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In a problem, I am given the aphelion and the perihelion of an elliptic orbit as well as the mas of the star at the center. Then I am asked two questions:

  1. Calculate the orbital period.
  2. Calculate the speed of the planet at the aphelion and the perihelion.

For the frist part, I calculated the mean radius and applied Kepler's third law, wich yielded the correct result. However, for the second part, I could not solve it using the orbital speed equation: $$ v_{orb} = \sqrt{\frac{G M}{r}} $$ Instead I solved it using conservation of momentum and enery.

In understand how to solve it using those two principles, but I cannot properly explain why the orbital speed equation is not correct. To reach that equation you simply combine Newton's law of gravitation with the centripetal force, and both apply at any point of the orbit in which the force and the velocity are perpendicular. How do I explain why I cannot use this approach?

Edit: I change the title to better reflect my question. The question is not how to solve the problem but why it cannot be solved another way.

jrglez
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2 Answers2

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The formula you use is only valid for circular orbits. The more general equation that should be used is the Vis-viva equation, which is simply a reformulation of conservation of energy: $$v = \sqrt{\mu\left(\frac2r - \frac1a\right)}$$ Where $\mu = GM$ is the gravitational parameter, and $a$ is the semi-major axis.
In a circle, the semi-major axis is the same as the radius, so the equation reduces to the form you use: $$v = \sqrt{\frac{\mu}{r}}$$ This of course only works if the orbit actually is a circle. There will be two points on an elliptical orbit where $a=r$ happens to be true, but in general the speed will be either lower or higher than the circular equivalent

Somnambulist
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I think that this is a possible explanation:

To derive the orbital speed equation you combine Newton's gravitational formula and the centripetal force. This is $$ | F_g | = G \frac{ M m } { r_S^2 } $$ and $$ | F_c | = m \frac{ v^2 } {r_c} $$ where $r_S$ us the distance to the star and $r_c$ the radius of curvature of the trajectory. Both forces are equal as long as the velocity is perpendicular to the force, but $r_S = r_c$ only if the orbit is circular. Therfore, you cannot simplify both $r$ to brain the desired formula.

jrglez
  • 332