For a particle projected in a transverse direction from a height `h` above earth's surface, find the 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 projection at point `A` is `v_(A)` & at point `B` the velocity of the particle is `v_(B)` then applying Newton's `2^(nd)` law at point `A&B` we get Where `r_(A) & r_(B)` are radius of curvature of the orbit at points `A&B` of the ellipse but `r_(A) = r_(B) = r(say)`
Now applying conservation of energy at point `A &B`
`(-GM_(e)m)/(R+h) + (1)/(2) mv_(A)^(2) = (-GM_(e)m)/(R) + (1)/(2)mv_(B)^(2)`
`rArrGM_(e)m((1)/(R)-(1)/((R+h)))=(1)/(2)(mv_(B)^(2)-mv_(A)^(2))=((1)/(2)rhoGM_(e)m((1)/(R^(2))-(1)/((R+h)^(2))))`
`orr=(2R(R+h))/(2R+h)=(2Rr)/(R+r):.V_(A)^(2)=(rGM_(e))/((R+h)^(2))=2GM_(e)(R)/(r(r+R))`
where r=distance of point of projection from earth's centre `=R +h`