Question 13.85: (a) Determine the kinetic energy per unit mass which a missi...

At the surface of the earth,                         g = 9.81 m/s²

$r_1=R=6370 \ km =6.37 \times 10^6  m$

Centric force at the surface of the earth,

Let position 1 be on the surface of the earth $\left(r_1=R\right)$ = and position 2 be at $r_2=O D$ . Apply the conservation of energy principle.

For the escape condition set $\quad \quad \frac{T_2}{m}=0$

$\frac{T_1}{m}=g R=\left(9.81 \ m / s ^2\right)\left(6.37 \times 10^6  m \right)=62.49 \times 10^6  m ^2 / s ^2$ 

(a) $\quad \quad \quad \quad \quad \quad \quad \quad \quad \frac{T_1}{m}=62.5 \ MJ / kg \blacktriangleleft$

$\begin{aligned}& \frac{1}{2} m \nu_{ esc }^2=g r \\& \nu_{ esc }=\sqrt{2 g R}\end{aligned}$

(b) $\quad \quad \quad\nu_{ esc }=\sqrt{(2)(9.81)\left(6.37 \times 10^6\right)}=11.18 \times 10^3  m / s \quad \nu_{ esc }=11.18 \ km / s \blacktriangleleft$

Note that the escape condition depends only on the speed in position 1 and is independent of the direction of the velocity (firing angle).