~/imallett (Ian Mallett)

Gravity is not "$9.8 \frac{m}{s^2}$"!
by Ian Mallett

Gravity is an attracting force that pulls, as far as we know, every two material particles together. The magnitude of this force in Newtonian mechanics is:$||\vec{F}|| = \frac{G m_1 m_2}{||\vec{r}||^2}$If you want to find the force itself, you can tweak the equation a little bit to make it vector valued ($\vec{p}_1$ and $\vec{p}_2$ are the positions of the masses):$\vec{F}_{1~on~2} = \frac{G m_1 m_2}{||\vec{p}_2 - \vec{p}_1||^2} \cdot \frac{\vec{p}_1 - \vec{p}_2}{|| \vec{p}_1 - \vec{p}_2 ||}\\ \vec{F}_{2~on~1} = \frac{G m_1 m_2}{||\vec{p}_1 - \vec{p}_2||^2} \cdot \frac{\vec{p}_2 - \vec{p}_1}{|| \vec{p}_2 - \vec{p}_1 ||}$

So, the forces exerted by you and the Earth on each other are (let's pretend the Earth is at the origin ($<0,0,0>m$) and that you are standing on its equator ($<6.3781370 \cdot 10^6,0,0>m$), and that you mass $75 ~kg$):$\vec{F}_{you~on~the~Earth} = \frac{ (6.67384 \cdot 10^{-11} ~N~m^2~kg^{-2}) (75 ~kg) (5.97219 \cdot 10^{24} ~kg) } { || <0,0,0>m - <6.3781370 \cdot 10^6,0,0>m ||^2 } \cdot \frac{<6.3781370 \cdot 10^6,0,0>m - <0,0,0>m}{|| <6.3781370 \cdot 10^6,0,0>m - <0,0,0>m ||}\\ \approx \frac{2.989308 \cdot 10^{16} ~N~m^2}{4.0680632 \cdot 10^{13} ~m^2} \cdot <1,0,0> ~\approx~ <7.34823408911 \cdot 10^2,0,0>N \\ \vec{F}_{the~Earth~on~you} = \frac{ (6.67384 \cdot 10^{-11} ~N~m^2~kg^{-2}) (75 ~kg) (5.97219 \cdot 10^{24} ~kg) } { || <6.3781370 \cdot 10^6,0,0>m - <0,0,0>m ||^2 } \cdot \frac{<0,0,0>m - <6.3781370 \cdot 10^6,0,0>m}{|| <0,0,0>m - <6.3781370 \cdot 10^6,0,0>m ||}\\ \approx \frac{2.989308 \cdot 10^{16} ~N~m^2}{4.0680632 \cdot 10^{13} ~m^2} \cdot <-1,0,0> ~\approx~ <-7.34823408911 \cdot 10^2,0,0>N$

If you plug these forces into Newton's Second Law ($\vec{F}=m \cdot \vec{a}$) and solve for acceleration $\vec{a}$, you get:$\vec{a}_{the~Earth} = \frac{<7.34823408911 \cdot 10^2,0,0>N}{5.97219 \cdot 10^{24} ~kg} ~\approx~ <1.230408625 \cdot 10^{-22},0,0> \frac{m}{s^2}\\ \vec{a}_{you} = \frac{<-7.34823408911 \cdot 10^2,0,0>N}{75 ~kg} ~\approx~ <-9.79764545,0,0> \frac{m}{s^2}$The step of converting for significant figures is omitted.

If you're standing on the Earth, then neither of you is moving. If you're falling, then it should be pretty clear from the above that the acceleration of the Earth is essentially zero. However, the acceleration on you is clearly not negligible (that's why you're falling, not floating). It turns out that the acceleration that we got ($9.79764545 \frac{m}{s^2}$ downward) is close to the value for Standard Gravity, which is defined to be $9.80665 \frac{m}{s^2}$ downward (our answer is a bit smaller since in the example we were standing on the equator, which due to the Earth's rotation has the largest radius, so it's the part where gravity is weakest).

In Physics, we'd like to keep things as simple as possible, so for most problems we do near the Earth's surface, we neglect the acceleration of the Earth and say that the acceleration of other objects is the standard gravity constant.

Say you're in the obvious 2D coordinate system at Earth's surface, looking at a falling object. What is standard gravity? It's $<0,-9.80665>\frac{m}{s^2}$. That's a vector quantity. And you'll notice something. The gravity is negative. That negative comes from the vector derivation above, and it makes sense. Gravity is an attracting force. If you look at any function like this (the Coulomb's Law electrostatic interaction comes to mind) you'll see reference to "source"s/"peak"s and "sink"s/"trough"s, with the former being positive and the latter being negative. These are referring to some energy landscape the function describes. And for gravity, every object is a sink. Every object produces a negative. Every object attracts.

You can, of course, use relativity to redefine the $\vec{y}$-axis to point into the ground. This makes the $\vec{y}$-component of gravity "positive", but it leads to uncomfortable results—like climbing upwards causes your "height" to get lower. Mathematics doesn't care—but why make it hard?

Ultimately, it all comes down to preference—but through teaching, I have found that picking the obvious basis and sticking with it leads to less confusion and fewer errors. Therefore I recommend:

• Instead of $9.8 \frac{m}{s^2}$ or $9.81 \frac{m}{s^2}$, use $9.80665 \frac{m}{s^2}$. This is the accepted standard in industry and research.
• Use vector-valued gravity (e.g. "gravity is $9.80665 \frac{m}{s^2}$ downward", "gravity is $<0,-9.80665>\frac{m}{s^2}$", "gravity has magnitude $9.80665 \frac{m}{s^2}$")
• I acknowledge that this is clumsy, but it helps. You should use vector-values when teaching.
• Pick a basis that has gravity pulling negative. This is preferred because it tells you something about energy, and it parallels the natural basis on Earth.
• If it's really obvious enough (that is, you're in a well-behaved reference system) and you simply must use a scalar value, at least use $-9.80665 \frac{m}{s^2}$, since this makes the most sense in a 2D reference system.