## NASA: We Will Try To Divert Binary Near-Earth Asteroid (65803) Didymos... | |

Astromut Senior Forum Moderator 07/26/2021 04:54 PM Report Abusive Post Report Copyright Violation | The raw data is available, but not the code or processes of how they composited it and created the website... at some point (not recorded anywhere that I can see), someone made some "choices" about how the images would look, including the gradient to make the "sphere" object look 3 dimensional. Quoting: Anonymous Coward 80630295 Wrong. That is a flat out lie. You want the code? Here you go, here's my code for processing the raw EPIC image data: [link to drive.google.com (secure)] Here is a photo of earth processed with my code: [link to drive.google.com (secure)] In your next post you will explain where in my code I made any decisions to make the "sphere" object look 3 dimensional. This is not a mosaic from multiple orbits of some low earth orbiting satellite, this is a single shot imager that takes images in a wide variety of wavelengths. The only "decision" I made was to grab a red, green and blue wavelength and load them as red, green and blue channels after converting the high bit depth data to 8 bits. |

Anonymous Coward User ID: 80266721 United States 07/26/2021 05:18 PM Report Abusive Post Report Copyright Violation | |

Astromut Senior Forum Moderator 07/26/2021 05:25 PM Report Abusive Post Report Copyright Violation | Always funny to see AstroMut get butthurt. You always have to ponder why self proclaimed intelligent people, need to defend their intelligence. Your wrong Astromut. You Are Wrong Anonymous Coward 80266721 No, I'm not. Astromut Lol, good answer. Anonymous Coward 80266721 It's the right answer. |

Anonymous Coward User ID: 80630295 Australia 07/26/2021 05:29 PM Report Abusive Post Report Copyright Violation | I'm a dipshit who failed to show where astro faked the image in his code, which is why I tried to quote irrelevant bullshit that has nothing to do with the photo he presented. Last Edited by Astromut on 07/26/2021 05:55 PM |

Astromut Senior Forum Moderator 07/26/2021 05:50 PM Report Abusive Post Report Copyright Violation | The raw data is available, but not the code or processes of how they composited it and created the website... at some point (not recorded anywhere that I can see), someone made some "choices" about how the images would look, including the gradient to make the "sphere" object look 3 dimensional. Quoting: Anonymous Coward 80630295 Wrong. That is a flat out lie. You want the code? Here you go, here's my code for processing the raw EPIC image data: [link to drive.google.com (secure)] Here is a photo of earth processed with my code: [link to drive.google.com (secure)] In your next post you will explain where in my code I made any decisions to make the "sphere" object look 3 dimensional. This is not a mosaic from multiple orbits of some low earth orbiting satellite, this is a single shot imager that takes images in a wide variety of wavelengths. The only "decision" I made was to grab a red, green and blue wavelength and load them as red, green and blue channels after converting the high bit depth data to 8 bits. Astromut lol... that digitally altered IMAGE (not a photo) is laughable. Anonymous Coward 80630295 You have failed to show where I altered the photo to make earth look like a sphere, as you claimed. The only thing I did was convert the raw data to an 8 bit image. It's still a photo of earth. From your website: Quoting: To avoid some currently unresolved geolocation issues at the edge of the disk, only pixels observed at viewing zenith angle (VZA) less than 75° have been corrected and plotted in the enhanced images. Pixels with VZA > 75° (3.7% of all pixels) have been plotted black.ACFunny, you just quoted the part for the enhanced color images, not the natural color images and definitely not the raw image data I linked to and processed myself using code I posted above. [link to epic.gsfc.nasa.gov (secure)] Show in my code where I did any of what you just described, or else you're done here. The "disc" is why it is a perfect sphere or globe shape on the screen, and not an oblate spheroid it is supposed to be. Quoting: ACWrong. If you measure photos taken near the equinox when the sun is directly over the equator, the earth measures about 1.003 times wider than it is tall in the image when north is oriented up. It's just a few pixels at this scale, but it is detectable. |

Anonymous Coward User ID: 77801268 United Kingdom 07/26/2021 07:10 PM Report Abusive Post Report Copyright Violation | |

Doc User ID: 80614448 United States 07/26/2021 09:49 PM Report Abusive Post Report Copyright Violation | ...I don't need to fool you, you're already a fool. I've polled GLP, about a third are unwilling to even watch my videos presenting the truth, same proportion of users who leave me red. It's held steady at about that percentage for years. Thread: Should I live stream ISS? Astromut He's tossed down the gauntlet Astro Anonymous Coward 78062180 What gauntlet? It's up to him to support his argument that it could redirect the moon to earth. So far he presented nothing. That which can be asserted without evidence can be dismissed without evidence. You are dismissed. Astromut Since some of you won't shut up about this, let's do the math. Then you both owe me an apology. We know the size of Didymos is 780 meters in diameter. Using the formula for the volume of a sphere we can approximate its volume as about 248,474,846 m^3. We can calculate its mass using the orbit of its moon, Dimorphos. For a given orbital period and distance, m = ((2*pi)^2 * distance^3)/(period^2*G). With period = 42837 seconds and distance = 1,180 m we get a mass of about 5.29632*10^11 kg. That means an overall density of about 2,132 kg/m^3 which is pretty typical for an asteroid. The moon is much smaller, only about 170 +/- 30 meters based on radar observations, so let's take the lower value of 140 m to be generous to the claim that DART can move this thing out of Didymos orbit and towards earth. Using our previously established density of about 2132 kg/m^3 we find a mass of Dimorphos of about 3.06 * 10^9 kilograms which is a little less than the official value because we've assumed the minimum size. We know the DART impactor will be 500 kg and traveling at about 6.6 km/s, so we can be very generous and assume an entirely elastic collision with no dissipation where the impactor transfers its entire kinetic energy into the moon as kinetic energy (which won't happen, but again we're trying to see the MAXIMUM change in velocity this impactor could impart). We can use the formula for an elastic collision and find out how much the moon will change in velocity from its initial velocity. (500kg*6600m/s)/3.06316175 * 10^9 kilograms = 0.001 m/s That's a tenth of a cm/s. That's barely anything at all. This is why they're doing this to a moon of a larger asteroid; the velocity of its orbit around its parent is small, so even a small change will be detectable. But how fast is it orbiting? Well for an average distance of about 1.18 km in a nearly circular orbit we get an orbital circumference of about 3.7 km and at an orbital period of 42,837 seconds, we find an orbital velocity of about 0.086 m/s. So a change of 0.001 m/s would be a change of about 1% to its orbital velocity around its parent asteroid. Is that even enough to escape from Didymos if applied in the right way to accelerate the moon? Well, do the math. Escape velocity = sqrt((2*G*M)/r). M is the mass of Didymos (5.29632*10^11 kg), r is the radius of Dimorphos' orbit (1.18 km) so plugging that in along with the gravitational constant gives an escape velocity of about 0.2448 m/s. 0.001 + 0.086 = 0.0861 which is way, way less than 0.2448 m/s. So no, even assuming the most generous things about the size and mass of the moon, even assuming a perfectly elastic collision with no losses to heat which brings the impactor to a dead stop and transfers all the kinetic energy to the moon, it still would not be nearly enough to push the moon out of Didymos orbit, let alone to earth. Astromut ERROR REPORT Energy is mass times velocity squared. You bust something up lots of smaller impacts produce the same total energy but spread out over time and space. |

Anonymous Coward User ID: 80476557 Singapore 07/26/2021 10:54 PM Report Abusive Post Report Copyright Violation | What gauntlet? It's up to him to support his argument that it could redirect the moon to earth. So far he presented nothing. That which can be asserted without evidence can be dismissed without evidence. You are dismissed. Astromut Since some of you won't shut up about this, let's do the math. Then you both owe me an apology. We know the size of Didymos is 780 meters in diameter. Using the formula for the volume of a sphere we can approximate its volume as about 248,474,846 m^3. We can calculate its mass using the orbit of its moon, Dimorphos. For a given orbital period and distance, m = ((2*pi)^2 * distance^3)/(period^2*G). With period = 42837 seconds and distance = 1,180 m we get a mass of about 5.29632*10^11 kg. That means an overall density of about 2,132 kg/m^3 which is pretty typical for an asteroid. The moon is much smaller, only about 170 +/- 30 meters based on radar observations, so let's take the lower value of 140 m to be generous to the claim that DART can move this thing out of Didymos orbit and towards earth. Using our previously established density of about 2132 kg/m^3 we find a mass of Dimorphos of about 3.06 * 10^9 kilograms which is a little less than the official value because we've assumed the minimum size. We know the DART impactor will be 500 kg and traveling at about 6.6 km/s, so we can be very generous and assume an entirely elastic collision with no dissipation where the impactor transfers its entire kinetic energy into the moon as kinetic energy (which won't happen, but again we're trying to see the MAXIMUM change in velocity this impactor could impart). We can use the formula for an elastic collision and find out how much the moon will change in velocity from its initial velocity. (500kg*6600m/s)/3.06316175 * 10^9 kilograms = 0.001 m/s That's a tenth of a cm/s. That's barely anything at all. This is why they're doing this to a moon of a larger asteroid; the velocity of its orbit around its parent is small, so even a small change will be detectable. But how fast is it orbiting? Well for an average distance of about 1.18 km in a nearly circular orbit we get an orbital circumference of about 3.7 km and at an orbital period of 42,837 seconds, we find an orbital velocity of about 0.086 m/s. So a change of 0.001 m/s would be a change of about 1% to its orbital velocity around its parent asteroid. Is that even enough to escape from Didymos if applied in the right way to accelerate the moon? Well, do the math. Escape velocity = sqrt((2*G*M)/r). M is the mass of Didymos (5.29632*10^11 kg), r is the radius of Dimorphos' orbit (1.18 km) so plugging that in along with the gravitational constant gives an escape velocity of about 0.2448 m/s. 0.001 + 0.086 = 0.0861 which is way, way less than 0.2448 m/s. So no, even assuming the most generous things about the size and mass of the moon, even assuming a perfectly elastic collision with no losses to heat which brings the impactor to a dead stop and transfers all the kinetic energy to the moon, it still would not be nearly enough to push the moon out of Didymos orbit, let alone to earth. Astromut ERROR REPORT Energy is mass times velocity squared. You bust something up lots of smaller impacts produce the same total energy but spread out over time and space. Doc 80614448 an Explosion is a whole bunch of molecules or atoms releasing energy at the same instant. As opposed to a Fire, which is a gradual release of energy over time. . |

Anonymous Coward User ID: 80630295 Australia 07/27/2021 12:23 AM Report Abusive Post Report Copyright Violation | I'm a cock sucking piece of shit who falsely accused astro of faking a photo of earth. Last Edited by Astromut on 07/27/2021 09:30 AM |

AZ40 User ID: 79621597 United States 07/27/2021 12:58 AM Report Abusive Post Report Copyright Violation | I've been watching this since hearing about it months ago, there's other threads about it... Quoting: My gut tells me this is going to be big trouble. 5 stars StellaBlue —— I’ve had the same intuitive, spiritual indicators as well. — Anonymous Coward 73478786 During the past week I came across prophetic predictions made by a young anonymous Mexican girl. Okay - there's an endless supply of these prediction people (mostly bogus) and this one has extra layers of farfetchedness. I mean, who would believe this? But one thing she said was that very soon Earth will have a close call with large celestial body. It would miss, but then a second smaller one associated with it would hit the planet, but land in a place where there are no people. (And after this the "Aviso" will occur.) I thought this was a pretty specific prediction and wondered about it. And now today I read this. If you are familiar with the "Aviso" it is said that it will occur when things are at their darkest. Lots of ifs, but "if" that prediction is correct, and "if" these asteroids are the ones, we also know the world will be in a very dark place by the fall of 2022. That might something to plan around or for. . |

Astromut Senior Forum Moderator 07/27/2021 09:28 AM Report Abusive Post Report Copyright Violation | What gauntlet? It's up to him to support his argument that it could redirect the moon to earth. So far he presented nothing. That which can be asserted without evidence can be dismissed without evidence. You are dismissed. Astromut Since some of you won't shut up about this, let's do the math. Then you both owe me an apology. We know the size of Didymos is 780 meters in diameter. Using the formula for the volume of a sphere we can approximate its volume as about 248,474,846 m^3. We can calculate its mass using the orbit of its moon, Dimorphos. For a given orbital period and distance, m = ((2*pi)^2 * distance^3)/(period^2*G). With period = 42837 seconds and distance = 1,180 m we get a mass of about 5.29632*10^11 kg. That means an overall density of about 2,132 kg/m^3 which is pretty typical for an asteroid. The moon is much smaller, only about 170 +/- 30 meters based on radar observations, so let's take the lower value of 140 m to be generous to the claim that DART can move this thing out of Didymos orbit and towards earth. Using our previously established density of about 2132 kg/m^3 we find a mass of Dimorphos of about 3.06 * 10^9 kilograms which is a little less than the official value because we've assumed the minimum size. We know the DART impactor will be 500 kg and traveling at about 6.6 km/s, so we can be very generous and assume an entirely elastic collision with no dissipation where the impactor transfers its entire kinetic energy into the moon as kinetic energy (which won't happen, but again we're trying to see the MAXIMUM change in velocity this impactor could impart). We can use the formula for an elastic collision and find out how much the moon will change in velocity from its initial velocity. (500kg*6600m/s)/3.06316175 * 10^9 kilograms = 0.001 m/s That's a tenth of a cm/s. That's barely anything at all. This is why they're doing this to a moon of a larger asteroid; the velocity of its orbit around its parent is small, so even a small change will be detectable. But how fast is it orbiting? Well for an average distance of about 1.18 km in a nearly circular orbit we get an orbital circumference of about 3.7 km and at an orbital period of 42,837 seconds, we find an orbital velocity of about 0.086 m/s. So a change of 0.001 m/s would be a change of about 1% to its orbital velocity around its parent asteroid. Is that even enough to escape from Didymos if applied in the right way to accelerate the moon? Well, do the math. Escape velocity = sqrt((2*G*M)/r). M is the mass of Didymos (5.29632*10^11 kg), r is the radius of Dimorphos' orbit (1.18 km) so plugging that in along with the gravitational constant gives an escape velocity of about 0.2448 m/s. 0.001 + 0.086 = 0.0861 which is way, way less than 0.2448 m/s. So no, even assuming the most generous things about the size and mass of the moon, even assuming a perfectly elastic collision with no losses to heat which brings the impactor to a dead stop and transfers all the kinetic energy to the moon, it still would not be nearly enough to push the moon out of Didymos orbit, let alone to earth. Astromut ERROR REPORT Energy is mass times velocity squared. Doc 80614448 And? I used the formula for an elastic collision, the starting and ending momentum is entirely preserved. If it busts up the moon to tiny pieces then those tiny pieces don't pose any threat anyway. |

Astromut Senior Forum Moderator 07/27/2021 09:30 AM Report Abusive Post Report Copyright Violation | A "photo" is a capture of a moment in time (there are thousands of "photos" of the moon for instance, taken with a variety of cameras), complete with context and unaltered. Quoting: Anonymous Coward 80630295 Which is what that raw data contains. An "image" (yes an photo is always an image... blah blah... that would be obfuscation) is a creative process, involving time and manipulation, and is altered in some way. Quoting: Your "images" are altered at multiple points throughout your process ACI provided the source code. You failed to back up your claims for the alterations you claimed. You are done here. |

Astromut Senior Forum Moderator 07/27/2021 09:40 AM Report Abusive Post Report Copyright Violation | ...What gauntlet? It's up to him to support his argument that it could redirect the moon to earth. So far he presented nothing. That which can be asserted without evidence can be dismissed without evidence. You are dismissed. Astromut Since some of you won't shut up about this, let's do the math. Then you both owe me an apology. We know the size of Didymos is 780 meters in diameter. Using the formula for the volume of a sphere we can approximate its volume as about 248,474,846 m^3. We can calculate its mass using the orbit of its moon, Dimorphos. For a given orbital period and distance, m = ((2*pi)^2 * distance^3)/(period^2*G). With period = 42837 seconds and distance = 1,180 m we get a mass of about 5.29632*10^11 kg. That means an overall density of about 2,132 kg/m^3 which is pretty typical for an asteroid. The moon is much smaller, only about 170 +/- 30 meters based on radar observations, so let's take the lower value of 140 m to be generous to the claim that DART can move this thing out of Didymos orbit and towards earth. Using our previously established density of about 2132 kg/m^3 we find a mass of Dimorphos of about 3.06 * 10^9 kilograms which is a little less than the official value because we've assumed the minimum size. We know the DART impactor will be 500 kg and traveling at about 6.6 km/s, so we can be very generous and assume an entirely elastic collision with no dissipation where the impactor transfers its entire kinetic energy into the moon as kinetic energy (which won't happen, but again we're trying to see the MAXIMUM change in velocity this impactor could impart). We can use the formula for an elastic collision and find out how much the moon will change in velocity from its initial velocity. (500kg*6600m/s)/3.06316175 * 10^9 kilograms = 0.001 m/s That's a tenth of a cm/s. That's barely anything at all. This is why they're doing this to a moon of a larger asteroid; the velocity of its orbit around its parent is small, so even a small change will be detectable. But how fast is it orbiting? Well for an average distance of about 1.18 km in a nearly circular orbit we get an orbital circumference of about 3.7 km and at an orbital period of 42,837 seconds, we find an orbital velocity of about 0.086 m/s. So a change of 0.001 m/s would be a change of about 1% to its orbital velocity around its parent asteroid. Is that even enough to escape from Didymos if applied in the right way to accelerate the moon? Well, do the math. Escape velocity = sqrt((2*G*M)/r). M is the mass of Didymos (5.29632*10^11 kg), r is the radius of Dimorphos' orbit (1.18 km) so plugging that in along with the gravitational constant gives an escape velocity of about 0.2448 m/s. 0.001 + 0.086 = 0.0861 which is way, way less than 0.2448 m/s. So no, even assuming the most generous things about the size and mass of the moon, even assuming a perfectly elastic collision with no losses to heat which brings the impactor to a dead stop and transfers all the kinetic energy to the moon, it still would not be nearly enough to push the moon out of Didymos orbit, let alone to earth. Astromut ERROR REPORT Energy is mass times velocity squared. Doc 80614448 And? I used the formula for an elastic collision, the starting and ending momentum is entirely preserved. If it busts up the moon to tiny pieces then those tiny pieces don't pose any threat anyway. Astromut Let's reverse the equation and find out how much mass the impactor could put on a parabolic orbit away from Didymos. m2 = (m1v1)/v2f 500*6600/0.2448 = 13,480,392 kg 13,480,392 kg / 2132 kg/m^3 = a remnant of 6322.88555 m^3 volume. That corresponds to a sphere with a radius of just 11 meters. And that's generously assuming that the impactor somehow conveys all of its kinetic energy to JUST that amount of mass, no other parts, nothing lost to friction or heat as it vaporizes on impact, that it's a perfect elastic collision etc. And with all of that you have a piece of the moon 11 meters in radius with just enough velocity to barely break orbit from Didymos but without enough delta-V to divert it to earth. That's just what it takes to get a piece of it away from Didymos permanently and into its own orbit around the sun. |

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