Scale of the Earth and Moon

The objective of this lab activity was to find the mass and radius of both the Earth and the moon and the distance from one's center to the other.  We were given a rock, a scale, a measuring cup half-full of water, and the following information:

• Boston and Los Angeles are 3000 miles apart from each other.
• Boston is 3 hours ahead of (or 21 hours behind) Los Angeles.
• There are about 6 kilometers in every 10 miles.  
• The Kepler Equation

With the knowledge of the distance between Boston and LA and the number of hours in a day, the first step was to set up this proportion:

$\frac{5000}{C\oplus = 2\pi R\oplus ^2} = \frac{3}{24}$

We rearranged the equation to solve for the radius of the Earth, which we found to be 6.7 x 10^8 cm.  We determined that any error would be due to rounding values during our calculations because a calculator was not listed among our tools.

The next step was to find the density of the rock by finding its mass using the scale and its volume based on how much water it displaced in the measuring cup.  Assuming that the density of the rock (2 grams per cubic cm) was comparable to the density of the Earth, we found the Earth's volume using its radius and multiplied it by the rock's density to find the mass.  We found Earth's mass to be 2.8 x 10^27 grams.  Error in this step would be due to the false assumption that the density of a rock found near the surface of the Earth is equal to the Earth's density.  We also continued to round in our calculations.  

We used Kepler's equation describing the relationship between the orbital period of an object, the mass of the object it orbits, and the distance between their centers to find the distance from the Earth to the moon.  Using a period of 28 days, converted to seconds, we found the distance to be 4 x 10^10 cm.  Error was once again attributed to rounding in our calculations. 

To find the radius of the moon, we used the hint that the thumb, when held at arm's length, takes up about 1 degree in the field of view and that the moon looks to be about half the size of a thumb.  We set up the following proportions:
 $\frac{.5}{360}=\frac{2 R_{moon}}{C_{orbit} = 2\pi a_{moon}}$
When we solved for R, we found that the radius of the moon is 1.7 x 10^8 cm.  We attributed any error in our answer to rounding in our calculation.

We found the mass of the moon by first assuming that the density of the moon is equal to that of the Earth, 2 grams per cubic cm.  Because we had already found the radius of the moon, we were able to use the following equation to find the volume of the moon: V = 4*pi*R^3 / 3
Rearranging the d = mV equation to solve for mass, we found that the mass of the moon is 6.4 x 10^25 grams.  Any error in our answer was attributed to the assumption that the moon's density was equal to that of the Earth and the rounding we did in our calculations.  

The first three problems were solved in collaboration with Dennis Lee, Delfina, and Vincent.