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Lately, I come across the following problem,

"The maximum and minimum speeds of a comet that orbits the Sun are 80 and 10 km/s respectively. The ratio of the aphelion distance of the comet to the radius of the Earth's orbit is ______. (Assume that Earth moves in a circular orbit of radius $1.5\times 10^8$ km with a speed of 30 km/s.)"

I couldn't find the aphelion distance from the information of max and min velocities alone. The eccentricity could be calculated using the formula $e = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$. Also from the given information about earth's orbit, the angular momentum of earth can be found out. If we have something like angular momentum per unit mass for orbit around a similar star(just saying) are similar, we can solve the problem. Is there any such relation between the orbits of planets around a common star?

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closed as off-topic by Aaron Stevens, G. Smith, John Rennie, Kyle Kanos, Jon Custer Jul 22 at 18:59

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  • $\begingroup$ Hint: What is the product of the minimum and maximum speed? $\endgroup$ – David Hammen Jul 21 at 11:40
  • $\begingroup$ From what i looked up i came across the formula $V_{max} = \sqrt{\frac{GM(1+e)}{a(1-e)}}$ Can you please tell me how this formula is derived? $\endgroup$ – Vibin Ram Narayan Jul 21 at 16:55
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A sudden intuition struck me! What I have overlooked was the information that gravitational force is conservative and the mechanical energy is conserved. That gives us: $$\frac{1}{2}mV^2_{max} - \frac{GMm}{r_p} = \frac{1}{2}mV^2_{min} - \frac{GMm}{r_a}$$ $$\frac{V^2_{max} - V^2_{min}}{2} = GM\left(\frac{1}{r_a} - \frac{1}{r_p}\right)$$ $V_{max}r_p = V_{min}r_a$ (Angular Momentum conservation) gives: $$\frac{(V_{max} + V_{min})(V_{max} - V_{min})}{2} = GM\left(\frac{1}{r_a} - \frac{V_{max}}{V_{min}r_a}\right)$$ $$r_a = \frac{2GM}{(V_{max} + V_{min})V_{min}}$$ Now $GM = V^2r$ (Where $V$ and $r$ are earth's velocity and radius) gives: $$r_a = \frac{2V^2r}{(V_{max} + V_{min})V_{min}}$$ Which is the required aphelion distance.

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