I have to show the acceleration due to gravity for the moon is one sixth the same for the earth. The question gives the formulae and the variables to plug in for each, thus:
- Gravitational constant: $G = 6.673 \times 10^{-11}$Nm$^2$kg$^{-2}$
- Acceleration due to gravity: $g = GMr^{-2}$
I'm given a formula to calculate the masses required:
$M = \frac{4}{3} \pi \rho \times r^3$
And given the means density and radius of the earth ( that's $\rho$ and $r$ ) as:
- $\rho = 5514$kg m$^{-3}$
- $r = 6.37 \times 10^6$m
For the moon:
- $\rho = 3322$kg m$^{-3}$
- $r = 1.74 \times 10^6$m
I've calculated the masses, that bit I can handle: I just plug the variables in. Now what? What I write below is guesswork, do feel free to correct me:
Something that's throwing me is it seems to say "Newtons per metre squared times the square root of the mass in kilos"?
Does this mean I must derive the density converting from kg per metre cubed to Newtons per square metre? I don't know.
And then convert mass in kg$^{-3}$ to mass in kg$^{-2}$?
As you can see, I'm confused.
Best Answer
Comparisons always give you a chance to avoid all constants of proportionality.
$g$ is proportional to $ M = k r^3 \rho$ and inversely proportional to $1/r^2,$ so you just need to compare with $ r *\rho $ product only. I.e.,
$$ \dfrac{g_{ Moon}}{g_{Earth}}= \dfrac {r_{Moon} \rho_{Moon}}{r_{Earth} \rho_{Earth} } . $$
It would be $ \approx \frac16. $