What is the distribution of expected changes in the period of Dimorphos' orbit around Didymos when the spacevehicle DART crashes against it?


2 Answers 2


Here's a preliminary attempt to answer.

It assumes that the effect of the collision is purely elastic or inelastic giving a change in orbit time of 7.5 to 15 minutes. This is different from the observed 32 minutes. That difference is because this computation does not take into account that DART will form a crater and rock fragments from the crater will result in additional momentum and kinetic energy transferred to Dimorphos. Or potentially there will be comet like ejection of material.

Crash speeds

DART has an orbit similar to earth and we might approximate the speed as being equal to earth's speed relative to the sun: 30 km/s

Dimorphos gas an elliptical orbit and we can use the distance $r$ and semi-major axis $a$ along with the vis-viva equation to compute the speed.

$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)$$

We get approximately 35 km/s.

So the relative speed will be 5 km/s.


These we look up DART 610 kg, Dimorphos 5 x 10^9 kg, Didymos 5 x 10^11 kg


The period of an orbit can be computed as

$$T = 2\pi \sqrt{\frac{a^3}{GM}}$$

where $a$ is the semi-major axis, $M$ the mass of Didymos (the effect of the mass od Dimorphos is negligible), and $G$ is the gravitational constant.

Velocity change

If the collision is inelastic and head on then we use

$$(0.174×5×10^9+5000×610)÷(5×10^9+610) \approx 0.17339$$

afterwards the orbital speed of Dimorphos is 0.17339 km/s instead of 0.174 km/s

Effect of velocity change on period.

We can express the period in terms of the semi-major axis, and the semi-major axis in terms of the velocity. So this allows us to express the period in terms of the velocity

$$a = \left(2/r - \frac{v^2}{GM}\right)^{-1} = \left(2\frac{v_{old}^2}{GM} - {\frac{v^2}{GM}}\right)^{-1} $$

where the second equality is based on assuming $a = r$ before impact.

Filling this into the equation for the period

$$T = 2\pi \mu^{-0.5}\left(2\frac{v_{old}^2}{GM} - {\frac{v^2}{GM}}\right)^{-1.5}$$

and the ratio will be

$$\frac{T_{new}}{T_{old}} = \left( 2- \frac{v^2}{v_{old}^2} \right)^{-1.5}$$

Which gives $T_{new} \approx 0.9896 \, T_{old}$ and the orbit period will be 447 seconds shorter for an inelastic collision (dimorphos moves slower but the orbit will be smaller making the orbit take less long).

For an elastic collision the change in speed will be approximately double and the change in the orbit time can be 894 seconds or approximately 15 minutes.

orbits of Dimorphos and dart

close up view

  • $\begingroup$ Whereas the overall approach looks fine something went wrong when comparing theory with reality. NASA has confiormed the the orbital period has been shortened by the impact. $\endgroup$
    – TomS
    Oct 13, 2022 at 15:52
  • $\begingroup$ ... The interesting discrepancy is the 32 minutes which is more than the 15 minutes computed here. There might be some simplifications, but a factor two difference seems a lot (and that is the optimal situation when the impact is elastic). Possibly there is some 'other' effect that gives Dimorphos an extra push? Like the jet? Or maybe material is lost and we observe a much smaller Dimorphos with lower weight? $\endgroup$ Oct 13, 2022 at 16:04
  • $\begingroup$ What do you mean by improve? The correct calculation for the head-on collision will always yield a shorter orbital period $ T < T_0$ after the impact compared to the orbital period $ T_0$ before the impact. $\endgroup$
    – TomS
    Oct 13, 2022 at 20:25
  • 2
    $\begingroup$ This computation doesn't account for the momentum of the mass ejected by the impact. $\endgroup$
    – John Doty
    Oct 13, 2022 at 23:18
  • 1
    $\begingroup$ @DanielSilverwood Whether the orbital speed of Dimorphos increases or decreases depends on which side of Dimorphos is being hit during it's rotation around Didymos. Dimorphos can be given a be given a push towards a higher orbit, but just as well a push to a lower orbit. $\endgroup$ Oct 16, 2022 at 19:01

Here's my approach.

We set $ \mu = G(M+m) $ and use Keplers 3rd law

$$ T^2 = 4\pi^2 \frac{a^3}{\mu} $$

and the vis-a-vis equation

$$ \frac{v^2}{\mu} = \frac{2}{r} - \frac{1}{a} $$

With $r = a_0$ where ${}_0$ means "before impact" we find

$$ \frac{v_0^2}{\mu_0} = \frac{1}{a_0} $$

$$ \frac{v^2}{\mu} = \frac{2v_0^2}{\mu_0} - \frac{1}{a} $$

and therefore

$$ \frac{1}{a} = \frac{2v_0^2}{\mu_0} - \frac{v^2}{\mu} $$

Using Kepler's law to calculate

$$ \frac{T_0^2}{T^2} = \frac{\mu}{\mu_0} \frac{a_0^3}{a^3} = \frac{\mu}{\mu_0} \left(2 - \frac{\mu_0}{\mu} \frac{v^2}{v_0^2} \right)^3 $$

I find agreement with your result (but one could also take changes in mass into account *).

For collisions with small changes

$$ v_0 \to v + \Delta v; \; |\Delta v| \ll v $$

a first order Taylor expansion yields shorter (longer) orbital periods for decreased (increased) speed.

*) The approach does not address the calculation of the amount of ejected mass $\Delta m$ and its recoil i.e. the effect on $\Delta v$. However, it does address the secondary effect on the orbit once $\Delta m$ and $\Delta v$ are known. One can determine $\Delta T$ based on observations - NASA did that - and from that one could derive a range of allowed $\Delta m, \Delta v$. To nail it down exactly further models for the impact itself are necessary.

  • $\begingroup$ For the ratio squared it's $6v/v_0$; taking the square root you get the $3v/v_0$. $\endgroup$
    – TomS
    Oct 14, 2022 at 7:30
  • $\begingroup$ I haven't found any data on the mass that has been ejected. Wikipedia says that m/M ~ 1/100, so the ejected mass delta m / M << 1/100, therefore << delta v / v_0. That means that the ejected mass may have a huge impact in terms of recoil and delta v, but not in terms of the secondary effect on the orbit. $\endgroup$
    – TomS
    Oct 14, 2022 at 7:38
  • $\begingroup$ The mass change does not need to be included into the equation derived here. The mass plays a role for the final velocity change, but the equation here is just expressing the orbital time in terms of the velocity. For this relationship the mass of Dimorphos, $m$, in Kepler's law is negligible $$T^2 = 4\pi^2 \frac{a^3}{G(M+m)} \approx 4\pi^2 \frac{a^3}{GM}$$ $\endgroup$ Oct 14, 2022 at 7:56
  • $\begingroup$ That's what my last comment says for Dimorphos, but this is not true in general. $\endgroup$
    – TomS
    Oct 14, 2022 at 8:07

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