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Universal Gravitational Potential Energy
Consider the case of dropping a ball from height ho near the surface of the earth. What if we want to use the conservation of energy approach someplace other than on the surface of the earth? What if we leave earth’s surface and head out into space? Clearly, we can’t use Ug = mgh because that formulation assumes the force of gravity is constant. Let’s follow the same approach we used above. m hf ho
mg
Assume that the force of gravity mg stays nearly constant. We say that the work done by gravity is the area under the Fgrav vs h graph, which is mgho – mghf. But in general: W = ΔK = - ΔUg and W = Area = mgho – mghf so – ΔUg = mgho – mghf or Ugo – Ugf = mgho – mghf. It makes sense to say Ugo = mgho and Ugf = mghf This leads to the familiar result that Ug = mgh where h is measured from wherever you want: you choose the reference level. 0 mg h Fgrav E A R T H
area = mgho - mghf
Strange side view chosen to go with the graph below. (Earth’s curvature is hugely exaggerated.) hf ho
Fgrav vs h
Consider the case of dropping a ball from an initial position r 0 pretty far out in space. Let the ball fall to a position rf. Let’s find the work done by gravity in going from r 0 to rf. We say that the work done by gravity is the area under the Fgrav vs r curve, which is: rf rf rf W = ∫ Fgrav dr = ∫ ( – GMm/r 2 )dr = GMm/r| ro ro ro
= + GMm/rf – GMm/ro
But W = ΔK = – ΔUg
So – (Ugf – Ugo) = GMm/rf – GMm/ro.
It makes sense to say Ugo = – GMm/ro , and Ugf = – GMm/rf
This leads in general to the idea that where r is measured from the center of mass M. Now let’s put this idea to use. Fgrav (^0) r W = |area| = GMm/rf - GMm/ro M m rf ro Ug = - GMm/r Note: Don’t worry too much about those pesky minus signs: they are a pain! Note: You may be concerned that the area from this formula ended up +. This is because we are going backwards toward M, making the work +. Now quit worrying about those darn – signs! Fg = – GMm/r 2 Note: If we start from infinite distance, ro = ∞, and Ugo = 0. What we are doing here is putting
our reference level (Ug= 0) at ∞.
Anything closer than ∞ will have to have gravitational potential energy less than 0, so that potential energy will have to be negative!
Prob 2: What if you were to drop the apple in the problem above from an infinite distance away from the earth? Find the impact speed on earth in this case. Follow the same steps you did in prob 1. Rearth ro = ∞ Mearth m
Prob 3: Escape speed. Calculate the minimum speed you would have to throw the apple with to launch it from the surface of the earth so it will never return. In other words, how fast will you have to throw the apple so that when it gets infinitely far from earth, it will just barely come to rest. This is called “escape speed.”
Rearth
rf = ∞
Mearth
vo =?
Write down the general formula for escape speed from an object in terms of the objects mass and radius:
vesc =
