# How much energy is required to remove Earth from it's orbit and exit the solar system under perfect conditions?

Ok, this is my first question on this site. But it's one I've been thinking about for a while.

Say through whatever means, we place a device capable of generating thrust/ kinetic energy on the surface of the Earth. The goal being to remove Earth from its orbit and exit the solar system. By "perfect conditions" I mean that hypothetically when this does happen, no other planets or space debris will be in its path. I'm thinking that first it would be easier to stop the rotation of the Earth, and then use our device, but I don't quite know how the Sun's gravitational field will affect all this. Also, there can be two versions of this: one where the device generates continuous thrust and the other where it's more of an impulse.

How much energy would the device need and, if any, what other conditions are needed in order to exit the solar system?

If the system contains only the sun and earth, then we can use conservation of energy to find the answer.

For the earth to escape the gravity of the sun, you would need to give it enough velocity so that by the time its velocity becomes zero (due to facing continuous deceleration from the sun's gravity), it has already covered a large distance from the sun. This then guarantees that the earth won't fall back towards the sun again as the sun's gravity at the location of the earth is now nearly zero.

Now, the kinetic energy of the earth is (when separated by $$r$$ from the sun)

$$K.E = \frac{1}{2}M_Ev^2 \,,$$ where $$v$$ is its orbital speed. For a circular orbit, this means that

$$\frac{M_Ev^2}{r} = \frac{GM_S M_E}{r^2}$$ where $$M_E, M_S$$ are the earth's and sun's mass. This means $$v = \sqrt{GM_S/r}$$

In addition, the potential energy of the earth-sun system is

$$P.E = - \frac{GM_S M_E}{r}.$$

Therefore, the total energy is $$K.E + P.E = - \frac{GM_S M_E}{2r}.$$.

Hence, to remove the earth from this orbit such that it escapes, we would need to feed the equivalent additional $$K.E$$ to the earth. Let us say we accelerate the earth to an outwards (away from the sun) speed of $$V$$; then we need

$$\frac{1}{2}M_E V^2 = \frac{GM_S M_E}{2r}$$

which gives $$V = \sqrt{\frac{GM_S}{2}}$$

Existing answers are missing the momentum conservation side of this.

Since escape velocity is $$\sqrt 2$$ times orbital velocity, the most efficient approach is to increase the Earth's speed by a factor of $$\sqrt 2$$ in one impulse. Due to the Oberth effect, that's more efficient than a continuous impulse. And the smallest velocity change is achieved if you accelerate along the direction that the Earth is already moving.

However, calculating the energy needed is much more complicated than simply calculating the velocity change, because you can't just insert energy to change a body's velocity. Momentum has to be conserved. Thus, the amount of energy that this maneuver will cost you depends on how much of the Earth's mass you are willing to shed. This is analogous to how $$\sim 90\%$$ of the mass of a typical rocket consists of propellant.

If you're willing to shed an arbitrary amount of mass, then you can escape the solar system using arbitrarily little energy. Just imagine launching a particle from the Earth (thereby shedding the whole Earth).

More generally, suppose the Earth has mass $$M$$ and you want it to retain mass $$M^\prime$$. The Earth has orbital velocity $$V$$, and you need it to have velocity $$\sqrt 2 V$$, so the Earth after the impulse moves at velocity $$\Delta V=(\sqrt 2-1)V$$ relative to the Earth before the impulse. The mass $$(M-M^\prime)$$ left behind is ejected with velocity $$v$$ relative to the pre-impulse Earth. By momentum conservation, $$M^\prime \Delta V = (M-M^\prime)v,$$ i.e. $$v=\frac{M^\prime}{M-M^\prime}\Delta V.$$ The energy required (in the frame of the pre-impulse Earth) is then \begin{align} E&=\frac{1}{2} M^\prime \Delta V^2 + \frac{1}{2}(M-M^\prime)v^2 \\ &=\frac{1}{2} \frac{M M^\prime}{M-M^\prime} \Delta V^2. \end{align} Notice that as the mass $$M^\prime$$ that you want to bring approaches 0, the energy required approaches 0. On the other hand, as $$M^\prime$$ approaches $$M$$ (you want to bring the whole Earth), the energy required approaches infinity.

One caveat is that this is a nonrelativistic calculation. In terms of relativity, the "high energy, low mass loss" limit is a photon rocket, which doesn't have to expel its mass directly. But that takes astronomical amounts of energy, which is equivalent to mass, so it has to lose mass in the end anyway.

Also, I've neglected the energy required to gravitationally unbind the propelled part of the Earth from the portion left behind.

From the famous Virial Theorem of statistical mechanics

for a bound state in a central potential, twice the mean kinetic energy equals the mean of r V'(r), yielding for the $$1/r$$ potential

$$2 \left< T \right> = - \left< V \right>$$

and

$$E=T+V=\frac{V}{2}$$ In order to get $$V=0,T=0,r=\infty$$ you have to add the half of the negative potential energy to the kinetic energy. This is independent of the form of the ellipse.

• You should define E, T, V, and be a bit more explicit about what the angle brackets mean. Commented Sep 10, 2023 at 20:59