Gravitation Potential Energy — Question 2

Question 2

A subescape radial launch has a finite turning radius

A particle is launched radially outward from radius $R$ with speed $.8v_{esc}$, where $v_{esc}=\sqrt{2GM/R}$. Neglect atmosphere and source recoil.

1. Find its maximum radius and maximum altitude.

2. Explain why setting final potential to zero would be incorrect for this launch.

Question 2 - Solution

A subescape radial launch has a finite turning radius

(a) Derivation and calculation

Let $v_0=.8\sqrt{2GM/R}$. The conserved specific energy is $$\epsilon=\frac{v_0^2}{2}-\frac{GM}{R}=(.64-1)\frac{GM}{R}=-.36\frac{GM}{R}.$$ At a radial turning point $v=0$: $$-\frac{GM}{r_{\max}}=-.36\frac{GM}{R},\qquad \boxed{r_{\max}=\frac{25R}{9}},\quad\boxed{h_{\max}=\frac{16R}{9}}.$$

(b) Result and interpretation

At infinity the potential is zero, which would require $$\frac{v_\infty^2}{2}=\epsilon=-.36\frac{GM}{R}<0.$$ This is impossible for a real speed. Instead, at the finite turning point, $$K_{\max r}=0,\qquad U_{\max r}=E=-.36\frac{GMm}{R}.$$ Negative potential energy must be retained.

Physical check

For a general subescape factor $k<1$, $r_{\max}=R/(1-k^2)$; it diverges as $k\to1^-$.