Kepler's Law: Error Analysis & Mass of Earth

[swickey]

hello everyone,
I have a couple of questions from my astonomy course that are confusing me.
1. Which do you think would cause the larger error: a 10% error in "p" (period of orbit) or "a" (length of semi-major axis in units)? I would think that the "p" would produce the larger error, because it could alter the position of planet. Is that right?

2. The orbit of Earth's moon has a period of 27.3 days and a radius (semi-major axis) of 2.56 x 10^-3 A.U. (=3.84 x 10^5 km). What is the mass of Earth? What are the units? Show your work?


I'm not sure which formula to use, can anyone give me some pointers?
Thanks a bunch.

 

[Laura1013]

Both your questions involve Kepler's third law of planetary motion.

1) Kepler's third law states that the period squared is proportional to the semi-major axis cubed. If you take 10% of any value and square it, then take 10% of that same value and cube it, which number is bigger? For example, let's assume your arbitrary value is 100 (forget units). Ten percent of 100 is 10. Compare 10-squared to 10-cubed. Which is larger?

2) This is a simple "plug-and-chug" problem. Use the general equation of Kepler's third law (it should be in your textbook), plug in the numbers you are given and plug in the constants to solve for your one unknown value. The variables in the equation are the period p, the semi-major axis a, the mass of the moon m, and the mass of the Earth M. You can look up the values for the gravitational constant G and pi. Solve for M.

 

[swickey]

I see where you going...
P^2 = a ^3
Comparing 10-squared to 10-cubed. Cubed would be greater. Thanks

M=4pi^2a^3 / P^2G

Thanks a lot. It seems so simple now.

 

[Kurdt]

Just to note that Kepler didn't say the period squared and the semi-major axis cubed were equal. He said they were proportional.