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If I'm standing at the equator, jump, and land 1 second later, the Earth does NOT move 1000mph (or .28 miles per second) relative to me, since my velocity while jumping is also 1000mph. However, the Earth is moving in a circle (albeit a very large one), while I, while jumping, am moving in a straight line. How much do I move relative to my starting point because of this? I realize it will be a miniscule amount, and not noticeable in practise, but I'd be interested in the theoretical answer. |
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While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun. However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces. The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored. Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during the jump you spend your time on average roughly one meter above earth's surface, your velocity lags 460 m/s times 1/6.4x10^6 (the denominator corresponding to earth's radius in meters) which equates to about 70 micrometer per second. So, when jumping exactly vertically, after a second you land roughly 70 micrometer west of where you started. |
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My answer: 816.126 nanometers. Someone please review my thought process before I accept my own answer + declare myself awesome. OK, I originally asked this question to save myself from doing some math, but this was clearly a bad idea, as it doubtless angered the Math Gods. I've now come up with an answer, which, of course, agrees with neither of the previous answers (one of which was deleted). I'm using a diagram, which of course makes my answer correct . Generalizing the problem a bit: if I travel distance 't' from the starting point, my new angle relative to the Earth's center is Atan[t/r], where r is the Earth's radius. I call this angle psi.
The hypotenuse of SOE is $\sqrt{r^2+t^2}$, so I am hovering above the surface: $$\sqrt{r^2+t^2}-r $$ If the Earth didn't rotate, I'd have moved $\psi r$ (along the Earth's surface) from my original position. If the Earth rotates theta in this time, I've moved $(\psi-\theta)r$. Plugging in some values (assuming earth circumference exactly 40Mm), where $s$ is how long I stay in the air: $$ t = 40*10^6/86400 s $$ $$ \psi = \arctan[t/r] = \arctan [(40*10^6/86400*s)/(40*10^6/2/\pi)] = $$ $$ \arctan[\pi*s/43200] $$ $$ \theta = 2*s*\pi/86400 $$ $$ \psi-\theta = \arctan(\pi s/43200) - 2 s \pi/86400 $$ $$ d = 40*10^6/2/\pi*(\arctan(\pi s/43200) - 2 s \pi/86400) $$ My hovering altitude is: $$ \sqrt{(40*10^6/2/\pi)^2 + (40*10^6/86400 s)^2} - (40*10^6/2/\pi) $$ For s=1, I get d=816.126 nanometers, and hover altitude of 1.68cm (the latter seems high). As a note $d$ appears to grow as $s^3$, so the longer you hover, the more you move per second (does this explain helicopter drift?) |
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