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I have the formula for a rocket's escape velocity from earth, $V$ being velocity, $v$ being initial velocity, and $r$ being the distance between the rocket and the center of the earth.

$$V = \sqrt{\frac{192000}{r}+v^2-48}$$

I am trying to find the value of $v$ for which an infinite limit for $r$ is obtained as $V$ approaches zero, this value of $v$ being the escape velocity for earth.

I have solved for $v$ (with $V$ being $0$), as $v = \sqrt{48-\frac{192000}{r}}$, but do not know how to continue solving the problem. I thought setting it up as the limit of the square root of $48-\frac{192000}{r}$ as $r$ approaches infinity (to give $v$), but that doesn't seem right.

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1 Answer 1

Let $r\to\infty$. Note that $\dfrac{192000}{r}\,$ approaches $0$. Thus since $$V=\sqrt{\frac{192000}{r}+v^2-48},$$ $V$ approaches $\sqrt{v^2-48}$. If we want $V$ to approach $0$, we want $v^2-48=0$. (Presumably we are measuring velocity in miles per second.)

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thanks. I just realized, "v" is actually squared, so I should then just take the square root of 48 to get v, correct? –  theialux Oct 2 '12 at 5:43

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