1. Centripetal acceleration
And you

2. Diagram the Force(s) acting on a bucket spinning at a constant velocity
Assume air resistance = 0
Did your diagram have only one force – the force of tension of the rope (my arm)?
Was that Ft toward the center of the circle?
You know where there is a net F there is a !
What direction was that acceleration?
That is correct!!! – in the same direction as the force!!! Towards the center of the circle!!!!

3. If:
I stop applying a F (I let go) in what direction will the bucket travel?

4. Well then:
What equation can we use to calculate the acceleration centripetal ( ac ) ?
Great question!
We will use the following equation:
ac = V2 / r
Did I just pull that one out of my Wozzoo?
No! Let me show you where that came from!

5. So: ac = v2 / r α + β = 90o α + θ = 90o So θ = β θ α
Similar. triangles β
So Δv /v = vΔt /r
d = v Δt
So Δv = v(vΔt) /r θ
Recall: ac = Δv /Δt v
Sub in: ac = v(vΔt) /r Δt
Δ v
So: ac = v2 / r

6. WOW

7. Where do we see Things moving in a circle?
Planets (req. Fg or movement will be in a straight line
Driving on an on ramp (req. Ffr or movement In straight line will be the result)
Swinging a bucket of water tied to a rope (req. Ft or movement will be in a straight line)
So all these forces are really Fc !!!!!!!!

8. Know the following: Speed = d/t = 2πr / T ac = v2/r = 4π2r / T2
Fc = m ac
Fc = m(v2 /r) (when we sub in from above) = m4π2r / T2
Recall that Fc = Fg or Ft or Ffr
Know how to derive what is in blue!!!!

9. The relationship of Fc to gravitation
Thanks to our friend Sir Issac Newton

10. Newton’s law of gravitation:
And you (really)
Fg= (Mass Earth)(Mass you) G r2
(r is distance between center of masses) (G = universal constant = X N * m2/ kg2 )
SO: Fg = G MEm / r2
So Fg provided Fc for objects in space
This breaks down for v-small objects (sub atomic)

11. How does this apply to satellites?
There is only one speed that a satellite can have in order to maintain an orbit with a constant r.
Set the Fg equation = to the Fc equation
GmME /r2 = mv2 / r
So: v = GME / r

12. v = GME / r Note: r is on the denominator so:
The closer the satellite is to Earth
(The smaller the r)
The larger the velocity needs to be!!!
Note: There is no mass of the satellite in the equation – so the mass of the satellite does not matter – only the velocity!

13. END HERE!!!!!!