Editing Mathematical coincidence (section)

==== Gravitational acceleration ====
{{See also|Seconds pendulum}}
While not constant but varying depending on [[latitude]] and [[altitude]], the numerical value of the [[gravitational acceleration|acceleration caused by Earth's gravity]] on the surface lies between 9.74 and 9.87 [[metre per second squared|m/s<sup>2</sup>]], which is quite close to 10. This means that as a result of [[Newton's laws of motion|Newton's second law]], the weight of a kilogram of mass on Earth's surface corresponds roughly to 10 [[Newton (unit)|newtons]] of force exerted on an object.<ref>{{cite book |title=Cracking the AP Physics B & C Exam, 2004–2005 Edition |page=25 |url=https://books.google.com/books?id=XcX_TvhNjK0C&q=approximation&pg=PA25 |isbn=978-0-375-76387-8 |year=2003 |publisher=Princeton Review Publishing}}</ref>

This is related to the aforementioned coincidence that the square of pi is close to 10. One of the early definitions of the metre was the length of a pendulum whose half swing had a period equal to one second. Since the period of the full swing of a pendulum is approximated by the equation below, algebra shows that if this definition was maintained, gravitational acceleration measured in metres per second per second would be exactly equal to ''π''<sup>2</sup>.<ref>{{cite journal|url=https://www.wired.com/2013/03/what-does-pi-have-to-do-with-gravity/|title=What Does Pi Have To Do With Gravity?|journal=[[Wired (website)|Wired]]|date=March 8, 2013|access-date=October 15, 2015}}</ref>
: <math>T \approx 2\pi \sqrt\frac{L}{g}</math>

The upper limit of gravity on Earth's surface (9.87 m/s<sup>2</sup>) is equal to π<sup>2</sup> m/s<sup>2</sup> to four significant figures. It is approximately 0.6% greater than [[standard gravity]] (9.80665 m/s<sup>2</sup>).