Near-Earth objects and close calls
- Thread starter Gawan
- Start date
Asteroid 2025 EF4 Makes a Closest Approach at approximately 65 meters in diameter and was recently discovered 2025 March 11.
Orbit Visualization
www.jpl.nasa.gov
March 19, 2024
Orbit Visualization
NASA Study: Asteroid’s Orbit, Shape Changed After DART Impact
After NASA’s historic Double Asteroid Redirection Test, a JPL-led study has shown that the shape of asteroid Dimorphos has changed and its orbit has shrunk.
After NASA’s historic Double Asteroid Redirection Test, a JPL-led study has shown that the shape of asteroid Dimorphos has changed and its orbit has shrunk.
When NASA’s DART (Double Asteroid Redirection Test) deliberately smashed into a 560-foot-wide (170-meter-wide) asteroid on Sept. 26, 2022, it made its mark in more ways than one. The demonstration showed that a kinetic impactor could deflect a hazardous asteroid should one ever be on a collision course with Earth. Now a new study published in the Planetary Science Journal shows the impact changed not only the motion of the asteroid, but also its shape.
DART’s target, the asteroid Dimorphos, orbits a larger near-Earth asteroid called Didymos. Before the impact, Dimorphos had a roughly symmetrical “oblate spheroid” shape – like a squashed ball that is wider than it is tall. With a well-defined, circular orbit at a distance of about 3,900 feet (1,189 meters) from Didymos, Dimorphos took 11 hours and 55 minutes to complete one loop around Didymos.
“When DART made impact, things got very interesting,” said Shantanu Naidu, a navigation engineer at NASA’s Jet Propulsion Laboratory in Southern California, who led the study. “Dimorphos’ orbit is no longer circular: Its orbital period” – the time it takes to complete a single orbit – “is now 33 minutes and 15 seconds shorter. And the entire shape of the asteroid has changed, from a relatively symmetrical object to a ‘triaxial ellipsoid’ – something more like an oblong watermelon.”
https://d2pn8kiwq2w21t.cloudfront.net/original_images/E_before-after-impact.png
This illustration shows the approximate shape change that the asteroid Dimorphos experienced after DART hit it. Before impact, left, the asteroid was shaped like a squashed ball; after impact it took on a more elongated shape, like a watermelon. Credit: NASA/JPL-Caltech
Dimorphos Damage Report
Naidu’s team used three data sources in their computer models to deduce what had happened to the asteroid after impact. The first source was aboard DART: The spacecraft captured images as it approached the asteroid and sent them back to Earth via NASA’s Deep Space Network (DSN). These images provided close-up measurements of the gap between Didymos and Dimorphos while also gauging the dimensions of both asteroids just prior to impact.
The second data source was the DSN’s Goldstone Solar System Radar, located near Barstow, California, which bounced radio waves off both asteroids to precisely measure the position and velocity of Dimorphos relative to Didymos after impact. Radar observations quickly helped NASA conclude that DART’s effect on the asteroid greatly exceeded the minimum expectations.
The third and most significant source of data: ground telescopes around the world that measured both asteroids’ “light curve,” or how the sunlight reflecting off the asteroids’ surfaces changed over time. By comparing the light curves before and after impact, the researchers could learn how DART altered Dimorphos’ motion.
As Dimorphos orbits, it periodically passes in front of and then behind Didymos. In these so-called “mutual events,” one asteroid can cast a shadow on the other, or block our view from Earth. In either case, a temporary dimming – a dip in the light curve – will be recorded by telescopes.
“We used the timing of this precise series of light-curve dips to deduce the shape of the orbit, and because our models were so sensitive, we could also figure out the shape of the asteroid,” said Steve Chesley, a senior research scientist at JPL and study co-author.
The team found Dimorphos’ orbit is now slightly elongated, or eccentric. “Before impact,” Chesley continued, “the times of the events occurred regularly, showing a circular orbit. After impact, there were very slight timing differences, showing something was askew. We never expected to get this kind of accuracy.”
The models are so precise, they even show that Dimorphos rocks back and forth as it orbits Didymos, Naidu said.
Orbital Evolution
The team’s models also calculated how Dimorphos’ orbital period evolved. Immediately after impact, DART reduced the average distance between the two asteroids, shortening Dimorphos’ orbital period by 32 minutes and 42 seconds, to 11 hours, 22 minutes, and 37 seconds.
Over the following weeks, the asteroid’s orbital period continued to shorten as Dimorphos lost more rocky material to space, finally settling at 11 hours, 22 minutes, and 3 seconds per orbit – 33 minutes and 15 seconds less time than before impact. This calculation is accurate to within 1 ½ seconds, Naidu said. Dimorphos now has a mean orbital distance from Didymos of about 3,780 feet (1,152 meters) – about 120 feet (37 meters) closer than before impact.
“The results of this study agree with others that are being published,” said Tom Statler, lead scientist for solar system small bodies at NASA Headquarters in Washington. “Seeing separate groups analyze the data and independently come to the same conclusions is a hallmark of a solid scientific result. DART is not only showing us the pathway to an asteroid-deflection technology, it’s revealing new fundamental understanding of what asteroids are and how they behave.”
These results and observations of the debris left after impact indicate that Dimorphos is a loosely packed “rubble pile” object, similar to asteroid Bennu. ESA’s (European Space Agency) Hera mission, planned to launch in October 2024, will travel to the asteroid pair to carry out a detailed survey and confirm how DART reshaped Dimorphos.
Meteor fireball over Ohio and 12 other states on March 18
We received 112 reports about a fireball seen over AL, GA, IL, IN, KY, MI, MO, NC, OH, PA, TN, VA and WV on Tuesday, March 18th 2025 around 03:23 UT. For this event, we received 4 videos and one photo.
www.sott.net
The CNEOS list includes the larger fireballs and bolides, but is it so that some areas of the Earth are hit more often? The question has intrigued me for some days, and when I came up with a model that might work, I gave it a try. To save readers time, let me say, the distribution appears rather random, but it was still a fun excercise, and maybe others can come up with much better approaches to investigate the problem and the available data, which in fact might not be as complete and correct as presented, but that is another story.
Here is the visual representation from the website of strikes equal to or larger than one kt. There are many other strikes, most are much smaller, down to about 0.073kt, but to avoid drowning in data, I reduced the complexity by taking just the somewhat larger strikes.
Skipping for now the calculations, for more on that later, next is the table based on the divisions according to latitude and grouped in areas that correspond to about 5 % of the surface of the earth, listing only the above strikes.
The distribution of fireballs at or above 1 kt according to latitude:
There were two more impact, both of 1.3 kt, but it was not mentioned where they took place though the dates have been noted.
It is clear, that the distribution is pretty equal between north and south, 54 to 56.
The patterns that might emerge will to some extend depend on the intervals chosen. In the table, there are 20 divisions, so with 110 strikes, one would expect around, 5.5 hits per zone of 5 %, or with a bit of leverage for the inexact divisions, say 4-7 hits for each. Indeed 14 zones fall within this range, while there is just one below and five above.
In the table, there is one zone with just 1 hit, (9-12 degrees), then seven times 4, three times 5, three times 6, one case of 7, four times 8, and one case of 9.
Is there any significance to 6-9, 12-15, 21-24, 52-57, and 57-63 having 8 strikes, the last 9? Together, they account for 25.41 % of the area and have 41/110 of the strikes. I don't know enough about statistics to say there is a pattern, I would even say it appears random, but on the other hand, it is also possible that windows do exist, but that the data is too small, or that windows do not follow the pattern, I have imposed using the division of the globe, by dividing it into neat slices.
The method used to calculate the distribution of the areas.
In case anyone is interested in how the table was created, here are som notes; I did not use any math that is not taught in highschool, or to be more specific, I used the concept of radians, the cosine function, and the formula for the circumference of a circle.
I worked on the assumption that the earth can be modelled as a perfect sphere, and that therefore I can use the Unit sphere as an approximation for calculation. What is handy is that the unit sphere has a radius of 1, which makes calculation easier.
First, I used an Excel sheet and listed the angles between 0 and 90,
Next I converted the degrees into radians, since that is what Excel prefers.
The cosine to the angle gives the radius for each degree, or shall we say the distance from a point on the sphere at a certain latitude to an imaginary rotational axis.
In a circle with the radius 1, the circumference is 2pi, the quater circle is then 0.5pi. If divided by 90 degrees we have the width of each degree. On the Earth for example, the distance from the equator to the poles is about 10,000 km, and therefore, the distance between two adjacent parallels, say 45 and 44, or 89 the pole, will be 10,000 km/90 or a little more than 110 km.
To calculate the surface of the area between 0 and 1 degree, I took the average of the cosine to 0 degrees, which is 1 and the cosine to 1 degree which is 0.99984... In this case the difference is small, but that was the principle working up to 90 degrees where the cosine is zero.
Given the average radius and width, I calculated the areas as the length of the strip, using that the circumference would be 2 pi*r and multiplying it by the width which was 0.01745... The reason, I could use the simple formula for the area of the rectangle was that I had averaged the radius. It is an approximation of course, but it would be a lot of calculating if it had to be done with pen and paper, or learning if done with more advanced mathematics. Fortunately, in Excel one can quickly generate results and adding all the small areas together gave a total surface area of 6.2830 which is a very good approximation, because the exact value for the area of half the unit sphere would be 2pi.
Next, I calculated the area occupied by each degree and divided it by the total area. Thus the area between 0 and 1 degree occupies little more than 1.7 % whereas the last area between 89 and 90 occupies only 0.015 % In the table I grouped the number of degrees that correspond to about 5 %. For a better model, it would be necessary to make the divisions more exact. After all the size of the areas vary between 4.31 and 5.5 %. On the other hand one could also group the small groups into larger groups of say 10 or 20 %.
An alternative approach would be to do a separation not according to latitude, but using longitude instead, and what would happen if instead of the limited number of just over 100 strikes, one included all the more than 1000 strikes in the list?
Here is the visual representation from the website of strikes equal to or larger than one kt. There are many other strikes, most are much smaller, down to about 0.073kt, but to avoid drowning in data, I reduced the complexity by taking just the somewhat larger strikes.
Skipping for now the calculations, for more on that later, next is the table based on the divisions according to latitude and grouped in areas that correspond to about 5 % of the surface of the earth, listing only the above strikes.
The distribution of fireballs at or above 1 kt according to latitude:
| Angle interval, latitude | Ap area in % of the earth | North impact | South impact | Sum |
| 0-3 | 5.23 | 4 | 2 | 6 |
| 3-6 | 5.22 | 2 | 2 | 4 |
| 6-9 | 5.19 | 3 | 5 | 8 |
| 9-12 | 5.15 | 0 | 1 | 1 |
| 12-15 | 5.09 | 6 | 2 | 8 |
| 15-18 | 5.02 | 2 | 2 | 4 |
| 18-21 | 4.94 | 3 | 3 | 6 |
| 21-24 | 4.84 | 3 | 5 | 8 |
| 24-27 | 4.73 | 3 | 2 | 5 |
| 27-30 | 4.60 | 2 | 4 | 6 |
| 30-33 | 4.46 | 3 | 1 | 4 |
| 33-36 | 4.31 | 2 | 2 | 4 |
| 36-40 | 5.50 | 1 | 3 | 4 |
| 40-44 | 5.19 | 4 | 3 | 7 |
| 44-48 | 4.85 | 2 | 3 | 5 |
| 48-52 | 4.49 | 1 | 3 | 4 |
| 52-57 | 5.07 | 5 | 3 | 8 |
| 57-63 | 5.23 | 5 | 4 | 9 |
| 63-71 | 5.45 | 1 | 4 | 5 |
| 71-90 | 5.45 | 2 | 2 | 4 |
| 54 | 56 | 110 |
It is clear, that the distribution is pretty equal between north and south, 54 to 56.
The patterns that might emerge will to some extend depend on the intervals chosen. In the table, there are 20 divisions, so with 110 strikes, one would expect around, 5.5 hits per zone of 5 %, or with a bit of leverage for the inexact divisions, say 4-7 hits for each. Indeed 14 zones fall within this range, while there is just one below and five above.
In the table, there is one zone with just 1 hit, (9-12 degrees), then seven times 4, three times 5, three times 6, one case of 7, four times 8, and one case of 9.
Is there any significance to 6-9, 12-15, 21-24, 52-57, and 57-63 having 8 strikes, the last 9? Together, they account for 25.41 % of the area and have 41/110 of the strikes. I don't know enough about statistics to say there is a pattern, I would even say it appears random, but on the other hand, it is also possible that windows do exist, but that the data is too small, or that windows do not follow the pattern, I have imposed using the division of the globe, by dividing it into neat slices.
The method used to calculate the distribution of the areas.
In case anyone is interested in how the table was created, here are som notes; I did not use any math that is not taught in highschool, or to be more specific, I used the concept of radians, the cosine function, and the formula for the circumference of a circle.
I worked on the assumption that the earth can be modelled as a perfect sphere, and that therefore I can use the Unit sphere as an approximation for calculation. What is handy is that the unit sphere has a radius of 1, which makes calculation easier.
First, I used an Excel sheet and listed the angles between 0 and 90,
Next I converted the degrees into radians, since that is what Excel prefers.
The cosine to the angle gives the radius for each degree, or shall we say the distance from a point on the sphere at a certain latitude to an imaginary rotational axis.
In a circle with the radius 1, the circumference is 2pi, the quater circle is then 0.5pi. If divided by 90 degrees we have the width of each degree. On the Earth for example, the distance from the equator to the poles is about 10,000 km, and therefore, the distance between two adjacent parallels, say 45 and 44, or 89 the pole, will be 10,000 km/90 or a little more than 110 km.
To calculate the surface of the area between 0 and 1 degree, I took the average of the cosine to 0 degrees, which is 1 and the cosine to 1 degree which is 0.99984... In this case the difference is small, but that was the principle working up to 90 degrees where the cosine is zero.
Given the average radius and width, I calculated the areas as the length of the strip, using that the circumference would be 2 pi*r and multiplying it by the width which was 0.01745... The reason, I could use the simple formula for the area of the rectangle was that I had averaged the radius. It is an approximation of course, but it would be a lot of calculating if it had to be done with pen and paper, or learning if done with more advanced mathematics. Fortunately, in Excel one can quickly generate results and adding all the small areas together gave a total surface area of 6.2830 which is a very good approximation, because the exact value for the area of half the unit sphere would be 2pi.
Next, I calculated the area occupied by each degree and divided it by the total area. Thus the area between 0 and 1 degree occupies little more than 1.7 % whereas the last area between 89 and 90 occupies only 0.015 % In the table I grouped the number of degrees that correspond to about 5 %. For a better model, it would be necessary to make the divisions more exact. After all the size of the areas vary between 4.31 and 5.5 %. On the other hand one could also group the small groups into larger groups of say 10 or 20 %.
An alternative approach would be to do a separation not according to latitude, but using longitude instead, and what would happen if instead of the limited number of just over 100 strikes, one included all the more than 1000 strikes in the list?
Session 30 August 2014
When I checked for the latest update to the CNEOS list, there were three new entries:
This means that there since the beginning of the year have been 12 new additions. In the same period of 2024 (4), 2023 (11), 2022 (12), 2021 (9), 2020 (14), 2019 (6), 2018 (9), 2017 (8), 2016 (9), 2015 (11), 2014 (7), 2013 (6), 2012 (10), and 2011 (8).
None of the above appears to have been captured by video on the ground, the only the one from March 21, was over land, and corresponds to Telfer in Western Australia, an isolated mining town.
That was more than ten years ago. Now, between the early months of 2024 and 2025 there has been a shift in activity from less to more:
When I checked for the latest update to the CNEOS list, there were three new entries:
This means that there since the beginning of the year have been 12 new additions. In the same period of 2024 (4), 2023 (11), 2022 (12), 2021 (9), 2020 (14), 2019 (6), 2018 (9), 2017 (8), 2016 (9), 2015 (11), 2014 (7), 2013 (6), 2012 (10), and 2011 (8).
None of the above appears to have been captured by video on the ground, the only the one from March 21, was over land, and corresponds to Telfer in Western Australia, an isolated mining town.
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