Turning point of ball, when throwing it into the atmosphere

Question

I am currently reading an introduction chapter about gravitation. One exercise was about calculating the escape velocity on earth. At one in the book it reached the following formular:

$ r_{max} = {R \over 1 - {{v_0}^2 \over 2Rg}} $.

If $ 1 - {{v_0}^2 \over 2Rg} = 0 $ , then $ v_0 $ of the thrown object has started off with the velocity needed to "escape" the gravitational field of earth.

However I found this imprecise, cause the acceleration g is continuously decreasing as the thrown object distances itself from earth. This also leads to the effect, that the velocity of the object doesn't decrease as much, because the g decreases...

Thus I looked for a way to calculate the escape velocity and the height of the points of no return are reached, (one when it stays in the orbit, and the other one, when escapes earth completely). My problem so far is probably the way I defined g:

I say: $ F = m*g $ (special case of Newton's law) and $ F = G * {m*M \over R^2 } $ At the moment I don't see, how I can bring those two pieces together without being constantly in the need of having knowing BOTH, $ g $ and $ v_0 $, at the same time. At least I don't see how I handle those terms/integrals well. Thus I think my approach is not good.

I would be happy, if someone could give me a hint how to progress at this point. Thx in advance.

Answer 1

The acceleration $g$ is constant and is equal to the acceleration due to gravity at the projectile's launch point. It is not updated as the projectile ascends. So, we just use the definition of $g$: $$g = \frac{F}{m} = G\frac{M}{R^2}.$$ Substituting this into the original equations results in $$r_{max} = \frac{R}{1-\frac{v_0^2}{2R}\frac{R^2}{GM}} == \frac{R}{1-\frac{v_0^2 R}{2GM}}$$ with escape velocity occuring when $$1-\frac{v_0^2 R}{2GM} = 0.$$ Solving this for $v_0$ results in the standard equation for escape velocity $$v_e = \sqrt{\frac{2GM}{R}}$$