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

(a) Determine the kinetic energy per unit mass which a missile must have after being fired from the surface of the earth if it is to to reach an infinite distance from the earth. (b) What is the initial velocity of the missile (called the escape velocity)? Give your answers in SI units and show that the answer to part b is independent of the firing angle.

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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,

\begin{aligned}F & =m g=\frac{G M m}{R^2} \\G M & =g R^2=(9.81)\left(6.37 \times 10^6\right)^2=398.06 \times 10^{12}  m ^3 / s ^2\end{aligned}

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.

\begin{aligned}& T_1+V_1=T_2+V_2 \\& \frac{1}{2} m ν_1^2-\frac{G M m}{r_1}=\frac{1}{2} m ν_2^2+\frac{G M m}{r_2} \\& T_1=T_2+\frac{G M m}{R}-\frac{G M m}{\infty} \\& \frac{T_1}{m}=\frac{T_2}{m}+\frac{G M}{R}=\frac{T_2}{m}+g R\end{aligned}

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).

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