# Gravitational Force - Newton Mechanics

Why do we use gravitational force in earth by relating just the mass of an object with the acceleration produced by the gravitational field: $$F_{g} = m\cdot \vec{g}$$And when we're dealing with planets, we use a relation defined by the masses of two planets, distance squared and gravitational constant: $$F_{g} = G \cdot \frac{M_{1} \cdot M_{2}}{d^{2}}$$

I really don't get why we use just the first relation here on earth, because we're dealing with a interction between two objects... It's because our mass is irrelevant??

Thanks!

The second equation is always correct, and you can derive the first equation from it.

Here on the surface of the Earth, $d$ is the radius of the Earth $r_e$ plus our height $h$. $$F = G \frac{M_e M_2}{(r_e + h)^2}$$ The radius of the Earth (6,371 km) is huge compared to our height above the surface (at least, when we're near the surface), so we can simplify the equation by assuming $r_e \gg h$ and therefore $r_e \approx r_e + h$. $$F = G \frac{M_e M_2}{r_e^2}$$ $G$, $M_e$ and $r_e$ are all constant, so we bundle them all into another constant $g = \frac{GM_e}{r_e^2}$ and voila $$F = gM_2$$

It's because our mass is irrelevant?

No, it's because the equation assumes that our height above Earth's surface is negligible compared to the radius of the Earth (which most of the time, for me at-least, it is).

Example: Suppose I'm a 70 kg man whose just spent the last week hiking up Mount Everest, which is 9 km above sea level. Using the correct equation we get $$F = G \frac{70 M_e}{(6,371,000+9,000)^2} = 685 N$$ Using the approximate equation we get $$F = 70g = 687 N$$ which is about $0.3$% different. Whether this is an acceptable error or not will depend on how precise you need your calculations to be, but for every-day purposes it's probably fine :)

• Really nice explanation, thank you! But you can't always consider our height above the surface irrelevant... Imagine someone in the top of everest for example (close to 9 km) it'll affect, even if it's just a small amount, the g... Commented Aug 31, 2016 at 19:28
• @BrunoReis try plugging in the numbers yourself. in the equation for $g$ if you replace $r_e$ with $r_e + 10$km or $r_e - 10$km, how much does $g$ change? The equation for $g$ is an approximation that only works near the surface of the Earth. Commented Aug 31, 2016 at 19:34
• @BrunoReis Naturally you "can't always consider our height above the surface irrelevant". That is the whole point when Judge uses the word "negligible". It is only irrelevant when considered negligible. And no, it isn't always negligible of course. So, this answer explains an approximation that fits for most cases, because that variable (the height) doesn't do any significant difference - and you are very right, that we of course need to be aware, within what the range it is valid. Commented Aug 31, 2016 at 20:06
• @BrunoReis You're very right :) I'll update my answer to clarify that this approximation is only valid when near the surface of the Earth. Commented Aug 31, 2016 at 20:10

We use the first formula for earth based calulations because $$G \cdot \frac{M_{1}}{d^{2}}=g$$to a good approximation. We used the second formula for planets because we do not have an easy simplification. So to put simply, they are the same formula but one just has a nice simplification.

• The approximation is based on $d$ being almost constant near the surface of the Earth: it is the radius of the planet. Commented Aug 31, 2016 at 18:26

In the equations you have given above, consider M2 as m.

Consider, m stays on earth's surface

M1= earth's mass and d= earth's radius

Find true values and substitute for G,M1 and d in your second equation. than you will see Fg=g (what you calculated) x M2 (m). So both equations tell the same.