Centripetal Acceleration Part II

[BitterSuites]

[SOLVED] Centripetal Acceleration Part II

Homework Statement

(From previous question) In order for a satellite to move in a stable circular orbit of radius 6689 km at a constant speed, its centripetal acceleration must be inversely proportional to the square of the radius of the orbit.

What is the speed of the satellite? The G is 6.67259e-11 and the mass of the Earth is 5.98e24kg.

The answer to this was 7723.55 m/s.

Part II:

Find the time required to complete one orbit. Answer in units of h.

Homework Equations

The book only gives one equation for T referencing orbits, which is v=2pir/T

The Attempt at a Solution

v=2pi/T
7723.55 = 2pi/T
7723.55T = 2pi
T = 2pi/7723.55
T = .000814

I think the homework system even laughed at me when I entered that answer :)

 

[Doc Al]

Homework Equations

The book only gives one equation for T referencing orbits, which is v=2pir/T

Right.

The Attempt at a Solution

v=2pi/T

Wrong.

Compare the two equations. (You left off the r.)

And don't forget to convert your answer to the desired units.

 

[BitterSuites]

*sigh* I keep making really silly mistakes.

So I input the correct equation but it came out correct.

v=2pir/t
7723.55 m/s = 2pi * 6689000m/T
7723.55T = 4.20282e7
T = 5221.57 seconds

Convert to hours

5221.57/60 = 90.6928 minutes
90.6928/60 = 1.51155 hours

Thank you so much.