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(D) Let $M$ be the mass of the Earth,$R$ be the radius of the Earth,and $r$ be the orbital radius of the satellite.The total energy of the satellite a

Vedclass · Accepted Answer

$\sqrt{v_e^2 - 2v_0^2}$

Vedclass · Answer

(D) Let $M$ be the mass of the Earth,$R$ be the radius of the Earth,and $r$ be the orbital radius of the satellite.The total energy of the satellite at distance $r$ is $E_i = -\frac{GMm}{r}$.When it hits the Earth's surface at distance $R$,its potential energy is $U_f = -\frac{GMm}{R}$ and kinetic energy is $K_f = \frac{1}{2}mv^2$.By the law of conservation of energy: $E_i = K_f + U_f$.$-\frac{GMm}{r} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{r}$.$v^2 = \frac{2GM}{R} - \frac{2GM}{r}$.Since $v_e^2 = \frac{2GM}{R}$ and $v_0^2 = \frac{GM}{r}$,we have $\frac{2GM}{r} = 2v_0^2$.Substituting these,$v^2 = v_e^2 - 2v_0^2$.Therefore,$v = \sqrt{v_e^2 - 2v_0^2}$.

$A$ satellite is orbiting the Earth with an orbital velocity $v_0$. If it is suddenly stopped and allowed to fall onto the Earth,what will be its velocity when it hits the Earth's surface? (Let $v_e$ be the escape velocity from the Earth's surface.)

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