Question Number 171359 by Mr.D.N. last updated on 13/Jun/22
$$\mathrm{A}\:\mathrm{heavenly}\:\mathrm{body}\:\mathrm{has}\:\mathrm{mass}\:\mathrm{one}\:\mathrm{third}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}\:\mathrm{and}\:\mathrm{its} \\ $$$$\:\mathrm{radius}\:\mathrm{is}\:\mathrm{half}\:\mathrm{as}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}.\:\mathrm{if}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{weights}\:\mathrm{200N}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{earth}'\mathrm{s}\:\mathrm{surface},\:\mathrm{find}\:\mathrm{its}\:\mathrm{weight}\:\mathrm{on}\:\mathrm{that}\:\mathrm{heavenly}\:\mathrm{body}. \\ $$
Answered by aleks041103 last updated on 13/Jun/22
$${W}=\frac{{GM}_{{pl}} {m}}{{R}^{\mathrm{2}} }\sim\frac{{M}}{{R}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{W}_{{earth}} }{{W}_{{planet}} }=\frac{{M}_{{earth}} }{{M}_{{pl}} }\:\left(\frac{{R}_{{pl}} }{{R}_{{earth}} }\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{W}_{{pl}} =\frac{\mathrm{4}}{\mathrm{3}}{W}_{{earth}} \approx\mathrm{266}.\mathrm{7}{N} \\ $$
Commented by Mr.D.N. last updated on 14/Jun/22
$$\mathrm{great}\:\mathrm{thank}\:\mathrm{you}. \\ $$
Commented by peter frank last updated on 23/Jun/22
$$\mathrm{thank}\:\mathrm{s} \\ $$