Use the exact values you enter to make later calculations. Satellites feel the g

Question

Use the exact values you enter to make later calculations.

Satellites feel the gravitational pull of the Earth. They remain in orbit because of their velocity, which acts to counteract gravity. (The satellite wants to fly out in a straight line, but gravity forces it to curve towards the Earth.) Consider a communications satellite that needs to be 38,000 km above the Earth's surface.

Use the exact values you enter to make later calculations. Satellites feel the gravitational pull of the Earth. They remain in orbit because of their velocity, which acts to counteract gravity. (The satellite wants to fly out in a straight line, but gravity forces it to curve towards the Earth.) Consider a communications satellite that needs to be 38,000 km above the Earth's surface. Assuming the satellite travels in a perfect circle, what is the radius of the satellite's travel? (The radius of the Earth is 6375 km.) At the satellite's altitude, the acceleration of gravity is 0.202 m/s2. What is the magnitude of the tangential velocity that the satellite must have to remain in orbit? How much time will the satellite take to orbit the Earth?

Explanation / Answer

a)

Just add the radius that the satellite needs to be at plus the radius of the earth.

30000 km + 6375 km = 36375 km = 36375000 m.

b)

You know that the only force acting on the satellite is gravity, and you know the acceleration is centripetal.

Fnet = ma
Fg = mv^2/r
mg = mv^2/r
g = v^2/r
v = sqrt(gr)
v = sqrt[(0.301 m/s^2)(36375000 m)]
v = 3309 m/s

c)

T = (2?r)/(v)
T = [2?(36375000 m)]/(3309 m/s)
T = 69069 s = 19.19 hours = 0.7994 days

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