Given potential is "V = -\\frac{k}{r}"
Let L be angular momentum then total effective potential will be
"V_0 = V + \\frac{L^2}{2Mr^2} = -\\frac{k}{r} + \\frac{L^2}{2Mr^2}"
Differentiate with respect to r and differentiation will be zero at "r = r_0"
Hence, "\\frac{dV}{dr} = \\frac{k}{r^2} - \\frac{L^2}{Mr^3}"
"\\frac{dV}{dr}|_{r=r_0} = \\frac{k}{r_0^2} - \\frac{L^2}{Mr_0^3} = 0"
Solving it we get,
"L = \\sqrt{kMr_0}"
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