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CNN Space and Defense Correspondent Kristin Fisher does a really good job of summarizing the current state of the Double Asteroid Redirection Test early observational results for general public consumption. See video and screenshot below.

The diagram shows that the impact was retrograde - in the direction opposite that of Dimorphos' orbital motion around 65803 Didymos.

At first I thought "Of course, maximize relative velocity." Hitting it when it was on the other side of its orbit would pack a smaller punch since it would be moving away, but then I found that its orbital velocity was only about 1.2 meters per second. Compared to the impact velocity of about 6,600 meters per second this is very small and so this would not likely be an overwhelming factor.

While the public is informed of the challenges to making the impact happen, the observation challenges are great as well! One aspect is to establish the new period and thus the momentum transferred, but another aspect is to characterize the impact and release of debris, as this recoil mass and its speed is an important part of understanding the kinematics and thus effectiveness for this technique for (potentially) future objects.

So I'd like to ask:

Question: Why did they decide to hit Dimorphos in the retrograde direction rather than prograde; was it a "coin-toss" or were there implications for observing?

Screenshot from the CNN October 13, 2022 NASA mission successfully changed the motion of an asteroid

screenshot from the CNN October 13, 2022 "NASA mission successfully changed the motion of an asteroid" https://youtu.be/cVWNsbOJ9LY

uhoh
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  • Why would it want to hit the asteroid prograde? Usually NASA performs these cuz its getting either too close to earth and posing a potential threat, so they would want to hit it in the opposite direction of where it's going, in this case retrograde. If they hit it prograde, they are increasing its speed, not much but aint doing anything to redirect its course – DialFrost Oct 12 '22 at 23:18
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    @DialFrost Both hitting it prograde and retrograde would be viable options for changing an asteroid's trajectory. It all comes down to the specifics of the geometry of the potential impact, and the orbits of the Earth and the asteroid, and how much lead time is available that determines whether slowing it down or speeding it up results in a miss with less energy expended by the redirector. – notovny Oct 12 '22 at 23:32
  • @DialFrost please re-read my question, especially the part "...its orbital velocity was only about 1.2 meters per second." This is obviously not the speed that the pair move in their orbit around the Sun, it's the speed that Dimorphos moves in its orbit around its companion, (and also the speed an average person moves in a shopping mall). Also changing an objects speed of course changes its orbit and its course! Solar system objects all orbit the Sun, whose gravity effects dominate all motion; any change in speed means the Sun's gravity will lead it on a new and different course. – uhoh Oct 13 '22 at 01:03
  • @DialFrost On average, a part-per-thousand change in speed either way (faster or slower) will deflect a threatening asteroid from hitting the Earth by roughly the same distance. – uhoh Oct 13 '22 at 01:06
  • I have a partial answer so I'm not posting this. NASA hasn't really provided much on this, but they wanted it to crash directly (retrogradely) into the asteroid to see what it would do. They are not finding the most efficient way to deflect it. It is just a test run. – DialFrost Oct 13 '22 at 01:25
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    A fact is waiting time to get the new orbit data is shorter than if orbit had been enlarged –  Oct 13 '22 at 13:45
  • @qqjkztd can that "fact" really be supported mathematically? If so, then it will certainly be the basis of a great supported answer, but right now I'm not convinced and probably won't be by prose alone. – uhoh Oct 13 '22 at 13:53
  • @uhoh it can be supported by Kepler's laws for instance. Still this point or fact (having less to wait to get a full orbital picture) is only one hypothetical reason among many other I have no clue about, to hit that thing retrograde instead of prograde, and therefore would not make a full answer either. –  Oct 13 '22 at 19:53
  • @qqjkztd "it can be supported by Kepler's laws for instance" How? In what way would the "waiting time to get the new orbit data" be affected in a meaningful and significant way? – uhoh Oct 13 '22 at 23:36
  • @uhoh not saying those few seconds are signigicant or meaningful, but if such a cyclic motion has to be fully described, the shorter the cycle the less you wait to compare it with previous cycle(s). FWIU the orbit instantly changed at impact time, yet orbital perturbations due to ejecta just begin and will last for many orbits. –  Oct 14 '22 at 05:20
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    @qqjkztd the early data looks like this https://i.stack.imgur.com/MrFG3.jpg (from here) I think your argument is that if over a month you successfully observe 31 eclipses instead of 29, that for a given S/N you get a better constraint on the momentum transfer. I can agree that those two extra eclipses give you a slightly better constraint on the period, but one has to work carefully through propagation of errors to show it constrains the momentum transfer better as well. Some math needed... – uhoh Oct 14 '22 at 06:08
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    Great out-of-context quote from CNN: https://youtu.be/cVWNsbOJ9LY?t=150 - "We've learned that humanity does not need Bruce Willis" ... slight pause as she realizes only saying one name would make it personal ... "and Ben Affleck". Way to dunk on their later careers, although slightly less funny because of Bruce Willis' aphasia contributing to some of his critically panned performances. :P But Armageddon's silliness is all on Michael Bay. – Peter Cordes Oct 16 '22 at 09:37
  • @PeterCordes yes I took note of that when I first listened to it :-) – uhoh Oct 16 '22 at 16:42
  • Which direction d'you think might produce the greatest result? You seem to have ruled out anything lateral, so what would work best for you? – Robbie Goodwin Oct 18 '22 at 20:01
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    @DialFrost DART did not target an asteroid that was dangerous to earth at the moment. DART was only a test to prove that we could move an asteroid. – Deko Revinio May 06 '23 at 14:58

4 Answers4

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The original orbital period of Dimorphos was approximately 11.92 hours. The DART scientists weren't sure how much momentum was going to be exchanged due to the collision since they weren't sure how much would be lost to ejecta.

There was a fear that if they hit Dimorphos from a prograde direction, that the orbital period would be increased to 12 hours. A 12 hour orbital period would not be ideal for studying the results since it would coincide with the Earth's diurnal period. In addition, the lighting during the collision was considered favorable for observation from Earth for a retrograde strike.

From McQuaide et al. [2021],

Dimorphos orbits Didymos with a period of 11.92 hours. The planned retrograde impact will reduce this orbit period. The retrograde impact was chosen over a prograde impact partly because the resulting increase in the orbit period for a prograde impact would cause Dimorphos’s orbit around Didymos to be nearly exactly 12 hours, synchronized with Earth’s diurnal period. Earth-based observations would see Dimorphos in the same configuration each observing opportunity. A retrograde impact and resultant orbit period reduction allows for more unique observations from night-tonight. Finally, the retrograde impact also provides for better lighting conditions, as it occurs on the sunlit side of Didymos. Secondary reflection from Didymos may allow imaging of the night side of Dimorphos.

Here are some geometry figures from the same paper:

enter image description here

As the OP notes, the difference in resultant orbital velocities between a prograde and retrograde strike would have been quite small, and so were not a factor in the decision of collision direction.

Connor Garcia
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    Intuitively it is always so confusing how hitting in prograde direction (increasing velocity) causes a longer orbital period by lifting to a higher orbit. – jpa Oct 14 '22 at 08:25
  • No momentum is "lost" to ejecta. That momentum is favourable to change the trajectory of the asteroid. – Rainald62 Oct 14 '22 at 17:04
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    Whether the momentum transferred to ejecta is favourable depends on which direction the ejecta go. – Pere Oct 14 '22 at 19:58
  • @Pere: Wouldn't the ejecta tend to be blown "upward" which increases the momentum transfer assuming the spacecraft hits the asteroid head on? – Kevin Kostlan Oct 16 '22 at 00:01
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    @KevinKostlan You and Rainald62 are correct about an ideal case, since the system momentum is conserved, but the "head-on collision" assumption is bad. The velocity vectors of the impactor and Dimorphos weren't collinear and the impactor didn't strike through Dimorphos's center of gravity. Also, Dimorphos is a loose rubble pile instead of a spherical planet. The mission planners ran monte carlo runs, but had a very loose idea of momentum transfer, which is one of the things they wanted to learn from the experiment. – Connor Garcia Oct 16 '22 at 00:29
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    @ConnorGarcia: If only we could split the spacecraft up and pepper the asteroid with a lot of smaller impacts and measure the effect of each impact (onboard radar?). To get some idea of a mean and standard deviation. – Kevin Kostlan Oct 16 '22 at 02:02
  • Wonderful answer, thank you! Here's a followup question: What does "Secondary reflection from Didymos may allow imaging of the night side of Dimorphos" mean? Why would this be useful? – uhoh Oct 17 '22 at 00:07
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Here's a thought that is too long to be shared in a comment.

In this question and answer, DART crash on Dimorphos: computation of orbital period change I tried to compute the change of the orbital period and it looks like this:

$$T_{new} = T_{old}(2-(v_{new}/v_{old})^2)^{-1.5}$$

It expresses the new period $T_{new}$ relative to the old period $T_{old}$ as a function of Dimorphos velocity before impact, $v_{old}$, and after impact, $v_{new}$.

For small changes in the velocity the effect on the change is nearly linear and reducing or increasing the impuls of Didymos is nearly the same.

But, if you would like to consider the non-linear behaviour, is it better to reduce or increase the speed? Here is a plot of the relationship between the speed and the velocity. (The red dot indicates the pre-hit velocity and orbital period)

enter image description here

What we can see in the image is that if we would consider non-linear effects, then it would have to be increasing the rotational speed of Dimorphos and increasing the rotational period which would have relatively a slightly stronger change.

As said before, this answer is more like a comment, it does not answer the question completely. But, at least it shows how non-linear effects in changes of the period were not a consideration, since in that case one should have opted to hit Dimorphos in prograde direction and increase it's speed, and increase the period rather than shortening it.

Nilay Ghosh
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Sextus Empiricus
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  • This shows that when modifying Dimorphos' velocity by some fixed amount, increasing the velocity will result in a bigger period change than decreasing it. But I feel like something's missing here, since the prograde and retrograde collisions don't change the velocity by the same fixed amount - the retrograde collision changes Dimorphos' velocity slightly more than the prograde collision due to greater relative velocity at impact. Maybe the effect is negligible due to the Dimorphos' low orbital speed, but the implicit assumption here of a fixed velocity change isn't quite correct. – Nuclear Hoagie Oct 14 '22 at 13:53
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    @NuclearHoagie the velocity if impact is of the order of 6 km/s whereas the orbital velocity was around 0.174 m/s. – Sextus Empiricus Oct 14 '22 at 14:13
  • @NuclearHoagie "This shows that when modifying Dimorphos' velocity by some fixed amount, increasing the velocity will result in a bigger period change than decreasing it" my main point was actually that it practically doesn't matter and the difference between prograde and retrograde is the same. But if you would really like to include higher order effects then increasing the velocity would have a stronger effect. – Sextus Empiricus Oct 14 '22 at 14:17
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It seems to me that it's simply due to the fact that hitting the asteroid in the retrograde direction would result in a larger transfer of momentum between the probe and the asteroid than if the asteroid was hit in the prograde direction. This would therefore have a larger effect on the motion of the asteroid.

Or am I completely off base here?

HippoMan
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    No, you're not "completely off base here", but note that the question says "At first I thought 'Of course, maximize relative velocity.' Hitting it when it was on the other side of its orbit would pack a smaller punch since it would be moving away, but then I found that its orbital velocity was only about 1.2 meters per second. Compared to the impact velocity of about 6,600 meters per second this is very small and so this would not likely be an overwhelming factor." – uhoh Oct 16 '22 at 03:10
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    Yes. The mass of Dimorphos is estimated at 5 x 10^9 kg. The probe's mass is 6.1 x 10^2 kg. The asteroid travels at 0.71 m/sec (not 1.2 m/sec), so its momentum is approximately 3.55 x 10^9 kg-m/sec. The probe travels at 6.6 x 10^3 m/sec, so its momentum is around 4 * 10^6 kg-m/sec, which is only around 0.11 percent of that of the asteroid. So yes, there's hardly a difference between retrograde and prograde. But maybe the small difference in momentum transfer between a prograde and retrograde collision could still be detectable and maybe slightly more desirable ... ??? – HippoMan Oct 16 '22 at 04:28
  • Ya I got the speed wrong; 1.2 is the semimajor axis in km but if I multiply that by 2 × 1000 and divide by 11.92 × 3600 I get only 0.17 m/s (not 0.71) – uhoh Oct 16 '22 at 04:39
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This is a speculation, not a sourced answer ... but it's a thought that seems obvious yet beneath publication.

From the layman's perspective, we could worry that a chunk of the asteroid would be knocked into a collision course with Earth. If we imagine DART bounced off Dimorphos or knocked a tiny chunk off the asteroid with a glancing blow, that may not have been a meaningful threat in a scientific sense - nonetheless, it would have been a black eye from the point of view of ongoing news articles about plotting the course of the wayward chunk of asteroid blasted out of its orbit by a NASA probe.

So it makes sense that they chose the collision course more likely to drop any unexpected debris into a lower orbit around Didymos, and avoid wagging tongues.

uhoh
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Mike Serfas
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    @uhoh - a lower orbit around the other asteroid, not the Sun or Earth. – Mike Serfas Oct 15 '22 at 23:40
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    From my question: "...but then I found that its orbital velocity was only about 1.2 meters per second. Compared to the impact velocity of about 6,600 meters per second this is very small and so this would not likely be an overwhelming factor." Any debris recoiling with a velocity more than a few meters per second will leave the asteroid system and enter an independent orbit around the Sun. Escape velocity from Dimporphos will likely be of order 10 centimeters per second, so completely inconcequential. – uhoh Oct 16 '22 at 00:03
  • @uhoh I understood that argument to be valid in terms of relative velocity (kinetic energy) at collision, but here I'm thinking of something a bit closer to the average momentum. For example, a chunk of the asteroid 5500 times the mass of the probe would plummet straight downward after a retrograde collision, but would double its orbital velocity from a collision in the other direction. (I haven't looked up whether that would escape from Didymos) – Mike Serfas Oct 16 '22 at 00:22
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    I think this answer could be improved by a citation. – Connor Garcia Oct 16 '22 at 00:35