Question #5ea25

1 Answer
Dec 6, 2017

v_{es}=\sqrt{(2GM_{\oplus})/R_{\oplus}} = 11.18\times^3\quad m.s^{-1} = 11.18\quad km.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_{\oplus}, R_{\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;
\DeltaK = K_f - K_i = 1/2mv^2

Mechanical Energy Conservation:

\Delta E = \Delta K + \Delta U = 0;\qquad \DeltaK = -\Delta U;
1/2cancel{m}v^2 = -(-G(M_{\oplus}m)/R_{\oplus}) = G(M_{\oplus}cancel{m})/R_{\oplus}

Escape Speed:

v_{es}=\sqrt{(2GM_{\oplus})/R_{\oplus}} = 11.18\times^3\quad m.s^{-1} = 11.18\quad km.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;