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I'm trying to help a student solve the following problem:

A 15 meter in diameter asteroid with a mass of approximately 10,000 metric tons going 20 km/s. The entry and time for disintegration was 28 seconds. Entry angle to horizontal is 18 degrees and the terminal art of the fireball was at about 25 km. What is the straight line distance the meteor traveled?

How is the angle of entry factored into the calculation of the straight line distance?

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    $\begingroup$ Carrie, this looks like a homework problem. If you have a specific question on how to solve this type of problem, please refine your question to ask about that specific detail. It may turn out that your question is off topic here, but it may be valid on another site, in which case we can migrate if it isn't already answered there. $\endgroup$ – called2voyage Dec 15 '16 at 14:41
  • $\begingroup$ I want to know if they angle comes into play. I am trying to help a student that has this on a NASA question. I have looked and looked for how to solve. We thought maybe to get the 320 Km that are in the atmosphere minus the 25 km. That would be our one side of the triangle. then x cos 72= 295. $\endgroup$ – Carrie Davis Dec 15 '16 at 16:33
  • $\begingroup$ Then I also wondered if it was much more simple. Jut to solve the distance traveled by using the 20km/s divided by the 28.5 sec. $\endgroup$ – Carrie Davis Dec 15 '16 at 16:37
  • $\begingroup$ I've revised your question. Let me know if you have any problem with my revision. Even if the angle of entry isn't meant to be factored in, our users should be able to answer from this perspective. $\endgroup$ – called2voyage Dec 15 '16 at 16:38
  • $\begingroup$ Also, welcome to Space Exploration Stack Exchange! I hope you find our site useful enough to stick around. If you have the time, you should definitely review our Tour and Help Center, if you haven't already. $\endgroup$ – called2voyage Dec 15 '16 at 16:40

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