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 Tue, 26 Jun 2007 00:20:56 -0700
<1182842456.180331.283850@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
On Jun 26, 1:06 am, The Ghost In The Machine
<e...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics.relativity, g...@xxxxxxxxxxx
<g...@xxxxxxxxxxx>
wrote
on Mon, 25 Jun 2007 20:14:03 -0700
<1182827643.282464.65...@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


One thing I don't understand exactly, Since E= F * r for both kinetic
and potential:

I can see I'm being a little sloppy. Potential is

E_p = -G*M_sun*M_earth/(1 AU).

Kinetic for a circular orbit is

E_k = 1/2 M_earth * v_earth^2;

this happens to equal 1/2 * G * M_sun * M_earth / (1 AU),
because

v_earth = sqrt(G * M_sun / (1 AU))

for a circular orbit.

Total energy of course is the algebraic sum of the two.


Where Fk = Fc= mv^2/r (Fc = Centrifugal Force) and Fg= GMm/r^2

thus using E= F*r therefore Ek= (mv^2/r) * r thus Ek =mv^2 and not
1/2mv^2??


No, Ek = 1/2 mv^2. Not sure of the F * r since I'm not sure what r is.
Of course E = F * d, though that's on a straightline path.

E = 1/2 I * omega^2, where I is the moment of inertia which
is equal to M_earth * r^2 (since we're modeling Earth here
as a point mass). Note that omega = v/r so
E = 1/2 * M_earth * v^2 as it should be.

For me, it's probably simplest to derive F by taking the second
derivative of the equation of motion:

P(t) = r cos (omega*t), r sin (omega*t)

yielding

F_c = -r * omega^2 * M_Earth = M_Earth * v^2 / r

for centripetal force for any circular path (an orbit
qualifies but so does whirling a sling, driving around a
curve, or riding on various carnival rides).



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).


Therefore the Earth would now be closer to the Sun and it's new
average distance (since it's an Ellipse) from the Sun is 4/7 AU (4/7
of the Earth original distance from the Sun, before the Asteroid
collision slowed it down).

No, the Earth will be in a highly elliptical orbit; on average, yes, it
will be closer to the Sun; the semimajor axis is 4/7 AU.

Of course, there are other considerations, such as the mass
and velocity of the impacting asteroid. Most likely the
velocity of the asteroid will be about
42 km/s = 30 km/s * sqrt(2) (the working assumption is that
the asteroid is from outside the solar system), which means
that relative to Earth one gets 72 km/s -- an issue for
impact energy considerations.

Since we're required to eliminate half of the Earth's
momentum (which is about 1.793 * 10^29 kg-m/s), that
requires an asteroid of about 2.135 * 10^24 kg -- about
1/3th the mass of the Earth, and more than three times
the mass of Mars.

"My, what a big asteroid you have there."

"The better to pulverize the Earth with." :-)



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, ewi...@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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#191, ewill3@xxxxxxxxxxxxx
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- allegedly said by Bill Gates, 1981, but somebody had to make this up!

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