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Question Number 11987 by tawa last updated on 08/Apr/17

The radius of the moon is (1/4), and its mass is  (1/(81))  that of the earth. If the  acceleration due to gravity on the surface of the earth is 9.8m/s^2 . What  is its value on the moon′s surface.

$$\mathrm{The}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{moon}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{and}\:\mathrm{its}\:\mathrm{mass}\:\mathrm{is}\:\:\frac{\mathrm{1}}{\mathrm{81}}\:\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\:\mathrm{on}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}\:\mathrm{is}\:\mathrm{9}.\mathrm{8m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{its}\:\mathrm{value}\:\mathrm{on}\:\mathrm{the}\:\mathrm{moon}'\mathrm{s}\:\mathrm{surface}. \\ $$

Answered by ajfour last updated on 09/Apr/17

g_e =((GM_e )/R_e ^2 )      ; g_m =((GM_m )/R_m ^2 )    g_m = g_e  ((M_m /M_e ))((R_e /R_m ))^2     g_m  = (9.8m/s^2 )((1/(81)))(16)           = 1.9358 m/s^2  .

$${g}_{{e}} =\frac{{GM}_{{e}} }{{R}_{{e}} ^{\mathrm{2}} }\:\:\:\:\:\:;\:{g}_{{m}} =\frac{{GM}_{{m}} }{{R}_{{m}} ^{\mathrm{2}} }\:\: \\ $$$${g}_{{m}} =\:{g}_{{e}} \:\left(\frac{{M}_{{m}} }{{M}_{{e}} }\right)\left(\frac{{R}_{{e}} }{{R}_{{m}} }\right)^{\mathrm{2}} \\ $$$$\:\:{g}_{{m}} \:=\:\left(\mathrm{9}.\mathrm{8}{m}/{s}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{\mathrm{81}}\right)\left(\mathrm{16}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{1}.\mathrm{9358}\:{m}/{s}^{\mathrm{2}} \:. \\ $$

Commented by tawa last updated on 09/Apr/17

God bless you sir.

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

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