What is the time of impact of a dropped item at a very high altitude [duplicate]

I'm trying to automate a landing at a specific spot in Kerbal Space Program and to do that I need to know when the estimated time of impact will be. In short, I'm trying to solve the following question:

Given a planet without an atmosphere with a gravitational parameter $\mu$, radius $r$ and an object with mass $m$, an initial velocity of $v=0$ and a distance $d$ between their centers, what will be the time of impact?

An example: Given a spherical earth without an atmosphere and a brick at an altitude of $100.000$km, how long after release will the brick hit the surface?

I've tried using conservation of energy but that left me with a differential equation I'm unable to solve: http://mathb.in/45230

marked as duplicate by John Rennie, Community♦Oct 24 '15 at 10:18

• This previous question should answer your question, if you replace $\mu$ by $GM$. I'm marking this as a duplicate since the two questions are pretty much the same. – eyqs Oct 24 '15 at 3:59

You can use the equation $$y=y_o + v_{oy} t - \frac{1}{2} gt^2$$ y_o is the original position (100.000 km) and y is the final position (0.000 km), v_o is the initial velocity, which is zero, and g is the acceleration of gravity, which is 9.8 m/s on Earth. Remember to match up the units!

• The problem I faced with this approach is that $g$ is defined in terms of $y_t$ – Thexa4 Oct 24 '15 at 2:28
• @Thexa4 how come? Can you give me the equation – TanMath Oct 24 '15 at 2:33
• $g = \frac \mu {y_t^2}$ – Thexa4 Oct 24 '15 at 2:38
• This is incorrect. $g$ isn't remotely constant over these these distances. You really need to use Universal Gravitation to solve this. 'Big' $G$ is constant but not 'little' $g$. – Gert Oct 24 '15 at 2:40
• @Gert g cannot be considered constant for this? How come? – TanMath Oct 24 '15 at 2:43