by davepotter » Tue Dec 23, 2014 11:58 pm
From an earlier post: "Doesn't it all depend upon in which direction you jump and how much effort you put in to it ?"
You are correct.
First, even a very small horizontal speed at the top of the cliff will result in a landing point far away from the cliff face (because you would keep moving horizontally at that speed for the 40 minutes it takes to drop to the surface)
Second, Coriolis effect would play a significant role. If you hung from a rope and dropped "vertically", your trajectory would be curved, because the bottom of the cliff is not moving laterally (due to the comet's rotation) as fast as the top of the cliff.
Given the 12.4 hour rotational period, the comet's mass (1e13[kg], and the height of this cliff (easily 3000 meters from the comet's center of mass), you might be approaching orbital speed just standing on top!
(vsurface=2*(pi)*r/12.4/3600) vorbit=sqrt(G*1e13/r). Set vsurface=vorbit, and solve for r. Result: if the top of the cliff is 3229 meters from the comet's center of mass, the surface is moving at orbital speed, meaning if you are there and drop a ball it would not fall! And if you threw the ball horizontally at even a centimeter per second in the same direction as the comet's rotation, it would rise into its own orbit. And if you were standing on a scale, it would read zero.
Strange!
From an earlier post: "Doesn't it all depend upon in which direction you jump and how much effort you put in to it ?"
You are correct.
First, even a very small horizontal speed at the top of the cliff will result in a landing point far away from the cliff face (because you would keep moving horizontally at that speed for the 40 minutes it takes to drop to the surface)
Second, Coriolis effect would play a significant role. If you hung from a rope and dropped "vertically", your trajectory would be curved, because the bottom of the cliff is not moving laterally (due to the comet's rotation) as fast as the top of the cliff.
Given the 12.4 hour rotational period, the comet's mass (1e13[kg], and the height of this cliff (easily 3000 meters from the comet's center of mass), you might be approaching orbital speed just standing on top!
(vsurface=2*(pi)*r/12.4/3600) vorbit=sqrt(G*1e13/r). Set vsurface=vorbit, and solve for r. Result: if the top of the cliff is 3229 meters from the comet's center of mass, the surface is moving at orbital speed, meaning if you are there and drop a ball it would not fall! And if you threw the ball horizontally at even a centimeter per second in the same direction as the comet's rotation, it would rise into its own orbit. And if you were standing on a scale, it would read zero.
Strange!