**Problem : **
In Solving the Orbits we derived the
equation:

= - + |

From this, derive the expression we stated for 1/

= - y^{2} + y + |

We can then complete the square on the right-hand side and we have:

= - y - + + |

Then we let

= - p^{2} + 1 + = - p^{2} + C |

where we have just defined

= dθ'âá’cos^{-1}(p'/)|^{p}_{p1} = θ - θ_{1}âá’p = cos(θ - θ_{1}) - cos^{-1}(p_{1}/) = cos(θ - θ_{0}) |

Finally, recalling

= 1 + cosθ |

The quantity under the radical is defined a

**Problem : **
Using the expression we derived for (1/*r*), show that this reduces to *x*^{2} = *y*^{2} = *k*^{2} -2*kεx* + *ε*^{2}*x*^{2}, where *k* = ,
*ε* = , and cos*θ* = *x*/*r*.

= (1 + εcosθ)âá’1 = (1 + ε)âá’k = r + εx |

We can solve for

x^{2} + y^{2} = k^{2}–2kxε + x^{2}ε^{2} |

which is the result we wanted.

**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?

x - - - = |

Rearranging this equation into the standard form for an ellipse we have:

+ = 1 |

This is an ellipse with one foci at the origin, the other at (, 0), semi-major axis length

**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.

**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)?

r_{min} = = (6.44×10^{-67})L^{2} |

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