Moon fall speed

  1. If the moon was to be stopped, what would its velocity be just before hitting the earth? Also, if you could throw in the force of the impact, that would be great too. And another thing, how long would it take?
     
    Last edited: Sep 10, 2007
  2. jcsd
  3. It would take a tremendous force to stop the moon. If you have a tremendous force at your disposal, why are you still so interested in some other force?
     
  4. Trying to figure out the effects of the moon falling on the earth. Not just general "big explosion happens," but the physics of it. I'm thinking it wouldn't be too fast, maybe a couple of km/s, but I could be wrong. Would you set the potential energy equal to the kinetic energy and solve for v?
     
  5. Janus

    Janus 2,387
    Staff Emeritus
    Science Advisor
    Gold Member

    velocity = 9.8 Km/sec
    force of impact: depends on the deceleration on impact
    Time = a little under 5 days.
     
  6. Thanks a lot Janus, but could you please show how you got those numbers?
     
  7. If we wanted to get maximum speed, would the speed of the earth affect it? Lets say the moon stops in front of the earth's orbit, so the earth would be moving towards the moon while the moon falls to the earth.
     
  8. Your talking about to relative velocities here. The Moons relative to the earth and the earth relative to the sun. So yes, If the moon stopped completely, (relative to the earth and the sun) directly in line with the earths trajectory, the earth would slam into the moon in about 3.6 hours, using average distances and velocities respectfully.

    This however is obviously a crude approximation as it does not take into consideration other relative velocities (solar system, milky way etc....)
     
  9. What does "stopped completely (relative to the earth and the sun)" mean?
     
  10. I think that would mean that it's velocity looks like it's 0 viewed from the earth/sun, but it's still moving relative to the universe. I think everything's moving within the universe.

    So it would make a noticible difference if the earth moved into the impact rather than if the moon was on the other side and fell while catching up to the earth? Or would these relative velocities make no difference?
     
  11. But the Earth has a different velocity from the sun... I don't see how the moon can be stationary relative to both.

    As for your question, I don't believe it would make any difference if the moon stated stationary relative to the Earth.
     
  12. I believe I misunderstood sskakam's question. I apologize. For some reason I visualized the moon orbiting the earth until it's orbit intersected the earths orbital path around the sun, at which point the moons velocity relative to the sun became 0. This constitutes my earlier response of the earth continuing to speed along (in orbit) at 29 kp/s toward a now stationary moon, impacting in about 3.6 hours.

    Apparently you meant the moons orbital velocity around earth becomes 0 as its orbit intersects the earth/sun orbital path. In this case It would not make a difference whether the moon was "catching up" or leading the earth in respect to its orbital velocity around the sun. If you consider that once the moon stops, in respect to its earth orbit. Its velocity becomes 0 relative to earth, thus the before mentioned gravitation scenario (speed, impact force/time) between the earth/moon system would not be affected by the moons position relative to earth.
     
  13. tony873004

    tony873004 1,562
    Science Advisor
    Gold Member

    I get 9.8 km/s for a perigee drop and 9.9 km/s for an apogee drop. This is in good agreement with Janus' answer. I numerically simulated it.
     
  14. Thanks for your replies.

    Going back to the force of impact, he said it depends on deceleration. What would be decelerating it? I would think that air friction would be negligable since the moon is so massive. Is there another decelerating force here or am I am I just wrong about ignoring air resistance?

    I found this neat website that calculates earth-impact effects (http://www.lpl.arizona.edu/impacteffects/) and it says that the force would be about 3.67 x 10^30 Joules. However this website limits the distance between the earth and the object at half the earth's circumference (about 20,000 m). I don't know if that affects it much, but 10^30 J is quite a blow.
     
  15. tony873004

    tony873004 1,562
    Science Advisor
    Gold Member

    One minute it's going 9.8 km/s. Shortly later, its velocity is 0. So it decelerated from 9.8 to 0 over some amount of time. This is not an easy problem. Consider dropping a glass on the floor. If you drop it on carpet it does not break. If you drop it on a hardwood floor, it might break. If you drop it on tile, it definately breaks. The difference is that on carpet, it takes longer to go from impact velocity to 0 hence a lower acceleration than on tile or hardwood. The larger the acceleration, the larger the force since Force = mass * acceleration.

    This is the concept behind air-bags. They want your acceleration to be spread out over a longer period of time, so they give you something soft to crash into rather than something hard. That way it lessens the force.

    How "soft" is the Earth? How soft is the Moon?

    Rather than force, it might make more sense to ask about energy.
     
    Last edited: Sep 11, 2007
  16. Ah ha, gotcha. So the force would change whether it landed on land or sea, and different types of rock and such. Thanks a lot for all your replies. Helped a bunch.
     
  17. Saladsamurai

    Saladsamurai 3,016
    Gold Member

    Wouldn't it be reasonable to treat the Moon and Earth as spheres of uniform density for something like this. I mean, whether it landed on land or sea, I'll be calling out of work that day. And I think the malls will be closed. Except for Starbuck's.

    Casey
     
  18. DaveC426913

    DaveC426913 16,226
    Gold Member

    Is this counting the fact that, at the Moon's distance, g is nowhere near 9.8m/s^2?
     
  19. DaveC426913

    DaveC426913 16,226
    Gold Member

    Using the JPL impact calculator I get:
    Energy:10^13 Megatons
    Transient crater: 2660 miles diameter with 939 miles depth
    Final crater: 7920 miles diameter with 3.16 miles depth

    Paradoxically, the Earth loses negligible mass, nor are orbit or inclination altered appreciably (despite having inexplicably transformed from a sphere to a lopsided dumbbell)
     
  20. tony873004

    tony873004 1,562
    Science Advisor
    Gold Member

    Whether you go to work or not that day, to compute the force you still need to provide some sort of math that describes the rate of decceleration.

    That's a good tip. When the Moon drops from the sky, run for Starbucks :)



    Yes.

    9.8 km/s^2 has nothing to do with 9.8 km/s, unless the fall time were 1 second. The Moon's initial acceleration towards Earth would be very small (G*6e24*1.0123/384000000^2 = 0.0000027 km/s^2).

    As I watched the simulation, I kept thinking "Janus is wrong... Janus is wrong". The Moon was moving way too slow, even after it crossed geosychronous altitude, for me to think it had a chance to build to 9.8. But as it got close to Earth's surface, Earth's gravity started pulling big time. At surface-to-surface contact, it was ~9.8- 9.9.

    The numerical method relies on the facts that:

    * The Moon starts with 0.0 km/s of velocity relative to Earth and at a distance (D) of either lunar perigee, lunar apogee, or somewhere inbetween..
    * And that acceleration due to Earth's gravity at that moment will be G(M+m)/D^2.
    * And that after delta t (t=time), this acceleration will produce a velocity of v(t)
    * And that after delta t, this velocity will produce a change of position of d(v,delta t).
    * And that acceleration due to Earths gravity at the new distance will be G(M+m)^newD^2
    * And that as delta t is made smaller and smaller, the simulated answer will asymptotically approach the true answer
    * And that a delta t of 1 second is sufficient enough to get an answer correct to > 0.1 km/s, while spending less than 2 minutes of computer time.
    * And if you make the computer repeat this process again and again until D < (Earth Radii + Moon Radii), you will have numerically computed v and t.
     
    Last edited: Sep 12, 2007
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?