# Change in planet's day length

1. Oct 28, 2006

### lizzyb

This is regarding that planet question. I set up the equation as:
$$L_{pi} = L_a + L_{pf} = L_a + I_p \omega_{pf} \Longleftrightarrow \omega_{pf} = \frac{L_{pi} - L_a}{I_p}$$
where $$I_p = \frac{2 M R^2}{5}$$
so $$\omega_{pf} = \frac{L_{pi} - L_a}{\frac{2 M R^2}{5}}$$

In the original problem, we're given T = 13 hours, so
$$T_i = \frac{13 "hours"}{"rev"} \cdot \frac{60 "min"}{1 "hour"} \cdot \frac{60 "sec"}{1 "min"} = \frac{46800 "sec"}{"rev"}$$

Using the final $$\omega$$ I came up with $$T_f = \frac{46801.3 "sec"}{"rev"}$$ - isn't that a longer day? Yet the question states "But, thanks to the asteroid's angular momentum, the planet rotates faster after the imapact than it did before."

2. Oct 28, 2006

### lizzyb

am I just being stupid? I've done this a few times now and I keep coming up with the same answer. If it takes longer to make a revolution, the day would be longer.

3. Oct 28, 2006

### lizzyb

I guess I set up the equation wrong. Using this:
$$L_{pi} + L_a = L_{pf} = I_p \omega_{pf} \Longleftrightarrow \omega_{pf} = \frac{L_{pi} + L_a}{I_p}$$
where $$I_p = \frac{2 M R^2}{5}$$
so $$\omega_{pf} = \frac{L_{pi} + L_a}{\frac{2 M R^2}{5}}$$

$$T_i = \frac{13 "hours"}{"rev"} \cdot \frac{60 "min"}{1 "hour"} \cdot \frac{60 "sec"}{1 "min"} = \frac{46800 "sec"}{"rev"}$$

Using this new method (and adding the mass of the asteroid to the mass of the planet), I got $$T_f = \frac{46798.7 "sec"}{"rev"}$$ which means a day is shorter.

Thanks for letting me figure it out!!! (I finally got the right answer!) Ya'll are great.

4. Oct 28, 2006

### DaveC426913

You're quite welcome.