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;#