Why do s orbital potentials have more nodes as principal quantum number increases?

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applestrudle
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... why are there more nodes/zeros?

If l = 0 then the angular momentum contribution to the effective potential is zero, and there is the coublomb potential only. So shouldn't it always go as ~ -k/r^2 (k = constant) like the n=1 s orbital?

Why is it that for n = 2 is there 1 zero, n= 3 there is 2 zeros, etc?

Thanks
 

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This has nothing to do with the potential. Even for a particle in a box where there is no potential, the excited wavefunctions have nodes. This relates to kinetic energy being the higher the more curved the wavefunction. Increasing curvature necessarily leads to the appearance of nodes.