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If we calculate the escape velocity of the not gaseous planets in solar system (mercury, venus, earth, mars...) is it possible to show that the escape velocity measured in Km/h is

$E. V. \sim 2 \pi \frac{r}{1 \mathrm{h}} $

where r is the radius of the planet measured in km and $1\mathrm{h}$ is one hour.

e,g. the E.V of the earth is 40.320 km/h $\sim \frac{\pi 12 756,27 \mathrm{km}}{1 \mathrm{h}}$ where 12 756,27 km is the diameter of the earth and $1 \mathrm{h}$ is still one hour.

Is this funny recurrence depending from how we have choose the definition of meter and second?

Edit My question is a little bit more subtle of what can look at first sight. Namely the earlier definition of the meter was 1/(4*10^6) part of earth circumference and the second was 1/86400 part of the solar day. So why not 1/(4.1*10^6) part of the earth or 1/86000 part of the day? Then is it conceivable that the definition of seconds or meters were chosen in order the get the e.v. on the earth equal to velocity of it circumference traveled in one hour only for sake of physical elegance?

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Can you give more info about your equation e.g. what is $h$ and does the 1 on the bottom of the fraction mean anything? The escape velocity for a planet of mass M and radius r is $\sqrt{2GM/r}$, which doesn't look anything like the expression you give. –  John Rennie Jul 13 '12 at 12:11
    
i am sorry, i guessed that 1h is one hour in all the world. This question is not on the calculation of escape velocity, but in the strange feature of it so i quit the basic formula $\sqrt{2G M/r}$. –  Emanuele Luzio Jul 13 '12 at 12:23
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2 Answers 2

The escape velocity is given by:

$$ v_e = \sqrt{\frac{2GM}{r}} $$

The mass of the Earth, $M$, is given by:

$$ M = \frac{4}{3} \pi r^3 \rho $$

where $\rho$ is the density, and if we substitute this into the expression for the escape velocity we get:

$$ v_e = r \sqrt{\frac{8}{3} \pi G \rho} $$

A quick Google gives G = $6.67^{-11}m^3kg^{-1}s^{-2}$ and the average density of the Earth is $5520kgm^{-3}$. If we substitute these values for $G$ and $\rho$ we get:

$$ v_e = 1.76^{-3}r $$

This speed is in SI units i.e. meters per second. Now look at the approximate expression you give (rearranged slightly) and assume that then "1h" means 3600 seconds, then you get:

$$ v_e \approx \frac{2\pi}{3600} r \approx 1.75^{-3}r $$

So your expression gives approximately the same result as the exact expression. There's no great significance to the result, it's just a numerical co-incidence and it only works because the densities of the inner planets are all roughly the same. Presumably it fails for the outer planets because they have a significantly lower density.

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If the planet's density is $d$, $M=\tfrac{4}{3}\pi d r^3$, and:

$$v_{\text{e}} = \sqrt{\frac{2G\tfrac{4}{3}\pi d r^3}{r}} = \sqrt{\tfrac{8}{3}G\pi d}\cdot r$$

So you suggest that:

$$\sqrt{\tfrac{8}{3}G\pi d} \sim 2\pi\;\text{h}^{-1}$$

or:

$$G d \sim \tfrac{3}{2}\pi\;\text{h}^{-2};\quad d \sim \frac{3G\pi}{2\;\mathrm{h}^2};\quad d\sim 5448\;\mathrm{kg/m^3}$$

And in fact, for Mercury $d=5427\;\mathrm{kg/m^3}$, for Venus $d=5204\;\mathrm{kg/m^3}$, for Earth $d=5515\;\mathrm{kg/m^3}$, for Mars $d=3934\;\mathrm{kg/m^3}$. So the approximation seems quite good (95%), except for Mars.

This holds for whatever definition of km you use, as long as you use it for both $r$ and $v_{\text{e}}$, but only for $1\;\mathrm{h}\simeq3600\;\mathrm{s}$.

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In other words we have (1) a coincidence in which you pick a time unit that makes this work out and (2) the observations that the rocky planets are made out of rock. –  dmckee Jul 13 '12 at 17:56
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