Question
Calculate mass of the Earth from given data, Acceleration due to gravity $g = 9.81 \ m/s^2,$ Radius of the Earth $R_E = 6.37 \times 10^6 m, G = 6.67 \times 10^{-11} N m^2/kg^2$
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Answer
Given: $g =9.81 \ m / s ^2, R_E=6.37 \times 10^6 m$
$G =6.67 \times 10^{-11} N m ^2 / \ kg ^2$
To find: Mass of the Earth $\left( M _{ E }\right)$
Formula: $g =\frac{ GM _{ E }}{ R _{ E }^2}$
Calculation: From formula,
$M_E=\frac{ gR _E^2}{ G }= \frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}}$
$= \frac{9.81 \times 6.37 \times 6.37}{6.67} \times 10^{23}$
$= \operatorname{antilog}\{\log (9.81)+\log (6.37)
+\log (6.37)-\log (6.67)\} \times 10^{23}$
$=\text { antilog }\{0.9912+0.8041+0.8041-0.8241) \times 10^{23}$
$=\text { antilog }\{1.7753\} \times 10^{23}$
$=59.61 \times 10^{23}$
$=5.961 \times 10^{24} \ kg$
Mass of the Earth is $5.961 \times 10^{24} \ kg$.
$G =6.67 \times 10^{-11} N m ^2 / \ kg ^2$
To find: Mass of the Earth $\left( M _{ E }\right)$
Formula: $g =\frac{ GM _{ E }}{ R _{ E }^2}$
Calculation: From formula,
$M_E=\frac{ gR _E^2}{ G }= \frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}}$
$= \frac{9.81 \times 6.37 \times 6.37}{6.67} \times 10^{23}$
$= \operatorname{antilog}\{\log (9.81)+\log (6.37)
+\log (6.37)-\log (6.67)\} \times 10^{23}$
$=\text { antilog }\{0.9912+0.8041+0.8041-0.8241) \times 10^{23}$
$=\text { antilog }\{1.7753\} \times 10^{23}$
$=59.61 \times 10^{23}$
$=5.961 \times 10^{24} \ kg$
Mass of the Earth is $5.961 \times 10^{24} \ kg$.
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