Ph
ys
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1110 F
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ll 2004
L34 - 1
Lecture 34 12 November 2004
Announcements
• Reading Assignment for Mon, Nov. 15: Review for Exam 3
• Exam 3, Tuesday Nov 16, 7:30-9pm
• Same rooms as Exam 2
• Covers Knight Ch 8-11
• Monday, Review in class, Q&A 5-7pm, Ramaley C250
Elliptical Orbits and Potential Energy
Now we cannot use our centripetal acceleration formula, since the
radius is not constant. Since the force will now change with distance
between the sun and a planet (for example) this means the
acceleration is not constant and we would need to solve differential
equations to derive a solution. However, we can do an energy
analysis if we use the potential energy U that comes from Newton’s
law for gravity. By considering the work when moving an object m
from ri to rf while feeling the force from a planet with mass M, we can
derive that the correct function is
U=GMm
r
.
This formula can be confusing. It just depends on the distance
between the centers so it extends in all directions from a planet. Also
it starts from a very negative number and grows with distance. Don’t
be confused by considering the magnitude of U. At small r, it is very
small (large magnitude and negative). Also, it appears to blow up to –
infinity when r=0. But in fact, this never happens; the objects hit each
other first and then we have additional contact forces to prevent them
from going to r=0. Since it is a potential energy, we can always just
imagine a ball “rolling” down the potential energy curve; it will roll
towards r = 0 as it should, and potential energy grows the “higher”
above the ground (greater r) you go.
The figure shows an energy diagram for this potential energy. Note
that if we also plot a particle’s energy E as a constant at some value,
there are 2 very different behaviors. If E > 0, then there is no turning
