A-body-moves-in-a-circular-orbit-of-radius-4R-round-the-earth-Express-the-acceleration-of-the-free-fall-due-to-gravity-of-the-body-in-terms-of-g-R-radius-if-the-earth-g-acceleration-due-to-gravi

Question Number 29144 by tawa tawa last updated on 04/Feb/18

$$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{orbit}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{4R}\:\mathrm{round}\:\mathrm{the}\:\mathrm{earth}.\:\:\mathrm{Express}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{free}\:\mathrm{fall}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{g} \\ $$$$\mathrm{R}\:=\:\mathrm{radius}\:\mathrm{if}\:\mathrm{the}\:\mathrm{earth} \\ $$$$\mathrm{g}\:=\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity} \\ $$

Answered by ajfour last updated on 04/Feb/18

((GMm)/((4R)^2 ))= ma ⇒ a =((GM)/(16R^2 )) = (g/(16)) .

$$\frac{{GMm}}{\left(\mathrm{4}{R}\right)^{\mathrm{2}} }=\:{ma} \\ $$$$\Rightarrow\:\:\:{a}\:=\frac{{GM}}{\mathrm{16}{R}^{\mathrm{2}} }\:=\:\frac{{g}}{\mathrm{16}}\:. \\ $$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by tawa tawa last updated on 04/Feb/18

Please how did it becomes (g/(16)) from ((GM)/(16R^2 ))

$$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{it}\:\mathrm{becomes}\:\:\:\frac{\mathrm{g}}{\mathrm{16}}\:\:\mathrm{from}\:\:\:\:\frac{\mathrm{GM}}{\mathrm{16R}^{\mathrm{2}} } \\ $$

Commented by tawa tawa last updated on 04/Feb/18