Problem : In Solving the Orbits we derived the equation:
|= - +|
From this, derive the expression we stated for 1/r. Hint: Define y = 1/r and use the fact that = - . Making the y substitution, we have:
|= - y2 + y +|
We can then complete the square on the right-hand side and we have:
|= - y - + +|
Then we let p = y - and we have:
|= - p2 + 1 + = - p2 + C|
where we have just defined C in terms of the constant E, G, M, m, L. We can now separate variables:
|= dθ'âácos-1(p'/)|pp1 = θ - θ1âáp = cos(θ - θ1) - cos-1(p1/) = cos(θ - θ0)|
Finally, recalling p = (1/r) - , and we can pick an axis such that θ0 = 0:
|= 1 + cosθ|
The quantity under the radical is defined a ε, the eccentricity.
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|
We can solve for r and then use r2 = x2 + y2:
|x2 + y2 = k22kxε + x2ε2|
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?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 - - - =|
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 a = and semi-minor axis length b = .
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|