Question #138088
A planet of mass M moves around the Sun along an ellipse so that its minimum distance from the Sun is equal to r and the maximum distance to R. Making use of Kepler s laws, find its period of revolution around the Sun.
1
Expert's answer
2020-10-16T11:00:13-0400

Explanations & Calculations


  • According to Kepler's 3rd law, T2a3\small T^2 \propto a^3 : a = the semi major axis
  • Accordingly the periodic time does not depend on the eccentricity as any time delay is caught up with the increased speeds on such location of the path.
  • Therefore, an average value for the semi major axis could be written as a=(r+R)2\small a = \frac{(r+R)}{2}
  • From the motion of a uniform circle by the application of Newton's second law towards center it could be written for the period, T2=4π2GM0r3T2r3\small T^2 = \large\frac{4\pi^2}{GM_0}r^3 \small \to T^2 \propto r^3
  • Therefore,

T2=4π2GM0×(r+R)38T=π(r+R)32GM0\qquad\qquad \begin{aligned} \small T^2 &= \small \frac{4\pi^2}{GM_0}\times \frac{(r+R)^3}{8}\\ \small T &= \small \pi \sqrt{\frac{(r+R)^3}{2GM_0}} \end{aligned} : M0 is the mass of the Sun


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