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No Fear Translations of Shakespeare’s plays (along with audio!) and other classic works
Flashcards
Mastery Quizzes
Infographics
Graphic Novels
AP® Test Prep PLUS
AP® Practice & Lessons
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Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with
understanding the crux of the text.
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H.
Testimonials from SparkNotes Customers
No Fear provides access to Shakespeare for students who normally couldn’t (or wouldn’t) read his plays. It’s also a very useful tool when trying to explain Shakespeare’s wordplay!
Erika M.
I tutor high school students in a variety of subjects. Having access to the literature translations helps me to stay informed about the various assignments. Your summaries and translations are invaluable.
Kathy B.
Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with understanding the crux of the text.
Kay H.
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We can write expressions for both the angular momentum and the total energy. If
pθ is the magnitude of the momentum in the tangential direction, then
since this perpendicular to , L = rpθ. But pθ = mvθ = m = mr = mr.
Hence L = r(mr) = mr2. Hence:
L = mr2
We can also write an expression for the total energy as a sum of the radial
kinetic energy term, the angular kinetic energy term and the potential term:
E = 1/2m + -
Rearranging and dividing through the left side by
m2r4 and the right by L2, and canceling factors
of dt2 we find:
= - -
To find the equations of motion we want to find r in terms of
θ. In principle we could take the square root of both sides of
the above equation, separate the variables, integrate to find θ(r), and then invert to find r(θ). This involves a lot of
messy algebra which is not very enlightening, so we will just state the
result:
= (1 + εcosθ)
where ε is the eccentricity that we saw in Ellipses
and Foci and is now given by:
ε =
This equation determines the motion of all orbital systems in the universe.
We can also find the maximum and minimum values of r. The minimum occurs
where the expression for 1/r is a maximum. This is when cosθ = 1 and
the maximum is therefore . Thus:
rmin =
The maximum occurs for the minimum of 1/r. There are two cases: first, when
ε < 1, the minimum of 1/r is .
Thus:
rmax =
When ε≥1, the expression for 1/r can take on the value zero when
cosθ = - 1/ε. Hence the maximum value r can take on is infinite
in this case.
We can also take the equation and using r2 = x2 + y2, and cosθ = x/r, we can write:
x2 + y2 = - xε
Solving the orbits
The orbits are determined by the various values that ε can take.
Circular orbits
When ε = 0, the expression for ε tells us that E = - . The negative value of the energy just means that the
potential energy is more negative than the kinetic energy is positive. In this
case we have rmin = rmax = . The particle is trapped
at the very bottom of a potential well, and the radius does not change as it
goes around the orbit, hence forming a circle. Substituting this value for r
into the energy we have E = - . Note that we could have derived
this directly by summing the potential energy we found for a circular orbit with
the kinetic energy (Gravitational Potential Energy).
E = 1/2mv2 + U = - = -
Figure %: Potential well for a circular orbit.
In the case ε = 0 we can see that this equation simplifies to
x2 + y2 = . This describes a circle with
radius .
Elliptical Orbits
Elliptical orbits occur when 0 < ε < 1. This means that - < E < 0. Again the particle is trapped in a potential
well, oscillating now between rmin and rmax.
Figure %: Potential well for a elliptical orbit.
We can also solve the equation for ε in this range. The algebra
turns out to be complicated, but we find that:
+ = 1
where a = and b = . This is an ellipse with its center at (- L2ε/GMm2(1 - ε2), 0), and with semimajor and semiminor axis length a and b
respectively. It can also be shown that one focus of this ellipse is at the
origin.