- #1

xpell

- 138

- 16

*g*is significantly lower than on the surface: let's say 5,000 km for instance, where

*g*would be 3.08 m/s

^{2}according to the equation g

_{height}= g

_{surface}· (Radius

_{Earth}/ (Radius

_{Earth}+ Height))

^{2}and will obviously increase during the fall. I have found this for time-dependent non-constant acceleration, but nothing for distance-dependent accelerations. Please, can you help with a couple of motion equations to calculate distance, time and velocity during the fall in this scenario? Thank you in advance!

PS. My knowledge of Math and Physics is close to negligible and I'm not a student, just a 50 year old curious person, but I'm still able to solve an equation if I can find it.

PS2. Please ignore drag and other atmospheric effects for simplicity. I assume that the final velocity in the atmosphere would be the terminal velocity, but not while still in outer space.