- #1

Geofleur

Science Advisor

Gold Member

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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?