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(A) The total energy of the rocket at the instant of launch must be less than or equal to zero for it to remain bound to the solar system.The orbital

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) The total energy of the rocket at the instant of launch must be less than or equal to zero for it to remain bound to the solar system.The orbital speed of the asteroid is given by $v_0 = \sqrt{\frac{GM}{r_0}}$,which implies $v_0^2 = \frac{GM}{r_0}$.The total energy $E$ of the rocket relative to the sun is the sum of its gravitational potential energy and its kinetic energy:$E = -\frac{GMm}{r_0} + \frac{1}{2}mv^2$Given that the speed of the rocket relative to the sun is $v = \alpha v_0$,we substitute this into the energy equation:$E = -\frac{GMm}{r_0} + \frac{1}{2}m(\alpha v_0)^2$For the rocket to remain bound,$E \leq 0$:$-\frac{GMm}{r_0} + \frac{1}{2}m \alpha^2 v_0^2 \leq 0$Substituting $v_0^2 = \frac{GM}{r_0}$:$-\frac{GMm}{r_0} + \frac{1}{2}m \alpha^2 \left(\frac{GM}{r_0}\right) \leq 0$Dividing by $\frac{GMm}{r_0}$:$-1 + \frac{1}{2} \alpha^2 \leq 0$$\frac{1}{2} \alpha^2 \leq 1$$\alpha^2 \leq 2$$\alpha \leq \sqrt{2}$Wait,re-evaluating the launch condition: If the rocket is launched *from* the asteroid,its initial velocity relative to the sun is the vector sum of the asteroid's orbital velocity and the launch velocity. However,the problem states $v = \alpha v_0$ is the speed relative to the sun. Thus,the condition for being bound is $E \leq 0$,leading to $\alpha \leq \sqrt{2}$. The highest value is $\sqrt{2}$.

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