Determining The Mass Of The Earth

[Bashyboy]

Homework Statement

(a) Given that the period of the Moon's orbit about the Earth is 27.32 days and the nearly constant distance between the center of the Earth and the center of the Moon is $3.84 \cdot 10^8 m$, use $T^2 = (\frac{4\pi^2}{GM_E})a^3$ to calculate the mass of the Earth.

(b) Why is the value you calculate a bit too large?

Homework Equations

The Attempt at a Solution

For (a), I solved for $M_E$ and got a value of $6.01 \cdot 10^{24}~kg$, of which I am certain is the correct answer, though not 100%. However, my main concern is with part (b); I am honestly don't know how to answer this question. Could someone possibly help me?

 

[mfb]

Hint: Your formula does not take the mass of moon into account.

 

[Bashyboy]

So, this formula treats the moon as if it were a point particle? What if the formula did take into account the mass of the moon; how would the formula look in that case?

 

[mfb]

The formula treats the moon as a massless point particle.
Perfect spheres and points are the same for gravity, and Earth and moon are very close to perfect spheres, so the size does not matter.

What if the formula did take into account the mass of the moon; how would the formula look in that case?

It should be easy to find that formula.

 

[Bashyboy]

Would it just be the equation to Newton's law of Universal gravitation?

 

[mfb]

You can derive the formula from Newton's laws if you like.
It looks very similar to the one you have, just with a modification to account for the mass of the second body.

 

[Chestermiller]

Hint: Your formula does not take the mass of moon into account.

Sure it does. The mass of the moon cancels out from both sides of the force balance equation.

 

[mfb]

Sure it does. The mass of the moon cancels out from both sides of the force balance equation.

Only with the assumption that the Earth does not move, or with a correction for the distance of moon. In both cases, you have to take the mass of moon into account, or neglect it, which is exactly the source of error which is interesting here.

 

[D H]

Sure it does. The mass of the moon cancels out from both sides of the force balance equation.

Both sides of what equation?

Imagine if the Moon was as massive as a neutron star. If this were the case, calculating the orbital period of the Earth-Moon system using just the mass of the Earth will obviously yield the wrong answer. Now it would be much better to just use the mass of the Moon.

In general, the masses of both objects come into play in determining the orbital period. You can ignore the mass of the smaller object only if it is much, much less massive than the larger object. Since the ratio of the Moon's mass to that of the Earth is 0.0123, ignoring the mass of the Moon will yield an answer that is observably too [strike]small[/strike] long.

 

[Chestermiller]

Very interesting. Thanks for the insight.