Taking $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, Earth's mass $6 \times 10^{24}$ kg and radius $6.4 \times 10^6$ m, the gravitational force of the Earth on a 75 kg person is closest to:

1. Use $F = \dfrac{G M m}{R^2}$ with $M = 6\times10^{24}$, $m = 75$, $R = 6.4\times10^6$. 2. Numerator $= 6.67\times10^{-11}\times75\times6\times10^{24} = 3.00\times10^{16}$. 3. Denominator $R^2 = (6.4\times10^6)^2 = 4.096\times10^{13}$. 4. $F = \dfrac{3.00\times10^{16}}{4.096\times10^{13}} \approx 733$ N. 5. This equals $mg$ with $g\approx9.77$; 9.8 N is the weight of just 1 kg, so it is the trap. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.16_