In this post a theoretical physicist says that one will weigh less at the point on the Earth closest to the Sun due to the Sun's gravity, which makes sense to me. But he also says that one will weigh less at the point on the Earth farthest from the Sun. This doesn't make sense to me, because I imagine that in this case the gravity of the Sun should add to the gravity of the Earth making one heavier. Is this correct or no?
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1$\begingroup$ Does this help? physics.stackexchange.com/q/649499/123208 $\endgroup$– PM 2RingCommented Aug 24, 2021 at 12:52
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3$\begingroup$ To figure this out, you also have to worry about the motion of the earth and the rotation of the earth, which both create centrifugal forces. $\endgroup$– Zo the RelativistCommented Aug 24, 2021 at 14:16
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2$\begingroup$ Related: en.wikipedia.org/wiki/Tidal_force $\endgroup$– Solomon SlowCommented Aug 24, 2021 at 14:21
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2$\begingroup$ Since there have been some mentions of tidal forces, I'll also link this: Moon's pull causes tides on far side of Earth: why? which IMHO is a great intuitive explanation. $\endgroup$– jng224Commented Aug 24, 2021 at 21:12
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3 Answers
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This doesn't make sense to me, because I imagine that in this case the gravity of the Sun should add to the gravity of the Earth making one heavier
What he means is that when one is at the point farthest from the sun, it also means that earth is closer to the sun than the person. So, the sun will pull on earth more than it pulls on the person. So, the person will be accelerated toward the sun slower than the earth is. The result is that he will be slightly lighter
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1$\begingroup$ I imagined that, like in the case of the Tower of Pisa, the two objects - the person and the Earth - will fall towards the Sun with the same speed. But yes, I had missed an important notion there. $\endgroup$– sequenceCommented Aug 24, 2021 at 19:42
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This is how tidal forces work, and why the tidal bulge appears on both the near and far sides of the body simultaneously.
Intuitive explanation--the Earth's center of gravity is in a weightless orbital path around the sun.
If you're on the noon side, your orbital velocity is slightly too slow for your orbital path (as you are traveling at about the same speed as the Earth's CG, but in a smaller ellipse). Therefore, you have a slight tendency to fall toward the sun, away from Earth, with the end result that you're slightly lighter.
If you're on the midnight side, your orbital speed is slightly too high for your orbital path, and you tend to be pulled away from the both the sun and Earth. Again, slightly lighter.
This analysis doesn't take into account the rotation of the Earth. Earth's orbital velocity, at about 30km/s, has more effect than its surface speed, <0.1km/s.
answered Aug 24, 2021 at 20:54
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After doing a little research, I believe that the gravity being weaker on the opposite side of the Sun is because of the earth being pulled towards the Sun more (by product of being closer) than a person standing on the earth's surface at midnight on the equator. Does this help?
