# Energy required to kick a planet orbiting the Sun from an elliptical to a parabolic path

I am trying to solve the following problem from Goldstein's Classical Mechanics:

A planet of mass $M$ is in orbit of eccentricity $e=1-\alpha$ where $\alpha<<1$, about the Sun. Assume that the motion of the Sun can be neglected and only gravitational forces act. When the planet is at its greatest distance from the Sun, it is struck by a comet of mass $m$, where $m<<M$, traveling in a tangential direction. Assuming the collision is completely inelastic, find the minimum kinetic energy the comet must have to change the new orbit to a parabola.

My reasoning:

Using the conservation of energy equation, we can find the energy of the planet in its elliptical orbit. Since the motion is purely tangential at this point, we have $$E = \frac{1}{2}M(r\dot\theta)^2 - \frac{k}{r}$$ Because it is an ellipse, we have $E<0$. Therefore, the energy the comet should have is $-E$ (since for a parabola, $E=0$), $$E_{comet}=\frac{k}{r}-\frac{1}{2}m(r\dot\theta)^2$$

All that now remains is to simplify the expression for $E_{comet}$ in accordance with the given data. Am I correct in my general reasoning?

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## 2 Answers

The expression for the energy of a parabolic orbit can be greatly simplified. A parabolic orbit is the lowest energy path that will allow the body to escape. So the sum of kinetic plus potential energy is exactly zero. It should be easiest to treat the elliptical orbit as a perturbative oscillation on a circular orbit. (The frequency of oscillation is exactly the period of the orbit.)

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Worth noting that this formulation works when potential energy is measured from zero at infinite remove (a very common, but not unique choice of gauge). –  dmckee Sep 7 '12 at 16:33

The orbital energy of two bodies is defined as the energy required to separate them to infinite distance: $$E = E_1 + E_2= \mu \frac{v^2}{2} - \frac{ G m_1 m_2 }{ r } = \mu \left( \frac{v^2}{2} - \frac{ G m_2 }{ r } - \frac{ G m_1 }{ r }\right) = - \frac{ G m_1 m_2 }{ 2 a } = - \frac{ G m_1 m_2 (1+e) }{ 2 d }$$ where: $\mu$ is the reduced mass, v is the relative (heliocentric) velocity, r is the relative separation, a is the semimajor axis, and d is the apoapsis distance. $E_1$ is the amount of energy applied to the 1st body, and $E_2$ is the amount of energy applied to the 2nd body.

So the energy necessary to knock a planet into a parabolic (escape) orbit is: $$\Delta\text{energy} = \frac{ G m_\text{sun} \, m_\text{planet}(1+e) }{ 2 d } = \frac{ G m_\text{sun} \, m_\text{planet} }{ 2 a }$$

Assuming the comet is of negligible mass, the amount of energy required is the same regardless of the position of the planet along its orbit. It is also not necessary to assume the eccentricity is nearly one.

Notice that the energy in your first equation is the energy of the planet ($E_2$) with respect to the center of mass of the system. An additional non-negligible amount of energy ($E_1$) is required to knock the Sun into a parabolic orbit. Also note that the r in your equations is defined as the barycentric speed of the planet (not the heliocentric speed).

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