Editing Escape velocity (section)

=== Height of lower-velocity trajectories ===
Ignoring all factors other than the gravitational force between the body and the object, an object projected vertically at speed ''v'' from the surface of a spherical body with escape velocity ''v''{{sub|e}} and radius ''R'' will attain a maximum height ''h'' satisfying the equation<ref>{{Cite book |last=Bajaj |first=N. K. |url=https://books.google.com/books?id=OLhyCgAAQBAJ |title=Complete Physics: JEE Main |publisher=[[McGraw-Hill Education]] |year=2015 |isbn=978-93-392-2032-7 |page=6.12}} [https://books.google.com/books?id=OLhyCgAAQBAJ&pg=SA6-PA12 Example 21, page 6.12]</ref>
: <math>v = v_\text{e} \sqrt{\frac{h}{R+h}} \ ,</math>
which, solving for ''h'' results in
: <math>h = \frac{x^2}{1-x^2} \ R \ ,</math>
where {{nowrap|1=''x'' = ''v''/''v''{{sub|e}}}} is the ratio of the original speed ''v'' to the escape velocity ''v''{{sub|e}}.

Unlike escape velocity, the direction (vertically up) is important to achieve maximum height.