\[ \definecolor{mycolor}{rgb}{0.3,0.7,0.7} \newcommand{\units}[1]{ { \color{mycolor} #1 } } \newcommand{\unitm}[]{ \text{m} } \newcommand{\unitnm}[]{ \text{nm} } \newcommand{\unitcm}[]{ \text{cm} } \newcommand{\unitkg}[]{ \text{kg} } \newcommand{\unitsec}[]{ \text{s} } \newcommand{\unitsr}[]{ \text{sr} } \newcommand{\unitcd}[]{ \text{cd} } \newcommand{\unitlm}[]{ \text{lm} } \newcommand{\unitlx}[]{ \text{lx} } \newcommand{\unitC}[]{ \text{C} } \newcommand{\unitW}[]{ \text{W} } \newcommand{\unitkW}[]{ \text{kW} } \newcommand{\unitJ}[]{ \text{J} } \newcommand{\unitK}[]{ \text{K} } \newcommand{\unitN}[]{ \text{N} } \newcommand{\unitspercent}[]{ \units{ \% } } \newcommand{\unitsaccel}[]{ \units{ \unitm \cdot \unitsec^{-2} } } \newcommand{\unitsE}[]{ \units{ \unitW / \unitm^2 } } \newcommand{\unitsL}[]{ \units{ \unitW/(\unitsr\cdot\unitm^2) } } \newcommand{\unitsnit}[]{ \units{ \text{nit} } } \newcommand{\unitsm}[]{ \units{ \unitm } } \newcommand{\unitsnm}[]{ \units{ \unitnm } } \newcommand{\unitsAU}[]{ \units{ \text{AU} } } \newcommand{\unitsA}[]{ \units{ \text{A} } } \newcommand{\unitsmA}[]{ \units{ \text{mA} } } \newcommand{\unitsW}[]{ \units{ \text{W} } } \newcommand{\unitsmW}[]{ \units{ \text{mW} } } \newcommand{\unitsHz}[]{ \units{ \text{Hz} } } \newcommand{\unitsTHz}[]{ \units{ \text{THz} } } \newcommand{\unitsJ}[]{ \units{ \text{J} } } \newcommand{\unitsN}[]{ \units{ \text{N} } } \newcommand{\unitsOmega}[]{ \units{ \text{Ω} } } \newcommand{\unitsV}[]{ \units{ \text{V} } } \newcommand{\unitsdeg}[]{ \units{ \deg } } \newcommand{\unitsDegC}[]{ \units{ \deg\text{C} } } \]
\[ \newcommand{\Gwithunits}[]{ 6.67430 \times 10^{-11}~\units{\unitm^3\cdot\unitkg^{-1}\cdot\unitsec^{-2}} } \]

Get Gravity Right Already

TL;DR

People are being sloppy with (1) the value, (2) the direction, and (3) the units of gravitational attraction at Earth's surface, and I'm mad about that. For the record, the value you should be using is Standard Gravity, which is defined to be:

\[ \vec{g}_0 \defeq 9.80665~\unitsaccel\text{ downward} \]

In a sane coordinate frame, where one direction is vertical, gravity is a negative scalar:

\[ g_0 = -9.80665~\unitsaccel \]

Unless you're in the very unusual case of having measured local gravity with specialized instruments, you should always use this for calculations!

Gravitational Force

In Newtonian mechanics, gravity is an attracting force that pulls, as far as we know, every two material particles together. The magnitude of this force is given by:

\[ ||\vec{F}|| = \frac{ G \cdot m_1 \cdot m_2 }{ ||\vec{r}||^2 } \]

\(\vec{F}\) is the (vector-valued) force, \(m_1\) and \(m_2\) are the masses of the two objects, \(\vec{r}\) is the displacement vector between them, and \(G \approx \Gwithunits\) is the gravitational constant. Taking \(\vec{p}_1\) and \(\vec{p}_2\) to be the positions of the masses \(m_1\) and \(m_2\), the vector-valued version is:

\begin{align*} \vec{F}_{1\text{ on }2} &= \frac{G \cdot m_1 \cdot m_2}{||\vec{p}_2 - \vec{p}_1||^2} \cdot \frac{\vec{p}_1 - \vec{p}_2}{|| \vec{p}_1 - \vec{p}_2 ||}\\[1em] \vec{F}_{2\text{ on }1} &= \frac{G \cdot m_1 \cdot m_2}{||\vec{p}_1 - \vec{p}_2||^2} \cdot \frac{\vec{p}_2 - \vec{p}_1}{|| \vec{p}_2 - \vec{p}_1 ||} \end{align*}

You can use this to work out the forces pretty easily.

\[ \newcommand{\massearth}[]{ 5.9722 \times 10^{24}~\units{\unitkg} } \newcommand{\massyou}{ 62.0~\units{\unitkg} } \newcommand{\origin}[]{ \vecinline{0,0,0}~\unitsm } \newcommand{\radpos}[]{ \vecinline{6.3781370 \times 10^6,0,0}~\unitsm } \]

For example, assuming the Earth at the origin (\(\vecinline{0,0,0}~\unitsm\)), you (average human mass, \(\massyou\)), standing on the equator (radius of \(\radpos\)), and an inertial reference frame, you can use the above to work out that:

\begin{align*} \vec{F}_{\text{you on the Earth}} &\approx \vecinline{607,0,0}~\unitsN\\[1em] \vec{F}_{\text{the Earth on you}} &\approx \vecinline{-607,0,0}~\unitsN\\[1em] \end{align*}
\begin{align*} \vec{F}_{\text{you on the Earth}} &= \frac{ (\Gwithunits) (\massyou) (\massearth) } { || \origin - \radpos ||^2 } \cdot \frac{ \radpos - \origin }{|| \radpos - \origin ||}\\[0.5em] &\approx \frac{ 2.47 \times 10^{16}~\units{\unitN\cdot\text{m}^2} }{ 4.07 \times 10^{13}~\units{\unitm^2} } \cdot \vecinline{1,0,0}\\[0.5em] &\approx \vecinline{607,0,0}~\unitsN \\[1em] \vec{F}_{\text{the Earth on you}} &= \frac{ (\Gwithunits) (\massyou) (\massearth) } { || \radpos - \origin ||^2 } \cdot \frac{ \origin - \radpos }{|| \origin - \radpos ||}\\[0.5em] &\approx \frac{ 2.47 \times 10^{16}~\units{\unitN\cdot\text{m}^2} }{ 4.07 \times 10^{13}~\units{\unitm^2} } \cdot \vecinline{-1,0,0}\\[0.5em] &\approx \vecinline{-607,0,0}~\unitsN \end{align*}

Gravitational Acceleration and Standard Gravity

If you plug these forces into Newton's Second Law (\(\vec{F}=m \cdot \vec{a}\)) and solve for acceleration (\(\vec{a}=\vec{F}/m\)), you get:

\begin{align*} \vec{a}_{\text{the Earth}} &= \frac{ \vecinline{607,0,0}~\unitsN }{ \massearth } \approx \vecinline{1.23 \times 10^{-22},0,0}~\unitsaccel\\[1em] \vec{a}_{\text{you}} &= \frac{ \vecinline{-607,0,0}~\unitsN }{ \massyou } \approx \vecinline{-9.80,0,0}~\unitsaccel \end{align*}

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 (in most cases, we just neglect it entirely). 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 (\(9.80~\unitsaccel\) downward) is close to the value for Standard Gravity, which is defined to be (see Resolution 2 03 of the 3rd General Conference on Weights and Measures (CGPM 1901)):

\[ \vec{g}_0 \defeq 9.80665~\unitsaccel\text{ downward} \]

Standard Gravity is the the constant to use when you're doing calculations. Not \(9.8\), not \(9.81\), but \(9.80665~\unitsaccel\), directed downward.

Vector-Valued

I want to stress that "directed downward" bit, too. \(9.80665~\unitsaccel\) is useless without a direction!

Say you're in a 2D coordinate system at Earth's surface, looking at a falling object. What is standard gravity? It's \(\vecinline{0,-9.80665}~\unitsaccel\). 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 'sources'/'peaks' are positive and the 'sinks'/'troughs' are negative. These are referring to some energy landscape the function describes. For gravity, every object is a sink. Every object produces a negative, a gravitational well. Every object attracts. The negatives should be there.

You can, of course, use Galilean invariance 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 unintuitive?

Return to the list of blog posts.