Question Number 106928 by aurpeyz last updated on 07/Aug/20
$${Imagine}\:{a}\:{planet}\:{having}\:{a}\:{mass}\:{twice}\:{that} \\ $$$${of}\:{the}\:{earth}\:{and}\:{a}\:{radius}\:{equal}\:{to}\:\mathrm{1}.\mathrm{414} \\ $$$${times}\:{that}\:{of}\:{the}\:{earth}.\:{Determine}\:{the} \\ $$$${acceleration}\:{due}\:{to}\:{gravity}\:{at}\:{its}\:{surface}. \\ $$
Answered by JDamian last updated on 07/Aug/20
$${g}_{{planet}} \:=\:{G}\:\frac{{M}_{{planet}} }{{R}_{{planet}} ^{\mathrm{2}} }\:=\:{G}\frac{\mathrm{2}{M}_{{Earth}} }{\left(\sqrt{\mathrm{2}}\:\centerdot{R}_{{Earth}} \right)^{\mathrm{2}} }\:= \\ $$$$=\:{G}\:\frac{\mathrm{2}{M}_{{Earth}} }{\mathrm{2}{R}_{{Earth}} ^{\mathrm{2}} }\:=\:{G}\:\frac{{M}_{{Earth}} }{{R}_{{Earth}} ^{\mathrm{2}} }\:=\:{g}_{{Earth}} \\ $$$$ \\ $$