# Deriving gravitational acceleration from the formula for the force of gravity [closed]

I'm trying to derive this $$g = 9.8 m/s^2$$ from this $$F = G\frac{m_E\cdot m_2}{(r)^2}$$

So I did:

$$g = G \frac{(5.972\cdot 10^{24}kg)}{(6.371km)^2} \Leftrightarrow \\ g = 6.67\cdot10^{-11}Nm^2/kg^2\cdot 0.147\cdot10^{24}kg/km^2 \Leftrightarrow \\ g = 0.98\cdot10^{13}\frac{Nm^2}{kg\cdot km^2} \\ \Leftrightarrow \\ g = 0.98\cdot 10^{13}\frac{m^3/s^2}{10^3m^2} \Leftrightarrow \\g = 0.98\cdot10^{10}m/s^2$$

What went wrong?

• Please be aware that check-my-work questions are off-topic on this site. Mar 21, 2020 at 22:16

These two formulas are in general valid:

1. $$F_g=G\frac{Mm}{r^2}$$
2. $$F_g=mg$$ So we can combine them into: $$F_g=G\frac{Mm}{r^2}\iff mg=G\frac{Mm}{r^2}\iff g=G\frac{M}{r^2}$$ And then we just calculate: $$g=G\frac{M}{r^2}=6.67\cdot 10^{-11}\frac{m^3}{kg\cdot s^2}\frac{5.972\cdot 10^{24}kg}{6371000^2m^2}=9.8136\frac{m}{s^2}\approx 9.81 \frac{m}{s^2}$$

where

1. $$G$$ is the gravitational constant, $$G = 6.67\cdot 10^{-11}\frac{m^3}{kg\cdot s^2}$$
2. $$M$$ is the mass of Earth, $$M = 5.972\cdot 10^{24}kg$$
3. $$r$$ is the radius of the Earth, $$r=6371km=6371000m$$
4. $$g$$ is final result - gravitational acceleration, $$g=9.81\frac{m}{s^2}$$

The actual value is $$g = 9.81\frac{m}{s^2}$$, so we got right.

• I have a question: Is gravity's acceleration a sort of centripetal acceleration? Mar 21, 2020 at 21:36
• @SilenceOnTheWire In the free fall not. You are just falling downwards. But you can take two stars (binary star). They are orbiting around each other with a help of gravity. They are accelerating in both magnitude and direction, so also centripetal. So, yes - gravity can make centripetal acceleration. Mar 21, 2020 at 21:40