How do I find way of comparing the density of the Earth and the Moon?

[DANIELWR1998]

Homework Statement

The radius of Earth is 4 times the radius of the moon. Estimate the acceleration due to gravity on the surface of the moon.

Do I use some orbital period information? I can't think of anything else that may be relevant.

Relevant Equations

M = density x sphere volume
g = GM/r²

Surface acceleration is proportional to density and radius of planet (as 2 powers of R cancel with the volume)
g(moon)/g(earth) = density(moon)*radius(moon)/density (earth)*radius(earth) = (1/4)*density(moon)/density(earth)

 

[Steve4Physics]

Hi @DANIELWR1998. Welcome to PF.

Note that question asks only for an 'Estimate'. This implies that an accurate answer is not required. If that's the case, you may be expected to assume that the densities of the earth and moon are equal. (If you do that, clearly state the assumption as part of your answer.)

 

[DANIELWR1998]

Hi @DANIELWR1998. Welcome to PF.

Note that question asks only for an 'Estimate'. This implies that an accurate answer is not required. If that's the case, you may be expected to assume that the densities of the earth and moon are equal. (If you do that, clearly state the assumption as part of your answer.)

I did initially think that, but I know that the acceleration on the moon is less than 1/6 of Earth's. This assumption would give 1/4 as the answer which I'm not sure is accurate enough, I don't actually know whether it just wanted a very rough approximation.

 

[Steve4Physics]

I did initially think that, but I know that the acceleration on the moon is less than 1/6 of Earth's. This assumption would give 1/4 as the answer which I'm not sure is accurate enough, I don't actually know whether it just wanted a very rough approximation.

Consider giving 2 answers to cover yourself. The first uses the (stated) assumption of equal densities. For the second, look-up the densities of the moon and earth (give the references) and use them.

Your answer should include a proper derivation (using symbols/algebra) of any formula you end-up using. A phrase such as '2 powers of R cancel with the volume' doesn't really do the job!

Edit. Note that using 4 for the ratio of the radii isn't particularly accurate.

 

[PeroK]

I did initially think that, but I know that the acceleration on the moon is less than 1/6 of Earth's. This assumption would give 1/4 as the answer which I'm not sure is accurate enough, I don't actually know whether it just wanted a very rough approximation.

Why not solve the whole problem? All the information you need is on the Internet. Namely, the radius and density of the Earth and Moon. And, you can check the answer as the surface gravity can be found online as well.

This should give you a quantitative corroboration of Newton's law of gravity.

I can't see the point in the age of the Internet of doing anything else.

 

[haruspex]

This should give you a quantitative corroboration of Newton's law of gravity.

Not really. That law would surely have been assumed in obtaining the density estimate found on the internet.

I can't see the point in the age of the Internet of doing anything else.

By that reasoning, just look up the moon's surface gravity.

Though it doesn’t quite say it, I read the question as "The radius of Earth is 4 times the radius of the moon. On that basis, estimate the acceleration due to gravity on the surface of the moon in terms of g". So I would just assume the densities are similar.

 

[DANIELWR1998]

Not really. That law would surely have been assumed in obtaining the density estimate found on the internet.

By that reasoning, just look up the moon's surface gravity.

Though it doesn’t quite say it, I read the question as "The radius of Earth is 4 times the radius of the moon. On that basis, estimate the acceleration due to gravity on the surface of the moon in terms of g". So I would just assume the densities are similar.

I think this is the right reasoning based off of the answers so far. It came at the end of a University of Cambridge Natural Sciences Interview Preparation paper so I thought it would require a bit more than this as it's just one step (and what I assumed to begin with).

 

[DANIELWR1998]

Why not solve the whole problem? All the information you need is on the Internet. Namely, the radius and density of the Earth and Moon. And, you can check the answer as the surface gravity can be found online as well.

This should give you a quantitative corroboration of Newton's law of gravity.

I can't see the point in the age of the Internet of doing anything else.

It's more just an estimation problem but I thought the densities could be related using some other phenomenon such as an orbital period or if I knew the distance between the moon and the Earth. I wasn't quite sure what I was allowed to assume and what would be taken as common knowledge.

 

[Frabjous]

I thought it was made of green cheese so roughly 1-1.1 gm/cc

Physics Today 67, 11, 32 (2014)

 

[jbriggs444]

One might attempt a calculation of the orbital period of the moon based on the distance to the moon, surface gravity of the earth and the earth's radius under the assumption that the moon is negligibly massive. Presumably this will give a slightly incorrect answer -- the moon is not negligibly massive.

Now repeat the calculation, computing the orbital period of the moon about the earth-moon barycenter, varying the mass of the moon (and the associated location of the barycenter) until the correct answer is obtained.

Congratulations. You have weighed the moon.

We have enough significant figures for earth size and surface gravity, lunar distance and lunar orbital period that this should be a feasible calculation. [I am not certain about how lunar distance was first calculated. Modern figures are based on laser reflection against a corner reflector placed for the purpose. By the time you can place a corner reflector, you can directly measure lunar surface gravity and eliminate the need for the above convoluted weighing approach. If it were me, I'd probably have used triangulatuion to measure the distance].