LesRhorer said:
Such as? The point of a gravity assist is to enable a spacecraft to achieve very high velocities without using much fuel. Attempting to do this in a 2 body system results in no increase in velocity at all.
This is simply false. From the wikipedia article on Elastic collisions:
https://en.wikipedia.org/wiki/Elastic_collision#Two-dimensional_collision_with_two_moving_objects
$$v_x=\frac{\cos (\phi ) (u (m-M) \cos
(\theta -\phi )+2 M U \cos (\phi ))}{m+M}-u \sin
(\phi ) \sin (\theta -\phi )$$ $$v_y=\frac{\sin
(\phi ) (u (m-M) \cos (\theta -\phi )+2 M U \cos
(\phi ))}{m+M}+u \cos (\phi ) \sin (\theta -\phi
)$$ $$V_x=\frac{\cos (\phi ) (U (M-m) \cos (\phi
)+2 m u \cos (\theta -\phi ))}{m+M}+U \sin ^2(\phi
)$$ $$V_y=\frac{\sin (\phi ) (U (M-m) \cos (\phi
)+2 m u \cos (\theta -\phi ))}{m+M}-U \sin (\phi )
\cos (\phi )$$
Where I have cleaned up the notation a bit using capital letters for the large mass and lower case letters for the small mass, and I have chosen the ##x## axis to be the direction of the initial velocity (##U##) of the large mass.
You can confirm that both momentum and energy are conserved for this system.
Note that this is a 2-body collision. There is no third body. From this expression you can obtain the increase in velocity, which you falsely claim is zero, by directly evaluating $$\Delta v=\sqrt{v_x^2+v_y^2} - u =
\sqrt{\frac{m^2 u^2-2 m M u^2 \cos (2 \theta -2 \phi
)+2 M u U (m-M) \cos (\theta -2 \phi )+2 M u U \cos
(\theta ) (m-M)+M^2 u^2+2 M^2 U^2 \cos (2 \phi )+2
M^2 U^2}{(m+M)^2}}-u$$
We can further simplify this by doing a Taylor series expansion for small ##m## to obtain $$\Delta v \approx 2U-2(U-u)\frac{m}{M} $$
So, contrary to your statement, there can be a large ##\Delta v## for a 2 body elastic collision provided that ##U## is large.
LesRhorer said:
The shape of an orbit which tends to zero at infinity is a hyperbola.
This is false, it is a parabola.
LesRhorer said:
If the craft does not slow all the way to zero in the limit as its distance tends toward infinity, then its orbit is not properly speaking a hyperbola
This is also false. Hyperbolic trajectories are precisely the ones that do not slow all the way to zero in the limit as distance tends toward infinity.
See:
https://en.wikipedia.org/wiki/Hyperbolic_trajectory "a body traveling along this trajectory will coast towards infinity, settling to a final excess velocity relative to the central body. ...
gravitational slingshots, can be described within the planet's
sphere of influence using hyperbolic trajectories."
LesRhorer said:
Just because there are an infinite number of frames in which the relative speed of the inbound path is lower than the relative speed of the outbound path in some FoR external to the barycenter of the system in no way means the large body has increased the KE of V.
In all of those frames the large body has indeed increased the KE of the small one. You cannot increase the speed of a mass without increasing its KE.
LesRhorer said:
there is no such things as an absolute inertial frame of reference, and that all linear motion is relative to any arbitrarily chosen FoR.
While this is true, it contradicts your previous statements. That you don't recognize that is unfortunate, but clearly not something you are interested in correcting.
At this point, this thread is done. Your initial question has been answered, we will not be providing you any material to support your argument as your argument is wrong on many, many points.
