The escape speed of an object on the surface | vedclass

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(B) Let $V_e = V$ be the escape speed of the object on the surface of the earth. The formula for escape speed is $V = \sqrt{\frac{2GM}{R}}$.By the law

Vedclass · Accepted Answer

$\sqrt{15}V$

Vedclass · Answer

(B) Let $V_e = V$ be the escape speed of the object on the surface of the earth. The formula for escape speed is $V = \sqrt{\frac{2GM}{R}}$.By the law of conservation of mechanical energy,the total energy at the surface equals the total energy at infinity:$K_i + U_i = K_f + U_f$$\frac{1}{2}m(4V)^2 - \frac{GMm}{R} = \frac{1}{2}mV_0^2 + 0$Since $V^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{V^2}{2}$.Substituting this into the energy equation:$\frac{1}{2}m(16V^2) - m(\frac{V^2}{2}) = \frac{1}{2}mV_0^2$$8mV^2 - 0.5mV^2 = 0.5mV_0^2$$7.5V^2 = 0.5V_0^2$$V_0^2 = 15V^2$$V_0 = \sqrt{15}V$

The escape speed of an object on the surface of the earth is $V$. If the object is thrown out with speed $4V$ from the surface of the earth,what will be the speed of the object far away from the earth?

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