Question Number 190814 by Spillover last updated on 12/Apr/23
Answered by Beginner last updated on 12/Apr/23
$${Attractive}\:{force}={mv}^{\mathrm{2}} /{r}={mrw}^{\mathrm{2}} \\ $$$${Similarly}\:{f}={q}_{\mathrm{1}} {q}_{\mathrm{2}} /\mathrm{4}\Pi\epsilon_{{o}} {r}^{\mathrm{2}} ={mrw}^{\mathrm{2}} \\ $$$${Recall}\:{that}\:{T}=\mathrm{2}\Pi/{w} \\ $$$${on}\:{substitution}\:{T}=\sqrt{\mathrm{16}\prod^{\mathrm{3}} {mr}^{\mathrm{3}} \epsilon_{{o}} /{q}_{\mathrm{1}} {q}_{\mathrm{2}} } \\ $$