# Find mass of the Earth

1. Nov 14, 2005

### NoMeGusta

The moon orbits the earth in an approximately circular path of radius 3.8 X 10^8 m. It takes about 27 days to complete one orbit. What is the mass of the earth as obtained from these data?

I started with

$$\frac {mv^2}{r} = G \frac {Mm}{r^2}$$

I did some simplification all the way to

$$\frac {v^2r}{G} = M$$

From here, the book then re-writes it as $$\frac {\Omega^2r^3}{G} = M$$. How did they do that?

Last edited: Nov 14, 2005
2. Nov 14, 2005

### Chi Meson

the book is using "angular velocity" instead of "tangential velocity."
$$\omega$$ is measured in radians per second. The tangential speed of a point at the outer radius is simply $$v_T = \omega r$$.

A more direct way to get orbital speed is to recognize speed as distance over time. The distance here is the circumference of the circle, and the time is the period of revolution. So, speed is $$2 \pi r /T$$. remember, time must be in seconds.

3. Nov 14, 2005

### NoMeGusta

Okay, so instead of v^2 you replace it as $$\Omega^2$$. How does r go from r to $$r^3$$ ?

4. Nov 14, 2005

### Chi Meson

No, do not just replace v with $$\omega$$. YOu are not even given the angular speed. Just replace v with $$2 \pi r /T$$. you have all of these.

And anyway it does not replace v with $$\Omega$$ , it replaces v with $$\Omega r$$ both of which get squared. The squared r joins with the r already there.

Does your book really use $$\Omega$$ and not $$\omega$$?

5. Nov 14, 2005

### NoMeGusta

That was a typo on my part, it should be $$\omega$$

6. Nov 14, 2005

### NoMeGusta

Okay, I worked it out and finally got it. I now see what you were saying, in terms of looking at it as orbital speed. That's what made it link. Thank you! I have an exam today and that was from my example problems sheet

7. Nov 14, 2005

### Chi Meson

Good luck.

(I need some more text).
Good luck again.