Editing Escape velocity (section)

== Deriving escape velocity using calculus ==
Let ''G'' be the [[gravitational constant]] and let ''M'' be the [[Earth mass|mass of the earth]] (or other gravitating body) and ''m'' be the mass of the escaping body or projectile. At a distance ''r'' from the centre of gravitation the body feels an attractive force
: <math>F = G\frac{Mm}{r^2}.</math>
The work needed to move the body over a small distance ''dr'' against this force is therefore given by
: <math>dW = F \, dr = G\frac{Mm}{r^2}\,dr.</math>
The total work needed to move the body from the surface ''r''<sub>0</sub> of the gravitating body to infinity is then<ref>{{cite book |title=A-level Physics |edition=illustrated |first1=Roger |last1=Muncaster |publisher=Nelson Thornes |year=1993 |isbn=978-0-7487-1584-8 |page=103 |url=https://books.google.com/books?id=Knov8XAyf2cC}} [https://books.google.com/books?id=Knov8XAyf2cC&pg=PA103 Extract of page 103]</ref>
: <math>W = \int_{r_0}^\infty G\frac{Mm}{r^2}\,dr  = G\frac{Mm}{r_0} = mgr_0.</math>
In order to do this work to reach infinity, the body's minimal kinetic energy at departure must match this work, so the escape velocity ''v''<sub>0</sub> satisfies
: <math> \frac{1}{2}m {v_0}^2 = G\frac{Mm}{r_0},</math>
which results in
: <math>v_0 = \sqrt\frac{2GM}{r_0} = \sqrt{2gr_0}.</math>