The Objects > Lunar Science

Changing Moon's Face?

(1/2) > >>

shade9:
Hi, I am in High School, and my Physics class is using different aspects of Zooniverse to have a research topic/paper. My topic is about changing the moon's face with a large impact. To explain better, I mean how large of an impact at what sort of angle would be needed to make the moon spin enough that the far side will become the near side, and near the far. Can anyone, I don't know, help a little? Give me something to go on, somewhere to start? I read it on a different article that this might have happened millions of years ago (article from universetoday.com, I believe). Thanks for anyone interested or willing to help!

kodemunkey:
Well, for starters the moon is tidally locked to the earth, so you'd need an impact large enough to put some spin into the moon, but not so large that it smashes to bits. I'll ask some of my astrophysics friends and see what they suggest.

[Edit, incoming science, moon data from wikipedia]


--- Quote ---Spin period of moon = P_s = 27.32days = 2360448s
Radius of moon = R_m = 1,738.14 km = 1.738x10^6m
Mass of moon = M_m = 7.35x10^22kg
Equatorial velocity of moon = v_e = (2*pi*R_m)/(P_s) = 4.626m/s
Asteroid mass (asteroid "1 Ceres" - one of the biggest asteroids) = M_a = 8.7x10^20kg
Asteroid velocity (currently velocity around sun) = v_a = 17,977km/s = 1.8x10^4m/s

Assume the asteroid collides perfectly and tangentially at the equator:

Resulting equatorial velocity of the moon = v_r = ((M_m)*(v_e) + (M_a)*(v_a)) / (M_m + M_a) = 215.141m/s

Differential velocity = v_r - v_e = 210.515m/s

Distance to spin half around is half the circumference of the moon = (1/2)*2*pi*R_m = 5.46x10^6m

Time taken to spin half around = (5.46x10^6)/(210.515) = 25936s = 7hrs 12mins 16secs

This is crazy fast, mainly because of the assumptions of a perfect collision, and therefore all the energy going into the spin. I may have gone wrong, but this is my first stab at it. Best to use a better sized asteroid and run the steps using the new mass. I got my asteroid data from here: http://burro.astr.cwru.edu/stu/asteroid.html

Hope this helps at least. The moon will keep spinning this fast until something slows it down over time. You'll need another comparable collision on the other side in the other direction for it to stop spinning that fast. if you use an asteroid mass of 1x10^15kg it'll take ~636yrs

if you use an asteroid mass of 5x10^5kg ,on wolfram alpha it'll come out as 1.274x10^12yrs = ~1 trillion years!!!

for the angle, just multiply the asteroid velocity by "cos(theta)", where "theta = 1" for the tangential case, and "theta = 0" if the asteroid hit it head on, i.e. radially (perpendicular to the tangent)

[amendment] Sorry, just noticed an error. The last comment should read:
...where "theta=0" for the tangential case, and "theta=90" of the asteroid...

--- End quote ---

shade9:
Oh jeez, I knew it was going to be complicated, but wow. Thanks for the information! Now I just have to work to understand it, lol.

bugiolacchi:
http://www.sciencedirect.com/science/article/pii/S0019103508004284

or the abstract

http://adsabs.harvard.edu/abs/2009Icar..200..358W


Did a large impact reorient the Moon?
Mark A. Wieczorek, , Mathieu Le Feuvre

Icarus
Volume 200, Issue 2, April 2009, Pages 358–366


maybe this is the paper you are referring to. All I can add is... there are no reasons why they should be 'wrong', it's a viable hypothesis..

Please let me know if you'll find it difficult to get to the paper or even the abstract. Also, if you find some of the jargon obtuse...  ;)

Good luck.

Geoff:
Thanks for posting this bugiolacchi and welcome to Moon Zoo  :)

Will read it later - have a Zooniverse meet-up to attend soon.

Navigation

[0] Message Index

[#] Next page