Moon fall speed


by sskakam
Tags: fall, moon, speed
sskakam
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#1
Sep10-07, 09:48 PM
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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?
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cesiumfrog
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#2
Sep10-07, 10:18 PM
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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?
sskakam
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#3
Sep10-07, 10:30 PM
P: 15
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?

Janus
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#4
Sep10-07, 10:37 PM
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Moon fall speed


velocity = 9.8 Km/sec
force of impact: depends on the deceleration on impact
Time = a little under 5 days.
sskakam
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#5
Sep10-07, 10:42 PM
P: 15
Thanks a lot Janus, but could you please show how you got those numbers?
sskakam
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#6
Sep11-07, 07:21 AM
P: 15
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.
BosonJaw
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#7
Sep11-07, 08:25 AM
P: 40
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....)
Capuchin
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#8
Sep11-07, 08:30 AM
P: 29
What does "stopped completely (relative to the earth and the sun)" mean?
sskakam
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#9
Sep11-07, 08:36 AM
P: 15
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?
Capuchin
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#10
Sep11-07, 08:40 AM
P: 29
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.
BosonJaw
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#11
Sep11-07, 11:06 AM
P: 40
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.
tony873004
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#12
Sep11-07, 11:35 AM
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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.
sskakam
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#13
Sep11-07, 11:44 AM
P: 15
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.
tony873004
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#14
Sep11-07, 02:47 PM
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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.
sskakam
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#15
Sep11-07, 04:09 PM
P: 15
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.
Saladsamurai
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#16
Sep11-07, 08:56 PM
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Quote Quote by tony873004 View Post
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.
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
DaveC426913
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#17
Sep11-07, 10:17 PM
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Quote Quote by tony873004 View Post
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.
Is this counting the fact that, at the Moon's distance, g is nowhere near 9.8m/s^2?
DaveC426913
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#18
Sep11-07, 10:31 PM
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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)


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