<p>The original orbital period of <a href="https://en.wikipedia.org/wiki/Dimorphos" rel="noreferrer">Dimorphos</a> 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.</p>
<p>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.</p>
<p>From <a href="https://ui.adsabs.harvard.edu/abs/2021plde.confE.251M/abstract" rel="noreferrer">McQuaide et al. [2021]</a>,</p>
<blockquote>
<p>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.</p>
</blockquote>
<p>Here are some geometry figures from the same paper:</p>
<p><a href="https://i.sstatic.net/Z8vi3.jpg" rel="noreferrer"><img src="https://i.sstatic.net/Z8vi3.jpg" alt="enter image description here" /></a></p>
<p>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.</p>

CC BY-SA 4.0
2022-10-14T08:25:56.507Z
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
CC BY-SA 4.0
2022-10-14T17:04:24.497Z
No momentum is "lost" to ejecta. That momentum is favourable to change the trajectory of the asteroid.
Rainald62
CC BY-SA 4.0
2022-10-14T19:58:27.690Z
Whether the momentum transferred to ejecta is favourable depends on which direction the ejecta go.
Pere
CC BY-SA 4.0
2022-10-16T00:01:36.550Z
@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
CC BY-SA 4.0
2022-10-16T00:29:55.063Z
@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
CC BY-SA 4.0
2022-10-16T02:02:36.180Z
@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
CC BY-SA 4.0
2022-10-17T00:07:32.457Z
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?](https://astronomy.stackexchange.com/q/50776/7982)
uhoh
CC BY-SA 4.0
2022-10-13T20:02:10.287Z
2022-10-15T02:38:11.650Z
2022-10-15T02:38:11.650Z
Connor Garcia
<p>Here's a thought that is too long to be shared in a comment.</p>
<p>In this question and answer, <a href="https://physics.stackexchange.com/questions/731677/dart-crash-on-dimorphos-computation-of-orbital-period-change">DART crash on Dimorphos: computation of orbital period change</a> I tried to compute the change of the orbital period and it looks like this:</p>
<p><span class="math-container">$$T_{new} = T_{old}(2-(v_{new}/v_{old})^2)^{-1.5}$$</span></p>
<p>It expresses the new period <span class="math-container">$T_{new}$</span> relative to the old period <span class="math-container">$T_{old}$</span> as a function of Dimorphos velocity before impact, <span class="math-container">$v_{old}$</span>, and after impact, <span class="math-container">$v_{new}$</span>.</p>
<p>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.</p>
<p>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)</p>
<p><a href="https://i.sstatic.net/Ydjlq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ydjlq.png" alt="enter image description here" /></a></p>
<p>What we can see in the image is that if we would consider non-linear effects, then it would have to be <em>increasing</em> the rotational speed of Dimorphos and <em>increasing</em> the rotational period which would have relatively a slightly stronger change.</p>
<hr />
<p>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 <em>not</em> 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.</p>

CC BY-SA 4.0
2022-10-14T13:53:15.737Z
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
CC BY-SA 4.0
2022-10-14T14:13:23.470Z
@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
CC BY-SA 4.0
2022-10-14T14:17:56.290Z
@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
CC BY-SA 4.0
2022-10-13T16:57:52.640Z
2023-05-06T12:57:54.700Z
2023-05-06T12:57:54.700Z
Sextus Empiricus
<p>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.</p>
<p>Or am I completely off base here?</p>

CC BY-SA 4.0
2022-10-16T03:10:38.913Z
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
CC BY-SA 4.0
2022-10-16T04:28:45.073Z
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
CC BY-SA 4.0
2022-10-16T04:39:33.733Z
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
CC BY-SA 4.0
2022-10-16T02:28:26.813Z
2022-10-16T02:28:26.813Z
HippoMan
<p>This is a speculation, not a sourced answer ... but it's a thought that seems obvious yet beneath publication.</p>
<p>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.</p>
<p>So it makes sense that they chose the collision course more likely to drop any unexpected debris into a <em>lower</em> orbit around Didymos, and avoid wagging tongues.</p>

CC BY-SA 4.0
2022-10-15T23:40:42.437Z
@uhoh - a lower orbit around the other asteroid, not the Sun or Earth.
Mike Serfas
CC BY-SA 4.0
2022-10-16T00:03:50.047Z
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
CC BY-SA 4.0
2022-10-16T00:22:32.030Z
@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
CC BY-SA 4.0
2022-10-16T00:35:37.460Z
I think this answer could be improved by a citation.
Connor Garcia
CC BY-SA 4.0
2022-10-15T23:11:06.217Z
2022-10-16T00:00:08.903Z
2022-10-16T00:00:08.903Z
Mike Serfas
<p>CNN Space and Defense Correspondent <a href="https://edition.cnn.com/profiles/kristin-fisher" rel="noreferrer">Kristin Fisher</a> does a really good job of summarizing the current state of the <a href="https://en.wikipedia.org/wiki/Double_Asteroid_Redirection_Test" rel="noreferrer">Double Asteroid Redirection Test</a> early observational results for general public consumption. See video and screenshot below.</p>
<p>The diagram shows that the impact was retrograde - in the direction opposite that of <a href="https://en.wikipedia.org/wiki/Dimorphos" rel="noreferrer">Dimorphos</a>' orbital motion around <a href="https://en.wikipedia.org/wiki/65803_Didymos" rel="noreferrer">65803 Didymos</a>.</p>
<p>At first I thought &quot;Of course, maximize relative velocity.&quot; 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.</p>
<p>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.</p>
<p>So I'd like to ask:</p>
<p><strong>Question:</strong> Why did they decide to hit Dimorphos in the retrograde direction rather than prograde; was it a &quot;coin-toss&quot; or were there implications for observing?</p>
<hr />
<p>Screenshot from the CNN October 13, 2022 <a href="https://youtu.be/cVWNsbOJ9LY" rel="noreferrer">NASA mission successfully changed the motion of an asteroid</a></p>
<p><a href="https://i.sstatic.net/uz6xa.jpg" rel="noreferrer"><img src="https://i.sstatic.net/uz6xa.jpg" alt="screenshot from the CNN October 13, 2022 &quot;NASA mission successfully changed the motion of an asteroid&quot; https://youtu.be/cVWNsbOJ9LY" /></a></p>

CC BY-SA 4.0
2022-10-12T23:18:08.927Z
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
CC BY-SA 4.0
2022-10-12T23:32:49.093Z
@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
CC BY-SA 4.0
2022-10-13T01:03:21.050Z
@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
CC BY-SA 4.0
2022-10-13T01:06:16.660Z
@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
CC BY-SA 4.0
2022-10-13T01:25:53.513Z
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
CC BY-SA 4.0
2022-10-13T13:53:01.560Z
@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
CC BY-SA 4.0
2022-10-13T23:36:05.303Z
@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
CC BY-SA 4.0
2022-10-14T06:08:37.737Z
@qqjkztd the early data looks like this https://i.sstatic.net/MrFG3.jpg (from [here](https://arstechnica.com/science/2022/10/dart-mission-successfully-shifted-its-targets-orbit/)) 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
CC BY-SA 4.0
2022-10-16T09:37:34.563Z
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](https://en.wikipedia.org/wiki/Bruce_Willis#2020s:_Critical_decline,_health_problems_and_retirement) contributing to some of his critically panned performances. :P  But Armageddon's silliness is all on Michael Bay.
Peter Cordes
CC BY-SA 4.0
2022-10-16T16:42:27.350Z
@PeterCordes yes I took note of that when I first listened to it :-)
uhoh
CC BY-SA 4.0
2022-10-18T20:01:32.207Z
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
CC BY-SA 4.0
2023-05-06T14:58:55.927Z
@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
<observational-astronomy><asteroids><collision>
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?
uhoh
# 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?

(Asked by: username_0 on 2022-10-12)

CNN Space and Defense Correspondent [Kristin Fisher](https://edition.cnn.com/profiles/kristin-fisher) does a really good job of summarizing the current state of the [Double Asteroid Redirection Test](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Dimorphos)' orbital motion around [65803 Didymos](https://en.wikipedia.org/wiki/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](https://youtu.be/cVWNsbOJ9LY)

[![screenshot from the CNN October 13, 2022 ](https://i.sstatic.net/uz6xa.jpg)](https://i.sstatic.net/uz6xa.jpg)

**Tags:** observational-astronomy, asteroids, collision

> **Comments:**
> **username_1**: 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  
> **username_2**: @username_1 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.  
> **username_0**: @username_1 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.  
> **username_0**: @username_1 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.  
> **username_1**: 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.  
> **username_0**: @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.  
> **username_0**: @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?  
> **username_0**: @qqjkztd the early data looks like this https://i.sstatic.net/MrFG3.jpg (from [here](https://arstechnica.com/science/2022/10/dart-mission-successfully-shifted-its-targets-orbit/)) 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...  
> **username_3**: 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](https://en.wikipedia.org/wiki/Bruce_Willis#2020s:_Critical_decline,_health_problems_and_retirement) contributing to some of his critically panned performances. :P  But Armageddon's silliness is all on Michael Bay.  
> **username_0**: @PeterCordes yes I took note of that when I first listened to it :-)  
> **username_4**: 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?  
> **username_5**: @username_1 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.  

---

## 4 answers

### Answer by username_6 on 2022-10-13. Score: 40

The original orbital period of [Dimorphos](https://en.wikipedia.org/wiki/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]](https://ui.adsabs.harvard.edu/abs/2021plde.confE.251M/abstract),

> 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](https://i.sstatic.net/Z8vi3.jpg)](https://i.sstatic.net/Z8vi3.jpg)

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.

> **Comments:**
> **username_7**: Intuitively it is always so confusing how hitting in prograde direction (increasing velocity) causes a longer orbital period by lifting to a higher orbit.  
> **username_8**: No momentum is "lost" to ejecta. That momentum is favourable to change the trajectory of the asteroid.  
> **username_9**: Whether the momentum transferred to ejecta is favourable depends on which direction the ejecta go.  
> **username_10**: @username_9: Wouldn't the ejecta tend to be blown "upward" which *increases* the momentum transfer assuming the spacecraft hits the asteroid head on?  
> **username_6**: @KevinKostlan You and username_8 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.  
> **username_10**: @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.  
> **username_0**: 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?](https://astronomy.stackexchange.com/q/50776/7982)  

---

### Answer by username_11 on 2022-10-13. Score: 8

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](https://physics.stackexchange.com/questions/731677/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](https://i.sstatic.net/Ydjlq.png)](https://i.sstatic.net/Ydjlq.png)

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.

> **Comments:**
> **username_12**: 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.  
> **username_11**: @NuclearHoagie the velocity if impact is of the order of 6 km/s whereas the orbital velocity was around 0.174 m/s.  
> **username_11**: @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.  

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### Answer by username_13 on 2022-10-16. Score: 2

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?

> **Comments:**
> **username_0**: 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."  
> **username_13**: 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 ... ???  
> **username_0**: 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)  

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### Answer by username_14 on 2022-10-15. Score: 1

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.

> **Comments:**
> **username_14**: @username_0 - a lower orbit around the other asteroid, not the Sun or Earth.  
> **username_0**: 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.  
> **username_14**: @username_0 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*)  
> **username_6**: I think this answer could be improved by a citation.  

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