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## Main Question or Discussion Point

I have been reading D.E. Littlewood's book "The Skeleton Key of Mathematics", and near the beginning he says that if a projectile weighing one (long) ton were given a velocity of 44 miles a second, this would be "sufficient to raise it to a height of 1000 miles above the earth's surface."

Naturally, I wanted to reproduce this number for myself, so I started with conservation of energy:

$\frac{1}{2}M_P V^2 - \frac{GM_P M_E}{R_E} = -\frac{GM_P M_E}{R_E+h}$,

where $M_E$ is the mass of the Earth, $M_P$ that of the projectile, $h$ is the height of the projectile above the surface, $G$ is the gravitation constant, and $V$ is the projectile's speed. Solving for the height gives

$h = \frac{R_E}{\frac{2GM_E}{V^2 R_E}-1}$.

First, note that the mass of the projectile does not appear in this formula. It appears, then, irrelevant that the projectile weighs a ton. Second, as the denominator approaches zero, $h \rightarrow \infty$; setting the denominator to zero and solving for $V = V_{esc}$ yields

$V_{esc} = \sqrt{2G M_E / R} \approx 7 \frac{mi}{s}$.

Thus, 44 mi/s seems like way more than you would need to lift the projectile 1000 miles above the surface. I didn't account for energy lost to air resistance in these calculations, but launching an object through the atmosphere into space like that seems unfeasible - wouldn't it just burn up? Am I missing something or is Littlewood's calculation off?

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Simon Bridge
Homework Helper
Well done - if you google "escape velocity in miles per second" what do you get?

Gold Member
Well, Google didn't give me what I expected, but Wolfram Alpha did - about 7 mi/s. For some reason, I always have a hard time believing that books could be wrong about things like this. I always think I must be missing something. Thanks!

Andy Resnick