There is a base-quadrilateraled pyramid ABCDS , which base is quadrilateral ABCD. Inscribed sphere is tangent in point P on the ABCD wall. Proof that
<)APB+<)CPD=180'
I don't like geometry and i really don't know how to start.
ok so, I read that mechanical energy(E): E=-MmG/R
R=(a+b)/2
a-semi-major axis
b-semi-minor axis
,of course kinetical Ek=mv^2/2
m-spaceship weight
Ep=-mMG/r
v^2/2-MG/r=-2MG/(a+b)
and
a-sqrt(a^2-b^2)=<r=<a+sqrt(a^2-b^2)
and v^2 for some r is less than 0
M>>m