Question Number 33824 by Nayon.Sm last updated on 25/Apr/18
$$\mathrm{two}\:\mathrm{sphere}\:\mathrm{with}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{radious}\:\mathrm{and} \\ $$$$\mathrm{1}\:\mathrm{kg}\:\mathrm{mass}\:\mathrm{and}\:\mathrm{distence}\:\mathrm{between} \\ $$$$\mathrm{this}\:\mathrm{two}\:\mathrm{is}\:\mathrm{1}\:\mathrm{m}\:.\mathrm{after}\:\mathrm{what}\:\mathrm{time}\:\mathrm{they} \\ $$$$\mathrm{will}\:\mathrm{touch}\:\mathrm{each}\:\mathrm{other}? \\ $$
Answered by Andrew Foxman last updated on 08/May/19
$${It}\:{gives}\:{us} \\ $$$${r}_{\mathrm{0}} =\mathrm{1}.\mathrm{2}{m} \\ $$$${r}=\mathrm{0}.\mathrm{2}{m} \\ $$$${m}=\mathrm{1}{kg} \\ $$$${t}=? \\ $$$${Do}\:{you}\:{remember}\:{formula}? \\ $$$${g}\left({r}\right)=\frac{{r}}{{t}^{\mathrm{2}} }\Rightarrow{t}=\sqrt{\frac{{r}}{{g}\left({r}\right)}}\:\left(\mathrm{1}\right){because}\:{the}\:{balls}\:{will} \\ $$$${fall}\:{into}\:{each}\:{other}.\:{Also}\:{g}\left({r}\right)={G}\frac{{m}}{{r}^{\mathrm{2}} }\:\left(\mathrm{2}\right). \\ $$$${Introducing}\:\left(\mathrm{1}\right)\:{in}\:\left(\mathrm{2}\right)\:{after}\:{using}\:{some}\:{algebra} \\ $$$${t}=\sqrt{\frac{{r}^{\mathrm{3}} }{{Gm}}}\:\left({r}=\mathrm{0}.\mathrm{2}{m}\right)\:{So}\:{use}\:{this}\:{formula}. \\ $$$${There}\:{is}\:{another}\:{complicated}\:{way}\:{that} \\ $$$${requires}\:{differential}.\:{I}\:{showed}\:{you}\:{a} \\ $$$${simple}\:{way}. \\ $$