Kepler Satellite
Physics / Kepler 's Law Question?
A satellite moves in a circular orbit around the Earth at a speed of 5000 m / s. Determine: a) The satellite altitude above Earth's surface b) The period of the satellite's orbit. The answers are: a) 9.58 * 10 ^ 6 mb) 5.57 pm I have no idea where to start, or equations to use. Please help?
For your information. M = mass of Earth RG = gravitational constant = Distance from the center of the Earth and the satellite. V = velocity of satellite Re = earth radius, H = height of satellite above the earth's surface, the acceleration of gravity: a = G * M / R ^ 2 For a circular orbit can match the acceleration of gravity with centripetal acceleration. acceleration centripetal a = V ^ 2 / RG * M / R ^ 2 = V ^ 2 / RG * M / V ^ 2 = R = Re + H = G * M / V ^ 2 – Re This formula will give the height of Satellite you can use one of Kepler's laws, I suppose, but the period can be easily found once you have the circumference of the orbit of R. = C = 2 * pi * r = Period C / V = 2 * pi * R / V These are the appropriate equations. I'll leave the details for you. If you have trouble adding some additional information your question and I will return to you.
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