Problem : In Solving the Orbits we derived the equation:
= - + |
= - y ^{2} + y + |
= - y - + + |
= - p ^{2} + 1 + = - p ^{2} + C |
= dθ'âá’cos ^{-1}(p'/)|^{p} _{p1 } = θ - θ _{1}âá’p = cos(θ - θ _{1}) - cos^{-1}(p _{1}/) = cos(θ - θ _{0}) |
= 1 + cosθ |
Problem : Using the expression we derived for (1/r) , show that this reduces to x ^{2} = y ^{2} = k ^{2} -2kεx + ε ^{2} x ^{2} , where k = , ε = , and cosθ = x/r .
We have:
= (1 + εcosθ)âá’1 = (1 + ε )âá’k = r + εx |
x ^{2} + y ^{2} = k ^{2}–2kxε + x ^{2} ε ^{2} |
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 - ε ^{2})x ^{2} +2kεx + y ^{2} = k ^{2} . We can divide through by (1 - ε ^{2}) and complete the square in x:
x - - - = |
+ = 1 |
Problem : What is the energy difference between a circular earth orbit of radius 7.0×10^{3} kilometers and an elliptical earth orbit with apogee 5.8×10^{3} kilometers and perigee 4.8×10^{3} kilometers. The mass of the satellite in question is 3500 kilograms and the mass of the earth is 5.98×10^{24} kilograms.
The energy of the circular orbit is given by E = - = 9.97×10^{10} 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 = = 5.3×10^{6} meters. Then E = - = 1.32×10^{11} Joules. The energy of the elliptical orbit is higher.Problem : If a comet of mass 6.0×10^{22} 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 1.99×10^{30} kilograms)?
Its closest approach is just r _{min} , which is given by:
r _{min} = = (6.44×10^{-67})L ^{2} |
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