jimrandomh comments on Rationality quotes: March 2010 - Less Wrong

3 Post author: Morendil 01 March 2010 10:26AM

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Comment author: RolfAndreassen 01 March 2010 09:47:17PM 3 points [-]

As you stand on the equator, with the Sun directly overhead, its gravity is pulling you away from the Earth's center. On the other side of the Earth, the Sun's gravity pulls you in towards its center. Consequently you weigh slightly less at noon than at midnight. However, since the force of the Sun's gravity on a 100-kg mass 1 AU distant is about 0.006 Newton, an average bathroom scale is not going to notice.

Comment author: jimrandomh 01 March 2010 09:57:45PM *  6 points [-]

No, because it pulls you, your scale and the Earth all (very close to) equally.

Comment author: RobinZ 01 March 2010 10:02:39PM *  1 point [-]

I feel like an idiot for not seeing this earlier: you're right; this is the tidal force problem.

More precisely, the lunar tidal acceleration (along the Moon-Earth axis, at the Earth's surface) is about 1.1 × 10−7 g, while the solar tidal acceleration (along the Sun-Earth axis, at the Earth's surface) is about 0.52 × 10−7 g, where g is the gravitational acceleration at the Earth's surface.

In other words, the measured weight of 100-kg human changes from Solar gravity by 5.2 [edit: milli]grams between equitorial solar noon or midnight and equitorial dawn or dusk.

Comment author: RolfAndreassen 01 March 2010 11:40:39PM 0 points [-]

This would only be relevant if you were accelerating relative to the Earth. The scale measures the normal force keeping you at rest relative to the Earth's center; the force being exerted on the Earth does not change that. (Modulo the orbital-velocity argument, which I'll respond to separately.)