Question

6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory is just its weight, so its centripetal acceleration is about 9.8 m/s2 (the acceleration due to gravity near Earth’s surface). If Earth’s radius is about 6375 km, how fast must the satellite be moving? How long will it take for the satellite to complete one trip around Earth?

Answer 1

Engular velocity:
`w=1,24*10^(-3)/s`

Linear velocity:
`V=7905 m/s`

The time it takes:

`t=5060s=84min`

Step-by-step explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1)
`a=r*w^(2)`

Solving for w:

(2)
`w=\sqrt{(a)/(r) }`

Replacing a=9,8m/s2 and r=6,375,000m:

(3)
`w=\sqrt{(9,8m/s^(2))/(6375000m) }=1,24*10^(-3)/s`

And the angular velocity relates to the linear velocity:

`V=w*r=1,24*10^(-3)/s*6375000m=7905 m/s`

The perimeter of the orbit is:

`P=\pi *2*r=\pi *2*6375000m=40.05*10^(6)m`

The time it takes:

`t=(P)/(V) =(40.05*10^(6)m)/(7905 m/s)=5060s=84min`