For a particle projected in a transverse direction from a height `h` above earth's surface, find te minimum initial velocity so that it just grazes the surface of earth such that path of this particle would be an ellipse with centre of earth as the farther focus, point of projection as the apogee and a diametrically opposite point on earth's surface as perigee.

Explanation

Suppose velocity of projection at point `A` is `v_(A)` and at point `B` the velocity of the particle is `v_(B)`

The applying Newton's second law at points `A` and `B`, we get
`(mv_(A)^(2))/(rho_(A))=(GM_(e)m)/((R+h)^(2)` and `(mv^(2))/(rho_(B)) =(GM_(e)m)/(R^(2))`
Where `rho_(A)` and `rho_(B)` are radii of curvature of the orbit at points `A` and `B` of the ellipe but `rho_(A)=rho_(B)=rho` (say)
Now applying conservation of energy at points `A` and `B`
`(-GM_(e)m)/(R+h)+1/2mv_(A)^(2)=-(GM_(e)m)/R+1/2mv_(B)^(2)`
`impliesGM_(e)m(1/R-1/((R+h)))=1/2(mv_(B)^(2)-mv_(A)^(2))`
`=(1/2rhoGM_(e)m(1/(R^(2))-1/((R+h)^(2))))`
`implies rho=(2R(R+h))/(2R+h)=(2Rr)/(R+r)`
Thus we get
`V_(A)^(2)=(rhoGM_(e))/((R+h)^(2))=2GM_(3)R/(r(r+R))`
where `r=` distance of point of projection from earth's centre `=R+h`