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Problem : In Solving the Orbits we derived the equation:
= - + |
= - y2 + y + |
= - y - + + |
= - p2 + 1 + = - p2 + C |
= dθ'âácos-1(p'/)|pp1 = θ - θ1âáp = cos(θ - θ1) - cos-1(p1/) = cos(θ - θ0) |
= 1 + cosθ |
Problem : Using the expression we derived for (1/r), show that this reduces to x2 = y2 = k2 -2kεx + ε2x2, where k = , ε = , and cosθ = x/r.
We have:= (1 + εcosθ)âá1 = (1 + ε)âák = r + εx |
x2 + y2 = k22kxε + x2ε2 |
Problem : For 0 < ε < 1, use the above equation to derive the equation for an elliptical orbit. What are the semi-major and semi-minor axis lengths? Where are the foci?
We can rearrange the equation to (1 - ε2)x2 +2kεx + y2 = k2. We can divide through by (1 - ε2) and complete the square in x:x - - - = |
+ = 1 |
Problem : What is the energy difference between a circular earth orbit of radius 7.0×103 kilometers and an elliptical earth orbit with apogee 5.8×103 kilometers and perigee 4.8×103 kilometers. The mass of the satellite in question is 3500 kilograms and the mass of the earth is 5.98×1024 kilograms.
The energy of the circular orbit is given by E = - = 9.97×1010 Joules. The equation used here can also be applied to elliptical orbits with r replaced by the semimajor axis length a. The semimajor axis length is found from a = = 5.3×106 meters. Then E = - = 1.32×1011 Joules. The energy of the elliptical orbit is higher.Problem : If a comet of mass 6.0×1022 kilograms has a hyperbolic orbit around the sun of eccentricity ε = 1.5, what is its closest distance of approach to the sun in terms of its angular momentum (the mass of the sun is 1.99×1030 kilograms)?
Its closest approach is just rmin, which is given by:rmin = = (6.44×10-67)L2 |
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