Using $g = \dfrac{GM}{R^2}$ with $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, $M = 6 \times 10^{24}$ kg and $R = 6.4 \times 10^6$ m, the value of $g$ at the Earth's surface is about:

1. Use $g = \dfrac{GM}{R^2}$. 2. Numerator $= 6.67\times10^{-11}\times6\times10^{24} = 4.00\times10^{14}$. 3. Denominator $R^2 = (6.4\times10^6)^2 = 4.096\times10^{13}$. 4. $g = \dfrac{4.00\times10^{14}}{4.096\times10^{13}} \approx 9.77$ m/s$^2$. 5. The other options misplace the decimal/power and are wrong. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.17_