[tex]|v| =\sqrt{ G \cdot M / r}[/tex], where
Explanation:
Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.
Equation 1 (see below) relates net force the object experiences, [tex]\Sigma F[/tex] to its orbit velocity [tex]v[/tex] and its mass [tex] m [/tex] required for it to stay in orbit :
[tex]\Sigma F = m \cdot v^{2} / r[/tex] (equation 1)
The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force [tex]\Sigma F[/tex] acting on the soccer ball shall equal to its weight, [tex]W = m \cdot g[/tex] where [tex]g[/tex] the gravitational acceleration constant. Thus
[tex]\Sigma F = W = m \cdot g[/tex] (equation 2)
Substitute equation 2 to the left hand side of equation 1 and solve for [tex]v[/tex]; note how the mass of the soccer ball, [tex]m[/tex], cancels out:
[tex]m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)[/tex] (equation 3)
Equation 4 gives the value of gravitational acceleration, [tex]g[/tex], a point of negligible mass experiences at a distance [tex]r[/tex] from a planet of mass [tex]M[/tex] (assuming no other stellar object were present)
[tex]g = G \cdot M/ r^{2}[/tex] (equation 4)
where the universal gravitation constant [tex]G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}[/tex]
Thus
[tex][/tex][tex]\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}[/tex]
