Menu Close

If-a-body-of-2kg-mass-is-at-a-distance-of-7200km-from-the-centre-of-the-earth-What-would-the-acceleration-due-to-gravity-be-at-this-point-in-the-Earths-field-a-9-6m-s-2-b-10m-s-2-c-11-3m-s




Question Number 12419 by tawa last updated on 21/Apr/17
If a body of 2kg mass is at a distance of 7200km from the centre of the  earth . What would the acceleration due to gravity be at this point in  the Earths field ?  (a) 9.6m/s^2  (b) 10m/s^2  (c) 11.3m/s^2  (d) 12.7m/s^2  (e) 15.6m/s^2
$$\mathrm{If}\:\mathrm{a}\:\mathrm{body}\:\mathrm{of}\:\mathrm{2kg}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{7200km}\:\mathrm{from}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{earth}\:.\:\mathrm{What}\:\mathrm{would}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\:\mathrm{be}\:\mathrm{at}\:\mathrm{this}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{Earths}\:\mathrm{field}\:? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{9}.\mathrm{6m}/\mathrm{s}^{\mathrm{2}} \:\left(\mathrm{b}\right)\:\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \:\left(\mathrm{c}\right)\:\mathrm{11}.\mathrm{3m}/\mathrm{s}^{\mathrm{2}} \:\left(\mathrm{d}\right)\:\mathrm{12}.\mathrm{7m}/\mathrm{s}^{\mathrm{2}} \:\left(\mathrm{e}\right)\:\mathrm{15}.\mathrm{6m}/\mathrm{s}^{\mathrm{2}} \\ $$
Answered by ajfour last updated on 22/Apr/17
((GM)/((9R/8)^2 )) = ((64g)/(81)) = ((64×9.81)/(81)) =7.75 m/s^2  .
$$\frac{{GM}}{\left(\mathrm{9}{R}/\mathrm{8}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{64}{g}}{\mathrm{81}}\:=\:\frac{\mathrm{64}×\mathrm{9}.\mathrm{81}}{\mathrm{81}}\:=\mathrm{7}.\mathrm{75}\:{m}/{s}^{\mathrm{2}} \:. \\ $$
Commented by tawa last updated on 22/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mrW1 last updated on 22/Apr/17
g=((Gm)/r^2 )  ⇒(g_1 /g)=((r/r_1 ))^2   ⇒g_1 =g((r/r_1 ))^2 =9.81×(((6371)/(7200)))^2 =7.68 m/s^2     ⇒none of the given answers is correct!
$${g}=\frac{{Gm}}{{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{g}_{\mathrm{1}} }{{g}}=\left(\frac{{r}}{{r}_{\mathrm{1}} }\right)^{\mathrm{2}} \\ $$$$\Rightarrow{g}_{\mathrm{1}} ={g}\left(\frac{{r}}{{r}_{\mathrm{1}} }\right)^{\mathrm{2}} =\mathrm{9}.\mathrm{81}×\left(\frac{\mathrm{6371}}{\mathrm{7200}}\right)^{\mathrm{2}} =\mathrm{7}.\mathrm{68}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{none}\:{of}\:{the}\:{given}\:{answers}\:{is}\:{correct}! \\ $$
Commented by tawa last updated on 22/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *