Question

Question #5ea25

Answer 1

`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;`