Using $g = \dfrac{GM}{R^2}$ with $G = 6.67 imes 10^{-11}$ N m$^2$/kg…

Question

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

Accepted Answer

A. 9.77 m/s$^2$

1. Use $g = \dfrac{GM}{R^2}$.
2. Numerator $= 6.67	imes10^{-11}	imes6	imes10^{24} = 4.00	imes10^{14}$.
3. Denominator $R^2 = (6.4	imes10^6)^2 = 4.096	imes10^{13}$.
4. $g = \dfrac{4.00	imes10^{14}}{4.096	imes10^{13}} \approx 9.77$ 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_

Answer

98.0 m/s$^2$

Answer

1.56 m/s$^2$

Answer

0.98 m/s$^2$

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: