Re: What would be Earth's new Orbit if an Asteroid slowed it's speed down by half?

In sci.physics.relativity, guskz@xxxxxxxxxxx
<guskz@xxxxxxxxxxx>
wrote
on Mon, 25 Jun 2007 20:14:03 -0700
<1182827643.282464.65610@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
On Jun 25, 6:25 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Jun 25, 1:54 pm, "g...@xxxxxxxxxxx" <g...@xxxxxxxxxxx> wrote:
[...]

http://www.amazon.com/Mechanics-3rd-Keith-R-Symon/dp/0201073927

Another boring problem thought up by an equally boring mind. You would
know how to do this *** if you just STUDIED a little.

Liar, do it.


I need the practice. :-)

Assuming the Earth's orbit is currently perfectly circular (which it's
not, but this is a back-of-the-envelope calc), we currently have an
orbital velocity of 30 km/s. The asteroid (which would have to be
pretty big!) would reduce that velocity to 15 km/s, at what will
essentially become the new aphelion.

It turns out that the old orbit energy is dependent solely on the
semimajor axis (the old radius). Since the old orbit was 1 AU,
we can express the energy as

E_old = -G*M_sun * M_earth / (2*1 AU) = -K/2

where K is some constant. The kinetic energy of this orbit turns
out to be +K/2; therefore

E = -K + K/8 = -7K/8 = -G*M_sun * M_earth / (2 * (4/7 AU))

With aphelion of 1 AU and semimajor axis of 4/7 AU, we get
4/7 * (1 + e) = 1 or e = 3/4; therefore perihelion is
4/7 * 1/4 = 1/7 AU.

Since T = 2*pi * a^(3/2) / sqrt(G*M_sun), and T_old = 1 year,
T_new = (4/7)^(3/2) = 157.8 old days (the new day, of course,
will depend on precisely where the asteroid hits).

Global warming will be the least of our problems. :-)

http://scienceworld.wolfram.com/physics/Orbit.html

http://en.wikipedia.org/wiki/Ellipse

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#191, ewill3@xxxxxxxxxxxxx
Useless C++ Programming Idea #12995733:
bool f(bool g, bool h) { if(g) h = true; else h = false; return h;}

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