Does an asteroid impact change a planet's day length?

[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."

 

[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.

 

[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.

 

[DaveC426913]

You're quite welcome.