Total energy of a planet in orbit

Question

Consider a planet in orbit. It has some kinetic energy and some gravitational potential energy. Now my question is, how much energy must we supply to move this planet's orbit to infinity? So by definition, the gravitational potential energy is zero at infinity, so we must supply the difference between zero and the planet's current gravitational potential energy. What is not clear to me, is why we must also provide the planet's kinetic energy. The question says the total energy required to move this planet's orbit to infinity equals the sum of kinetic and potential energy. However, can't there still be kinetic energy at infinity? This relies on the idea that the total energy at infinity is zero, but why is this the case?

Answer 1

There can still be kinetic energy at infinity, but if there is, then you've wasted energy in your goal of getting to infinity. The goal is to calculate the amount of energy needed to just barely escape to infinity. So first, convert all of the planet's kinetic energy into potential energy. Then, supply whatever additional energy is needed to bring it to infinity. Since the potential energy $V$ is negative, this means that the total energy needed to escape to infinity is $-V - K = -(V + K)$, where $K$ is the kinetic energy.

Answer 2

If the question concerns the minimum energy to send a gravitating body to infinity, the answer is a direct consequence of energy conservation.

Reading carefully, the question asks about sending the orbit to infinity. If I interpret it correctly, it becomes equivalent to asking why the limit of the orbit to infinity, keeping the same form, should correspond to vanishing kinetic energy. For example, why should the kinetic energy go to zero in a circular orbit whose radius is infinity?

The reason can be traced to the particular relation, holding for closed (limited) orbits, between the average kinetic and potential energy in the gravitational case: $$ \langle K \rangle = -\frac12 \langle V \rangle\tag{1}\label{eq:1} $$ that can be easily derived from the virial theorem (see, for instance, this Wikipedia page, but take into account that, at variance with the text, the theorem can be proved as a theorem of Mechanics without Statistical Mechanics assumptions).

As a consequence of the relation $\eqref{eq:1}$, the kinetic energy must vanish in the infinite limit of similar orbits of increasing size, even in the special case of a circular orbit.

Answer 3

A slightly mathematical answer:

The motion of planets around the sun traces out a conic section (ellipse). At the same time, the eccentricity of any body exhibiting central force motion has the value: $$\epsilon=\sqrt{1+\frac{2EL^2}{\mu C^2}},$$

where

• $E$ = total energy of the body
• $L$ = (constant) angular momentum of the body
• $\mu$ = reduced mass (in the case of planetary motion this is just $M_{\text{planet}}$
• $C$ = central force motion constant (for gravitation, $GM_s$).

For the path to be unbounded, we must have $\epsilon \geq 1$.
For this, the total energy must be $\geq 0$.
The extra energy to be provided is $E_{\large \epsilon \ \geq \ 1}-E_{\text{ellipse}}$.

Since we have to provide the minimum energy, we have to choose $\epsilon$ such that $E_{\large \epsilon \ \geq \ 1}$ is minimum.

This corresponds to $\epsilon=1$, or $E_{\large \epsilon \ \geq \ 1}=0$ (parabolic path).

Hence, the total energy at infinity must be $0$.

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Edit 1: $$E_{\text{ellipse}}=-\frac{GMm}{2a},$$ where

• $M$ = Mass of the Sun
• $2a$ = major axis of ellipse
• $m$ = mass of the body.