Catchy Title

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Catchy Title

• By Brandon Cathey and Keaten Holley

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Keplers Laws(As Shown in Calc Text)

• 1. A planet revolves around the Sun in an
  elliptical orbit with the Sun at one focus.
• 2. The line joining the Sun to a planet sweeps
  out equal areas in equal times.
• 3. The square of the period of revolution of a
  planet is proportional to the cube of the
  length of the major axis of its orbit.

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Proving Keplers Second Law

• GIVEN
• R in two dimensional cartesian coordinates from
  polar coordinates
• r (rcos?)i (rsin?)j
• h is defined by
• h r x v

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1.a) Prove
1.b) Deduce that

• If
• Which yields,

• As h is in k direction, both sides of equation
  can be divided by a unit vector in k direction,

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1.c) Show that

• If AA(t) is the area swept out by the radius
  vector rr(t) in time interval to,t

• A dA can be defined by

• Where,

• For small amount of time, dt,

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1.d) Deduce that

• and,
• Therefore,

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2.a) Show that

• Start with
• Where pab is the total area of the ellipse, the
  same area swept out by the line from one focus in
  the duration of one period of revolution.
• This yields

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2.b) Show that

• P.878 of text defines the following variables
• c is the magnitude of a constant vector, c, G is
  the gravitational constant, and M is the mass of
  the larger body (in this case the sun).
• Simply multiplying e and d gives the first part
  of the equation.

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• For the second part, we look to the following
  equation derived in the text for Keplers 1st
  Law
• Looking at the minimum and maximum values of r,
  we can derive equations for a and b using the
  geometry of an ellipse
• a the arithmetic mean of the two, and b the
  geometric mean

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• Therefore,

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2.c) Show that

• Looking at results from parts 2.a and 2.b,
• Solving for h2 for both equations,

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3.) Use Keplers 3rd Law to find the length of
the major axis of the Earth

• Period of Earth is approximately 365.25 days,
  mass of the Sun is 1.99 X 1030kg, and G is equal
  to 6.67 X 10-11 Nm2/kg2.
• From part 2.c, we found
• With the Earths period in seconds 3.16 X 107,

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4.) Find the altitude necessary to create a
geosynchronous satellite.

• Period of satellite is 24 hours, G is equal to
  6.67 X 10-11 Nm2/kg2, the mass of the earth (now
  M) is 5.98 X 1024 kg, and its radius is 6.37 X
  106 m.
• From part 2.c, we found
• With a being the distance to the satellite from
  Earths center of mass
• There fore,
• Satellite must be at an altitude of 3.59 X 107m,
  which is about 240,000 miles (over 5X larger than
  Earths radius).