# Acceleration due to gravity: what's the derivation? [closed]

Modelling the Earth as a symmetric, spherical body (and by using the law of gravitation), we come up with the equation $$w = F_g = \frac{Gm_Em}{R_E^2}$$ How do we arrive to the equation to get the acceleration due to gravity at the earth's surface? $$g = \frac{GM_E}{R_E^2}$$ Where:

• $G = \text{gravitational constant} = 6.67\times10^{-11}\ \mathrm{N}$
• $M_E = \text{mass of Earth} = 5.98\times10^{24}\ \mathrm{kg}$
• $R_E = \text{Earth's radius} = 6380\ \mathrm{km}$

One thing I know of is the fact that we use Newton's second law, but how?

• Oh, come on! What wrong did I do? Sep 9 '15 at 10:24
• My guess: this is such an easy question that it seems you can't possibly have made any effort to figure it out yourself before posting here. That might account for the downvote. Sep 9 '15 at 11:43
• I quickly answered it the moment I realised. I posted the answer as I thought it may help someone in the future. Sadly, I am no expert in Physics. Sep 9 '15 at 11:48
• Yeah, I figured that's what happened. Don't worry about it as a one-time thing. I'd just say, don't be so hasty to post a question in the future. If you have something you think would be good to post, take a little while (minutes, hours, days, depending on the question) to think about it first and try a few things yourself before posting here. Sep 9 '15 at 14:15
• @DavidZ :) Cool! I thought that it'd help someone with a similar problem, so I didn't remove the entire question! I'll keep that in mind! Sep 9 '15 at 16:16

So, it turns out that I am out of coffee.

Anyway, since we already know that:

$w = m \times g$ (Newton's Second Law)

We know that $w = \frac{Gm_Em}{R_E^2}$

Dividing the entire equation by $m$, we get:

$\frac{w}{m} = \frac{mg}{m}$

Which is equal to:

$\frac{Gm_Em}{R_E^2m} = g$ (Substituting the values)

And finally:

$\frac{Gm_E}{R_E^2} = g$

• @weirdpanda, you can use "double dollar signs" (i.e. $$...$$) to get your equations on a separate line.
– Danu
Sep 9 '15 at 10:26