# Question 5ea25

Dec 6, 2017

${v}_{e s} = \setminus \sqrt{\frac{2 G {M}_{\setminus \oplus}}{R} _ \left\{\setminus \oplus\right\}} = 11.18 \setminus {\times}^{3} \setminus \quad m . {s}^{- 1} = 11.18 \setminus \quad k m . {s}^{- 1} .$

#### Explanation:

U_i = U_{\infty} = 0;\qquad U_f = -G(M_{\oplus}m)/R_{\oplus};
\DeltaU = U_f - U_i = -G(M_{\oplus}m)/R_{\oplus};

${M}_{\setminus \oplus} , {R}_{\setminus \oplus}$ are the mass and radius of the Earth.
$m$ is the mass of the falling object.

K_i = 0; \qquad K_f = 1/2mv^2;
$\setminus \Delta K = {K}_{f} - {K}_{i} = \frac{1}{2} m {v}^{2}$

Mechanical Energy Conservation:

\Delta E = \Delta K + \Delta U = 0;\qquad \DeltaK = -\Delta U;
$\frac{1}{2} \cancel{m} {v}^{2} = - \left(- G \frac{{M}_{\setminus \oplus} m}{R} _ \left\{\setminus \oplus\right\}\right) = G \frac{{M}_{\setminus \oplus} \cancel{m}}{R} _ \left\{\setminus \oplus\right\}$

Escape Speed:

${v}_{e s} = \setminus \sqrt{\frac{2 G {M}_{\setminus \oplus}}{R} _ \left\{\setminus \oplus\right\}} = 11.18 \setminus {\times}^{3} \setminus \quad m . {s}^{- 1} = 11.18 \setminus \quad k m . {s}^{- 1} .$

G=6.674\times10^{-11}\quad (N.m^2)/(kg^2); \qquad M_{\oplus} = 5.972\times10^{24}\quad kg;
R_{\oplus} = 6371\quad km = 6.371\times10^{6}\quad m;#