the satellites altitude, the acceleration of gravity is0.222m/s^2, what is teh tangential velocity that the satellite musthave to remain in orbit/ c. How long will the satellite take to orbit the earth? If a communications satellite needs to be 36,000 km above theequator, how fast must it travel to remain at thatheight? a. Assuming the satellite travels in a perfect circle, what isthe radius of the satellite's travel. b. At the satellites altitude, the acceleration of gravity is0.222m/s^2, what is teh tangential velocity that the satellite musthave to remain in orbit/ c. How long will the satellite take to orbit the earth?

Explanation / Answer

A.) Since the satellite must travel 36,000km above theearth's surface, and the satellite travels in a perfect circlearound the earth. The radius of the satellite's travel isgoing to be the radius of the earth, plus the distance between thesatellite and the earth. The radius of the earth at theequator is roughly 6,378km. So the total radius of thesatellite's orbit is 36,000 + 6,378 = 42,378km B.) It is given that the acceleration due to gravity atthe altitude of the satellite is 0.222m/s^2. If the satelliteis traveling in a circular path (and it is) it experiencescentripetal acceleration, which acts against the pull ofgravity. If the satellite were to remain at that altitude,the centripetal acceleration has to be equal to the accelerationdue to gravity (gravity pulls in, centripetal accel. pushesout). The equation for centripetal acceleration is: ac = V2 / r Where ac is the centripetal acceleration, V is thetangential velocity, and "r" is the radius of the object'spath. Since we have a constant radius (42,378km), and we know theacceleration that must be produced to oppose gravity (0.222m/s^2),then it is just a matter of solving for V. First, sinceour radius is in km, and our acceleration is m, we should convertone of the two. 42378km = 42378000m. Using the equation above: v = .222 * 42378000 =3067m/s C.) Since we know the radius of the path, we can findthe circumference. c = 2r = 2(42378000) =2.66x108m and we know our velocity, 3067m/s. we can use theequation v = d/t. t = d/v = 2.66x108/3067 = 86729.7 seconds. which isabout 24.1 hours. Note: I didn't really pay attention to sig figs, andprobably rounded a little weird so the numbers might vary slightly,but i think you get the idea.

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