it wouldnt i would have eaten on you before then![]()
Hypothetically speaking, if I was at the very center of the earth and dropped an apple, which way would it fall?
it wouldnt i would have eaten on you before then![]()
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If you could find the "centre" of the Earth from a gravity perspective (the earth is not round) and somehow you could get there and drop an apple? I expect it would move first toward you and then continue on in that direction (assuming no air currents (you weren't breathing)). If you were able to "teleport" an apple into a sphere containing a vacuum in the gravitational centre of the Earth? It would initially travel in the direction of the sun which exerts 177x more force on the Earth than the moon. Subsequently it would continue to travel in the same direction as the Earth's gravitational force would take over.
To be sure about point #1 you would have to run the numbers to identify the gravitational force you exert on an apple.
The numbers:
Force you would exert on apple 1m from your body: 6.00534E-10 N
Force of the sun on the same apple: 0.0008852316N
So yes, it would fall towards you and slightly towards the Sun.
The apple would not fall in any direction relative to the Earth, since it is already at the point of lowest potential energy.
It would of course continue to "fall" towards (i.e. orbit around) the Sun, like the Earth and everything on and in it were doing before you released it. Since it was already doing that before you released it, the net effect from your point of view is no motion.
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Totally correct. I was imagining that the Earth and apple where frozen in space which of course is not the case. I also (embarrassingly) drew the wrong conclusion from my above numbers anyway.
However, would not an apple, teleported into a sphere containing a vacuum in the gravitational center of the Earth not "fall" towards the moon (in the same way the Earth's oceans do) then?
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Main: Orion 200mm f/4 Newtonian Astrograph; HEQ5 pro; KWIQ/QHY5 guide scope; Televue Paracorr 2; Siebert 3x Telecentric Barlow
Imaging Cameras: ATIK 383L+, EFW2 filter wheel, Astrodon LRGB filters (waiting for a break in the clouds); Canon 350D (modified/Baader);
EPs: 27mm TeleVue Panoptic; 8-24mm Baader Hyperion Mk III Zoom; 15mm, 6mm Antares W70;
Other: Celestron C-90 (old orange tube); Celestron 20x80 binos;
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Seems like the only new force in that frame of reference would be the gravitational attraction between the apple and the observer. Everything else would be the same as what is already acting on the Earth.
But, if I understand the intent of the question correctly, the Earth would have zero gravitational attraction for the apple. One of the things I still remember striking me from college physics oh so many years ago was that the gravitational force of a body was only from the mass within the radius at which it is calculated. I.e., if your were half way between the surface and Earth's center, the gravitational force on you could be calculated from the mass of the earth that was contained within 1/2 the radius of the Earth.
Of course, my mind is getting older and isn't what it used to be, so I could be wrong.