Lunar effect on weight on earth

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How much does the phase and position of the moon affect the weight of an object on the surface of the earth? For example, if a person at the equator on the equinox weighs 80 kg at noon when the moon is new, how much does he or she weigh at midnight that day? Conceptually, it would seem that it would have to be significant; otherwise, the height of the tides would not be significant. Also conceptually, it would seem that it would have to be non-zero, otherwise Lagrangian points wouldn’t work. On the other hand, if it is significant, how come it doessn’t come into play when jockeys and wrestlers try to make their weight?
 

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The gravitational force between two massive bodies (i.e. moon and ocean or moon and human) is directly proportional to the product of the two body's masses. If the hydrosphere is, say, 10^20 times more massive than a human, then you can expect the gravitational force of the moon on a human body to be 10^20 times smaller than the moons pull on the hydrosphere.
 
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On the other hand, if it is significant, how come it doessn’t come into play when jockeys and wrestlers try to make their weight?
Won't it depend on the scales they use to weigh them.
Jockeys use balance scales or they did when I last looked.
So the balance weight would be effected by the same amount as the jockey by the moons pull or lack of it.
If there was any discrepency caused by the moons gravity it would not show up using balance scales.
 
  • #4
russ_watters
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How much does the phase and position of the moon affect the weight of an object on the surface of the earth?
Not at all: the earth and moon are in orbit around their common center of gravity.
Conceptually, it would seem that it would have to be significant; otherwise, the height of the tides would not be significant.
The tides are the same height on the far side as on the near-side...
On the other hand, if it is significant, how come it doessn’t come into play when jockeys and wrestlers try to make their weight?
Well even if the moon were stationary above the earth (in which case, it would make a difference), the moon is pretty small and pretty far away. You could always use Newton's equation for gravity to calculate just how significant it would be.
 
  • #5
Janus
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How much does the phase and position of the moon affect the weight of an object on the surface of the earth? For example, if a person at the equator on the equinox weighs 80 kg at noon when the moon is new, how much does he or she weigh at midnight that day? Conceptually, it would seem that it would have to be significant; otherwise, the height of the tides would not be significant. Also conceptually, it would seem that it would have to be non-zero, otherwise Lagrangian points wouldn’t work. On the other hand, if it is significant, how come it doessn’t come into play when jockeys and wrestlers try to make their weight?
Since the Earth and Moon orbit their barycenter, The difference in weight for someone standing directly under the Moon vs No moon, is determined by the difference in the Lunar gravity at the distance of the surface of the Earth from the moon compared to it at the distance of the center of the Earth.

For a 150 man, this works out to about 0.00028 oz.

As far as the tides being significant, it depends as to what you compare it to. If there were no landmasses and the ocean was of uniform depth, the Lunar tide would be about 54cm high.( The reason the tides we measure are different from this is that we do have landmasses and varying depths which effect the height of local tides)
This is only 8.5e-8 times that of the radius of the Earth. In other words, when considered by how much the tides deforms the ocean surface from that of a perfect sphere, it is very little.
 

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