[geocentrism] slingshot
- From: "Philip" <joyphil@xxxxxxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Sat, 12 Feb 2005 16:08:18 +1000
Despite my questions, and my answers, and with reference to Mikes off list
inputs, I am not yet convinced that Neville is wrong. He has not convinced me
I'm wrong, but I need to play with it some more.
Remember my analogy of the ball hitting the moon! Thats where I'm so superior,
lol I am not attached to my knowledge.. I have to keep an open mind... despite
the dangers.
Well when we get captured by the "moons" gravity, it is like an elastic
slingshot, and that elastic never breaks, or at least till it does, it keeps
pulling back, just as Neville said. So thus I went a googlin for what is meant
by escape velocity. I had used the term glibly without knowing... WOW... None
of the experts could agree. Look at this little bit.
I have to solve a problem regarding the minimum velocity required to escape
from the solar system while taking into account the gravitational force of the
earth as well as the sun...........? What I am supposed to do is calculate
the velocity of an object after escaping from earth. Then I am supposed to go
to a sun fixed frame and calculate the minimum velocity for escape from the
sun. The answer that I am striving for is the following:
v^2= v(escape from earth)^2 +v1^2
v1=v(escape from sun) - v(orbit)
I am not sure how these equations above are derived. I can calculate the escape
velocity for the sun and earth, but I am not sure how to relate these two
separate frame.
The answer...
I will give you a derivation of the formula, but I am not completely sure
whether it is correct.
Take v as the required escape velocity, m the mass of the rocket or object
which is going to escape, Me the mass of the earth, Ms the mass of the sun, re
and rs the radius of earth and the distance from the sun to the earth
respectively, v_es escape velocity from sun, v_e escape velocity from earth and
v_o orbital velocity of earth.
As already mentioned, escape velocity is escape velocity, and it should not be
dependent on trajectories and so on.
Thus, consider an rocket escaping from the earth and going to a point in the
earth's orbit, which is far from the influence of the earth's gravitational
field. Consider also velocities only relative to earth. In such a situation the
required velocity to escape the sun is (v_es - v_o).
Aplying energy conservation to this situation in relation to the launch of the
rocket:
½ m v^2 - (G Me m) / re - (G Ms m) / rs = ½ m (v_es - v_o)^2 - (G Ms m) / rs
½ m v^2 - (G Me m) / re = ½ m (v_es - v_o)^2
Since (2 G Me) / re = v_e^2, then
v^2 = v_e^2 + (v_es - v_o)^2
Regards.
Well like I said... maybe nothing ever really escapes. What goes up must come
down , eventually, the elastic never breaks, unless another giant lassoos it.
still loohing for the real escape velocity.. Got 41 thousand sites to check....
Phil.
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