Applications and Extensions
Exoplanets As a result of NASA's Kepler mission, astronomers have discovered hundreds of planetary systems beyond our own. In one such planetary system, the planet GJ 667Cc has a circular orbit of radius \(1.8\times 10^{10}{\textrm m}\) that takes \(28.1\) days to revolve about its central star GJ 667C.
Suppose another planet in circular orbit about central star GJ 667C is discovered. If this second planet is twice as far from the central star as planet GJ 667Cc, how many days would it take for the second planet to orbit the central star?
Source: http://www.kepler.nasa.gov.
Satellites of Jupiter The giant planet Jupiter has many satellites. The four largest satellites were discovered by Galileo in 1609. One of them, Io, is the most volcanically active body in the solar system. It is 422,000 km from Jupiter and takes 1.77 days for each orbit. Another satellite, Europa, probably has an ocean (and life?) beneath its icy surface. It is 671,000 km from Jupiter. A third satellite, Ganymede, is larger than the planet Mercury and takes 7.16 days for each orbit. Assuming the orbits of each of these satellites is circular,
Equation (2) on p. 802 states that \(\mathbf{r}\times \mathbf{v}\) is a constant. Explain why this means the motion of the planet lies in a plane.
Explain why \(\dfrac{dA}{dt}\) equal to a constant results in the statement given for Kepler's second law.
Challenge Problems
Kepler's Third Law
From the equation \(\left\Vert \mathbf{r}\right\Vert\,{=}\,\dfrac{\left\Vert \mathbf{D}\right\Vert ^{2}}{GM\,{+}\,\left\Vert \mathbf{H}\right\Vert \cos \theta}\,{=}\,\dfrac{\left\Vert \mathbf{D}\right\Vert ^{2}}{GM}\!\left(\!\!\dfrac{1}{1\,{+}\,e\cos \theta }\!\!\right)\), deduce the following: \begin{eqnarray*} \hbox{Length of the semimajor axis} &\;=\;&\dfrac{\left\Vert \mathbf{D}\right\Vert ^{2}}{GM} \left( \dfrac{1}{1-e^{2}}\right) \\ \hbox{Length of the semiminor axis} &\;=\;&\dfrac{\left\Vert \mathbf{D}\right\Vert ^{2}}{GM} \left( \dfrac{1}{\sqrt{1-e^{2}}}\right) \\ \hbox{Area of the ellipse} &\;=\;&\dfrac{\pi \left\Vert \mathbf{D}\right\Vert ^{4}}{G^{2}M^{2}}\left[ \dfrac{1}{( 1-e^{2}) ^{3/2}}\right] \end{eqnarray*}
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