One of the standard theories behind the formation of our Moon is the giant impact hypothesis, according to which Earth was struck by a Mars-sized body (about $10\%$ the mass of Earth) early in its history. The violent collision threw a large amount of material into orbit, and this coalesced to become the Moon.

Now, we previously addressed the question of whether this would destabilize Earth's orbit around the Sun. The answer of course is that, no, orbits can't destabilize in Newtonian physics.

But one might wonder how eccentric the Earth+Moon orbit around the Sun would be. After all, having a planet slam into you at several kilometers per second seems like it may have consequences.

For simplicity, let's assume the early Earth is in a perfectly circular orbit at $1\ \mathrm{AU}$. It is then hit by an object with the mass of Mars. What would be the new eccentricity (and, why not, semi-major axis) after the collision?


  • There are two important variables left unspecified - the impact velocity and the angle of impact. For the former, Wikipedia quotes $4\ \mathrm{km}/\mathrm{s}$, but an answer that shows how the result scales with this value would be best, since there is a great deal of uncertainty in it anyway. The same holds for the impact angle being $45^\circ$. That said, it seems reasonable to restrict ourselves to a single plane.

  • Intuitively I feel a head-on collision will be a local maximum in post-collision eccentricity, since it will make the location of impact at $1\ \mathrm{AU}$ the aphelion. Similarly, a direct-from-behind collision will make that location the new perihelion and thus also will induce a greater eccentricity change than nearby angles. Is this intuition borne out by the math?

  • Bonus points for justifying (or providing references that justify) the impact speed. One could ask how much energy is needed to separate all that mass, and how much of the pre-collision kinetic energy goes into this versus melting the surface and heating the mantle. This is mostly to put an upper bound on how much the impact could have perturbed Earth's orbit - it should be easy enough to argue that the collision wasn't any faster than, say, $30\ \mathrm{km}/\mathrm{s}$.

  • When it comes to angular momentum (and energy), the situation is somewhat complicated by the fact that our orbiting object is not a classical point mass. The Earth can rotate, and the Earth+Moon system will clearly also have angular momentum. Can these extra degrees of freedom relieve the burden of the impact, leaving Earth in a still rather circular orbit?

  • For the record, Earth's eccentricity around the Sun is only $0.0167$ today. Basically, the question is whether a Moon-forming impact could impart no more than about this much eccentricity. Of course, this value can change over time via long-range interactions with other planets - let's ignore that complication for now.

References, graphs, equations, and good old-fashioned order-of-magnitude physics all welcome.

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    $\begingroup$ Orbits tend to become circular if there is significant tidal dissipation, and the tidal forces from the Sun on the Earth are significant. It's possible that the originally elliptical orbit relaxed to the current low eccentricity. NB you say "this value can change over time via long-range interactions with other planets" but it is the tidal forces from the Sun that matters, not the other planets, and these would act quite quickly on planetary timescales. $\endgroup$ Commented Jun 17, 2013 at 6:09
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    $\begingroup$ Chris, can I just point out that any post that could fully answer this and provide everything you ask for could also probably be published in a scholarly journal. I absolutely love this question, but I think it's too ambitious. If you want a nice answer, try scaling it back; ask for fewer inclusions, let people use the wikipedia values, etc. IMO, I'm not going to answer a question like this and not expect a PhD out of it. $\endgroup$
    – Jim
    Commented Jun 17, 2013 at 13:15
  • $\begingroup$ @JohnRennie Well, I was looking at the plot in this part of a wiki article at the time. If you believe it (which I'm not saying I do), it seems planet-planet effects can be significant. I suppose the magnitude of the Earth-Sun tidal dissipation could be another whole question, but I guess I was assuming without proof that it fell off too steeply with $r$ to matter. $\endgroup$
    – user10851
    Commented Jun 17, 2013 at 19:53
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    $\begingroup$ @ChrisWhite: the existance of spring and neap tides suggests otherwise. $\endgroup$ Commented Jun 18, 2013 at 5:39
  • $\begingroup$ @ChrisWhite the impact velocity of 4 km/s in the wiki article refers to the velocity of the asteroid outside the Earth's gravitational field. As the object falls into our potential well, its velocity will increase with ~11 km/s, i.e. the Earth's escape velocity. $\endgroup$
    – Pulsar
    Commented Jun 18, 2013 at 15:56

1 Answer 1


The Kepler orbit of the Earth around the Sun is determined by two constants: the specific orbital energy $E$ and the specific relative angular momentum $h$: $$ \begin{align} E &= \frac{1}{2}v_{r,\oplus}^2 + \frac{1}{2}v_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\ h^2 &= r^2\,v^2_{T,\oplus} = \mu a(1-e^2), \end{align} $$ where $\mu = G(M_\odot + M_\oplus)$, $r$ is the distance Earth-Sun (at the moment of impact), $a$ is the semi-major axis, $e$ is the orbital eccentricity, $v_{r,\oplus}$ is the radial orbital velocity of the Earth, and $v_{T,\oplus}$ the tangential velocity. Now, suppose that a large asteroid collides with the Earth, with orbital velocity $(v_{T,A},v_{r,A})$ and mass $M_A$. Its relative velocity is then $$ \begin{align} v_{T,A}' &= v_{T,A} - v_{T,\oplus},\\ v_{r,A}' &= v_{r,A} - v_{r,\oplus}. \end{align} $$ We can express these relative velocities in terms of the total impact velocity $v_\text{i}$ and the impact angle $\theta$: $$ \begin{align} v_{T,A}' &= v_\text{i}\cos\theta,\\ v_{r,A}' &= -v_\text{i}\sin\theta, \end{align} $$ where I defined $\theta$ as in Fig. 1 of this article. So we obtain $$ \begin{align} v_{T,A} &= v_{T,\oplus} + v_\text{i}\cos\theta,\\ v_{r,A} &= v_{r,\oplus} - v_\text{i}\sin\theta. \end{align} $$ If we assume that the collision is central, that heat loss is negligible and that the debris remains gravitationally bound to the Earth, then conservation of momentum implies $$ \begin{align} M_\oplus\,v_{T,\oplus} + M_A\,v_{T,A} &= (M_\oplus+M_A)u_{T,\oplus}\\ M_\oplus\,v_{r,\oplus} + M_A\,v_{r,A} &= (M_\oplus+M_A)u_{r,\oplus}, \end{align} $$ with $(u_{T,\oplus},u_{r,\oplus})$ the new orbital velocity of the Earth (and the gravitationally bound debris) after the impact. We get $$ \begin{align} u_{T,\oplus} &= v_{T,\oplus} + \frac{M_A}{M_\oplus+M_A}v_\text{i}\cos\theta,\\ u_{r,\oplus} &= v_{r,\oplus} - \frac{M_A}{M_\oplus+M_A}v_\text{i}\sin\theta. \end{align} $$ So the orbital energy and angular momentum will have changed into $$ \begin{align} E' &= \frac{1}{2}u_{r,\oplus}^2 + \frac{1}{2}u_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a'},\\ h'^2 &= r^2\,u^2_{T,\oplus} = \mu a'(1-e'^2). \end{align} $$ (the change in $\mu$ is negligible). Right, let's plug in some numbers. Suppose we start with a circular orbit, with a radius equal to the present-day semi-major axis: $$ \begin{align} \mu &= 1.32712838\times 10^{11}\;\text{km}^3\,\text{s}^{-2},\\ r &= a = 1.49598261\times 10^{8}\;\text{km},\\ e &= 0. \end{align} $$ For a circular orbit, it follows that $$ \begin{align} v_{T,\oplus} &= \sqrt{\frac{\mu}{r}}= 29.785\;\text{km}\,\text{s}^{-1},\\ v_{r,\oplus} &=0\;\text{km}\,\text{s}^{-1}. \end{align} $$ The impact velocity of an asteroid will always be at least equal to the Earth's escape velocity $11.2\,\text{km/s}$, which is the speed it takes up as it falls into the Earth's gravitational potential well. The article that I already linked to states that typical asteroid impact velocities are in the range of $12-20\,\text{km/s}$. In theory, the impact velocity can be as large as $72\,\text{km/s}$ in the case of a head-on collision, when the Earth and the asteroid have opposite orbital velocities, thus a relative velocity of ~$60\,\text{km/s}$, augmented with the escape velocity as the asteroid falls into our gravitational potential well. This is very unlikely for asteroids, but it is possible for comets.

So, let us assume a typical impact velocity $v_\text{i}=16\,\text{km/s}$, a mass $M_A = 0.1M_\oplus$ and an impact angle $\theta=45^\circ$. We find $$ \begin{align} u_{T,\oplus} &= 30.813\;\text{km}\,\text{s}^{-1},\\ u_{r,\oplus} &= -1.0285\;\text{km}\,\text{s}^{-1},\\ E' &= -411.87\;\text{km}^2\,\text{s}^{-2},\\ h'^2 &= 2.1248\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\ a' = -\frac{\mu}{2E'} &= 1.61109\times 10^8\;\text{km},\\ e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0788,\\ r_\text{p} = a'(1-e') &= 1.48411\times 10^8\;\text{km},\\ r_\text{a} = a'(1+e') &= 1.73807\times 10^8\;\text{km}, \end{align} $$ with $r_\text{p}$ and $r_\text{a}$ perihelion and aphelion. Evidently, the influence on the Earth's orbit is substantial.

In the case of a direct-from-behind collision, we get $\theta=0^\circ$, $v_\text{i}=11.2\,\text{km/s}$, so that $$ \begin{align} u_{T,\oplus} &= 30.803\;\text{km}\,\text{s}^{-1},\\ u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\ E' &= -412.72\;\text{km}^2\,\text{s}^{-2},\\ h'^2 &= 2.1234\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\ a' = -\frac{\mu}{2E'} &= 1.60778\times 10^8\;\text{km},\\ e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0695,\\ r_\text{p} = a'(1-e') &= 1.49598\times 10^8\;\text{km},\\ r_\text{a} = a'(1+e') &= 1.71958\times 10^8\;\text{km}. \end{align} $$ As expected, the radius at impact has become the perihelion, and the change in eccentricity is lowest.

And just for fun, let's try the worst-case scenario: $\theta=180^\circ$, $v_\text{i}=72\,\text{km/s}$: $$ \begin{align} u_{T,\oplus} &= 23.239\;\text{km}\,\text{s}^{-1},\\ u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\ E' &= -617.10\;\text{km}^2\,\text{s}^{-2},\\ h'^2 &= 1.2086\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\ a' = -\frac{\mu}{2E'} &= 1.07530\times 10^8\;\text{km},\\ e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.391,\\ r_\text{p} = a'(1-e') &= 0.65462\times 10^8\;\text{km},\\ r_\text{a} = a'(1+e') &= 1.49598\times 10^8\;\text{km}, \end{align} $$ so that the radius at impact has become the aphelion, and the change in eccentricity is highest. Although I wonder how much would be left of the Earth after such an apocalyptic event...

If the collision isn't central, then part of the energy will be transferred to the axial rotation of the Earth, which should reduce the effect on the orbit. But that will be more difficult to quantify.


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