A meteor of mass m having a speed v at infini | vedclass

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(A) According to the law of conservation of energy,the total energy of the meteor at infinity is equal to the total energy at the earth's surface.At i

Vedclass · Accepted Answer

$\sqrt{v^2 + v_e^2}$

Vedclass · Answer

(A) According to the law of conservation of energy,the total energy of the meteor at infinity is equal to the total energy at the earth's surface.At infinity,the potential energy is $0$ and the kinetic energy is $\frac{1}{2}mv^2$.At the earth's surface,the potential energy is $-\frac{GMm}{R}$ and the kinetic energy is $\frac{1}{2}mv_f^2$,where $v_f$ is the final speed.So,$\frac{1}{2}mv^2 + 0 = \frac{1}{2}mv_f^2 - \frac{GMm}{R}$.We know that the escape speed $v_e = \sqrt{\frac{2GM}{R}}$,which implies $v_e^2 = \frac{2GM}{R}$,or $\frac{GM}{R} = \frac{v_e^2}{2}$.Substituting this into the energy equation: $\frac{1}{2}mv^2 = \frac{1}{2}mv_f^2 - m(\frac{v_e^2}{2})$.Dividing by $\frac{m}{2}$ gives $v^2 = v_f^2 - v_e^2$.Therefore,$v_f^2 = v^2 + v_e^2$,which means $v_f = \sqrt{v^2 + v_e^2}$.

$A$ meteor of mass $m$ having a speed $v$ at infinity reaches the surface of the earth with a speed of (where $v_e$ is the escape speed from the earth's surface).

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