Question

A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force exerted by the Earth on the satellite? b) the centripetal acceleration of the satellite? c) the speed of the satellite? d) the period of the satellite’s rotation around the Earth?

Answer 1

a)
`F = 660.576\,N`, b)
`a_(c) = 6.953\,(m)/(s^(2))`, c)
`v \approx 7255.423\,(m)/(s)`,
`\omega = 9.583* 10^(-4)\,(rad)/(s)`, d)
`T \approx 1.821\,h`

Step-by-step explanation:

a) The gravitational force exerted by the Earth on the satellite is:

`F = G\cdot (m\cdot M)/(r^(2))`

`F = \left(6.674* 10^(-11)\,(m^(3))/(kg\cdot s^(2)) \right)\cdot ((95\,kg)\cdot (5.972* 10^(24)\,kg))/((7.571* 10^(6)\,m)^(2))`

`F = 660.576\,N`

b) The centripetal acceleration of the satellite is:

`a_(c) = (660.576\,N)/(95\,kg)`

`a_(c) = 6.953\,(m)/(s^(2))`

c) The speed of the satellite is:

`v = \sqrt{a_(c)\cdot R}`

`v = \sqrt{\left(6.953\,(m)/(s^(2)) \right)\cdot (7.571* 10^(6)\,m)}`

`v \approx 7255.423\,(m)/(s)`

Likewise, the angular speed is:

`\omega = (7255.423\,(m)/(s) )/(7.571* 10^(6)\,m)`

`\omega = 9.583* 10^(-4)\,(rad)/(s)`

d) The period of the satellite's rotation around the Earth is:

`T = (2\pi)/(\left(9.583* 10^(-4)\,(rad)/(s) \right)) \cdot \left((1\,hour)/(3600\,s) \right)`

`T \approx 1.821\,h`