Earth's mass is $6 \times 10^{24}$ kg and its distance from the Sun is $1.5 \times 10^{11}$ m. If the gravitational force between them is $3.5 \times 10^{22}$ N and $G = 6.7 \times 10^{-11}$ N m$^2$ kg$^{-2}$, the mass of the Sun is about:

1. From $F = \dfrac{G M_{sun} M_{earth}}{r^2}$, solve $M_{sun} = \dfrac{F r^2}{G M_{earth}}$. 2. $r^2 = (1.5\times10^{11})^2 = 2.25\times10^{22}$. 3. Numerator $F r^2 = 3.5\times10^{22}\times2.25\times10^{22} = 7.875\times10^{44}$. 4. Denominator $G M_{earth} = 6.7\times10^{-11}\times6\times10^{24} = 4.02\times10^{14}$. 5. $M_{sun} = \dfrac{7.875\times10^{44}}{4.02\times10^{14}} \approx 1.96\times10^{30}$ kg. 6. $6\times10^{24}$ kg is Earth's own mass, so it is the trap. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.24_