Comments by "WhatAboutTheBee" (@WhatAboutTheBee) on "Astrum" channel.

  1. Puts paid to the idea of deflecting an incoming asteroid with nuclear explosions. Rock pile becomes rock shower when dispersed by an explosion.
    11
  2.  @ppsarrakis  Respectfully disagree. Yes, the sum of the energy of the parts is essentially the energy of the whole. We agree there. Yet the effect of a shower strike being less than the effect of a pile strike is entirely debatable. We need to consider alternative methods of deflecting asteroids. Any substantive strike can be existentially dangerous. I'd prefer a solution that avoids the strike entirely, instead of a pile strike or a shower of rock. Popping off nuclear weapons may be all we have right now, but frankly, it isn't a good solution.
    10
  3. Formula yielding peak velocity under constant acceleration, when v0 =0 is Vpeak=sqrt(2×A×D). A=0.08 meters/second squared. D = 20,000 meters. Therefore Vpeak=56.568 meters/second. Or 126.53 mph. That's plenty enough speed. Yet it is the deceleration at the end that kills you, not the speed. For example, 43 earth gravities (Gs) are sufficient to cause your skull to fracture away from your spine (for the curious, basilar skull fracture). Deceleration is inversely proportional to time. Let us assume it takes two seconds to decelerate from 57 meters/second to 0 meters/second. Deceleration is V/T, yielding 28.5 m/s^2. That's only 2.9 Gs and may be survivable, particularly if you land on your feet. If, however, we decrease the deceleration time to 0.135 seconds, we get V/T = 422m/s^2. That, of course, yields 43 earth gravities and will result in massive damage to your skeletal system. Your video glossed over the real cause of injury. It is not the velocity. It is the sudden stop at the end, to wit, deceleration.
    7
  4. Had the same thought. The wavelengths a person can perceive does not change. The actual wavelengths we perceive shift in the spectrum.
    3
  5. The weapon would have to accelerate the projectile by some other means. Compressed gas, springs, etc. Once accelerated, it will obey the nominal rules of ballistic trajectory. The air resistance slows the bullet down. As there is no air on Miranda, this does not occur. After the acceleration phase, the projectile velocity remains constant. However, the gravity vector (0.08m/s^2) on Miranda pulls the projectile down under a constant acceleration regime. Depending upon the velocity of the projectile, and its initial vector, escape velocity may be achieved. Hope that answered your question. I can clarify anything you like
    3
  6. At 8:57 Are we seeing the shadow of Cassini on Enceladus?
    2
  7. I think its a perception problem. I struggled with the "instantaneous" remark until I realized the following: Traveling to a point 100,000 LY away, at the speed of light, will still take 100,000 years. However, to the person traveling, it will be perceived to be 0 years. That is my understanding, your milage may vary.
    2
  8. More Mercury please!
    2
  9. Not quite. On Earth, terminal velocity is ~53m/s. For the instant problem, V peak is actually 56.5685 m/s.
    1
  10. @john doe   Hello John. The explanation is for those who seek the truth, and not a glossed over video. Astrum provides wonderful content, yet sometimes he misses the mark. For example, race cars routinely go at speeds well in excess of the peak velocity indicated by the cliff, and then crash, yet they don't routinely die. Why? Because the deceleration time is extended, decreasing the shock, increasing survivability. But if you strictly went by the video, they would all be dead. Don't believe everything you see in a video, even if it appears plausible
    1
  11.  @chrisdavis4230  Hi Chris. Let us Assume you shot straight up. Here on Earth, the bullet eventually slows to a stop due to gravity and air resistance. It then tumbles to the earth from that height, under a constant acceleration regime, until terminal velocity due to air resistance for the projectile is reached. On Miranda, there is no atmosphere. The projectile is only slowed by gravity. Let us assume that the projectile never reaches escape velocity. At some point, then, the projectile comes to a stop. The height above the surface is the initial 20Km plus the displacement over the projectiles travel. Let us further assume it took an additional 30Km to decelerate to zero. So at 50Km above the surface, we begin to fall towards the surface, under a constant 0.08m/s^2 regime. Vpeak = sqrt(2×A×D) ao we have Vpeak = sqrt(2×.08×50,000) 89.44 meters per second instead of 56.5685 m/s for the 20 Km fall. As there is no atmosphere, there is no terminal velocity. Once we achieve escape velicity, orbital mechanics apply, and I think, not really your question.
    1
  12. In more general terms, the projectile will follow a ballistic trajectory. The horizontal component of motion is independent of the vertical component. Whenever the vertical velocity upwards reaches zero, we can apply a constant acceleration regime to find the peak vertical velocity. To wit sqrt(2×A×D). A=0.08m/s^2. D is the height when the vertical velocity goes to zero. The horizontal component of motion just describes how far away. The projectile never slows due to atmosphere. So if shot perfectly horizontally, in a plane, it will strike the ground after the gravity accelerates there. Assume you are firing at 1.6 meters. The projectile must travel 1.6 meters down, strictly by gravity. The time to do so is sqrt(2x/A). That is 6.324 seconds. If the projectile has a horizontal component of 1200 meters per second, then it will fly 6.324 seconds at a rate of 1200 meters/second. It will touch down 7.589 kilometers from you, still traveling at 1200 meters/second (no air resistance) horizontally, but also with a 0.505 meter/second vertical component.
    1