# Homework Help: Find the Period

1. Sep 15, 2007

Of an asteroid whose Mass is 2.0*10^-2 times that of earth's and whose distance from the sun is twice the earth's distance from the sun. Find the period in years.

I am supposed to use the concept of gravitational F=centripital force$$=m\frac{v^2}{r}$$ and the fact that $$v=\frac{2\pi r}{T}$$

so this is my attempt:

$$F_g=\frac{GMm}{r^2}=m(\frac{v^2}{r})$$

implies $$\frac{GMm}{r^2}=\frac{m4\pi^2r}{T^2}$$

implies$$T=\sqrt \frac{4\pi^2r^3}{GM}$$

This M though is the mass of the sun correct?

Casey

2. Sep 15, 2007

Maybe just the acting of posting it helps, cause I got this one too. 2.8 years.

Thanks,
Casey

3. Sep 15, 2007

### G01

Yes, M is the mass of the sun and m is the asteroid in this case.

4. Sep 15, 2007

### G01

5. Sep 16, 2007

### dynamicsolo

Properly speaking, you will probably learn in your course about Newton's generalization of Kepler's Third Law, in which case M is actually the total mass of the "system". But since the asteroid's mass is negligibly small compared to the Sun's, you get entirely satisfactory precision using the Sun's mass alone. (You'd see a slight discrepancy if we used Jupiter as the second mass; for binary stars, you *must* use the sum of the stars' masses.)

You can avoid the annoyance of needing to know G and M (or even pi!) or even worrying about units by solving a comparison ratio. Both the Earth and this asteroid orbit the Sun, so we have

T^2 for asteroid 4(pi^2)(r for asteroid ^3)/GM
______________ = _________________________

T^2 for Earth 4(pi^2)(r for Earth ^3)/GM

(this doesn't show as neatly as I'd like and I don't have a TeX manual)

or just

(T for ast. / T for Earth)^2 = (r for ast. / r for Earth)^3 .

The asteroid orbits twice as far from the Sun as the Earth does, so

(T for ast. / T for Earth)^2 = 2^3 = 8 ;

thus the asteroid's orbital period is sqrt(8) times the Earth's (which is one year) or

The comparison ratio approach is very tidy when the problem provides the right sorts of information. The only extent to which you even deal with units is in making sure that quantities to be compared in ratios are in the same units.

6. Sep 16, 2007