Question
If the mass of the sun is $2 \times 10^{30} kg$, the distance of the earth from the sun is $1.5 \times 10^{11} m$ and period of revolution of the earth around the sun is one year ( $=365.3$ days), calculate the value of gravitational constant.

Answer

$
\begin{aligned}
& G \text { (gravitational constant) }=6.673 \times 10^{-11} Nm^2 kg^{-2} \\
& M(\text { mass of sun })=2 \times 10^{30} kg \\
& m \text { (mass of earth) }=6 \times 10^{24} kg
\end{aligned}
$
d (average distance between sun and earth) $=1.5 \times 10^8 km=1.5 \times 10^8 \times 1000=1.5 \times 10^{11} m$
$
\begin{aligned}
& F=\frac{G M m}{d^2} \\
& =\frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^2} \\
& =\frac{6.673 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{1.5 \times 1.510^{22}}\left(10^{24+30-11}=10^{43}\right) \\
& =\frac{6.673 \times 2 \times 6 \times 10^{43}}{1.5 \times 1.5 \times 10^{22}}\left(10^{43-22}=10^{21}\right) \\
& =\frac{6.673 \times 2 \times 6 \times 10^{21}}{1.5 \times 1.5}\left(10^{43-22}=10^{21}\right) \\
& =\frac{80.076 \times 10^{21}}{1.5 \times 1.5} \\
& =35.59 \times 10^{21} N=3.57 \times 10^{22} N
\end{aligned}
$
Force of gravity $=3.57 \times 10^{22} N$

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