Subject:
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Re: Falling thru earth revisited
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 8 Jun 2005 21:41:35 GMT
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Viewed:
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2641 times
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In lugnet.off-topic.geek, David Koudys wrote:
> My idea is that basically the accelleration of an object going towards the
> center due to gravity and the decelleration of an object from the center to the
> other side basically cancels out, so we would just have to measure the
> distance/time from the entry point to the exit and figure in the thrust of the
> rocket across that distance and we'd have a pretty good idea of how fast it'l
> lbe going.
Sounds right to me. Acceleration adds linearly, so when you integrate to get
the velocity, you have the integral of the thrust plus the integral of the
gravitational acceleration. Assuming that the change in mass of the rocket due
to depleted fuel is negligible, the gravitational portion will just be zero.
--Bram
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Message is in Reply To:
 | | Falling thru earth revisited
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| 'cause I've been ruminating about it.... So let's take this from the base model-- Base model ignores the following-- Rotation AIr Resistance you have the earth and a hole from where you are, going straight thru the center and right out to the other (...) (19 years ago, 8-Jun-05, to lugnet.off-topic.geek)
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