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?

Answer 1

a) The radius of the satellite`s travel is 47375 km

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

c) The time will be 102792.29 s

Step-by-step explanation:

a) Given data:

h = height above the Earth = 41000 km

Re = radius of the Earth = 6375 km

The radius of the satellite`s 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=√(gr) =√(0.177*47375000) =2895.8m/s`

c) The time is:

`t=(2\pi r)/(v) =(2*\pi *47375000)/(2895.8) =102792.29s`