## Satellites feel the gravitational pull of the earth. They remain in orbit because of their velocity, which acts to counteract gravity. (The

Question

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 41000 km above the Earth’s surface.

(a) 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.)?

(b) At the satellite’s altitude, the acceleration of gravity is 0.177 m/s2. What is the magnitude of the tangential velocity that the satellite must have to remain in orbit?

c) How much time will the satellite take to orbit the earth?

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1 year 2021-07-27T16:13:53+00:00 1 Answers 13 views 0

## Answers ( )

a) The radius of the satellites travel is 47375 km

b) The magnitude of the tangential velocity is 2895.8 m/s

c) The time will be 102792.29 s

Explanation:

a) Given data:

h = height above the Earth = 41000 km

Re = radius of the Earth = 6375 km

The radius of the satellites travel is:

rs = Re + h = 6375 + 41000 = 47375 km = 47375000 m

b) The acceleration is produced to opposite gravity, thus:

Fg = Fc

mg = (m*v²)/r

g = v²/r

Clearing v:

$$v=\sqrt{gr} =\sqrt{0.177*47375000} =2895.8m/s$$

c) The time is:

$$t=\frac{2\pi r}{v} =\frac{2*\pi *47375000}{2895.8} =102792.29s$$