If the gravitational attraction of the moon is strong enough to draw countless tons of seawater to one side of the earth, then why don’t people weigh significantly less during high tide?
As the Master has so cogently pointed out ( http://www.straightdope.com/classics/a881216.html ; hey, Jill, do I get a prize for this?), tides are not caused by gravitational attractation per se, but by differential gravitatonal attraction.
The Earth is nearly 8,000 miles in diameter; most of the people that I know aren’t. Even with the Moon a considerable away, the difference between 234,000 miles and 242,000 miles is significant; the difference between a person’s head and toes is not.
“Kings die, and leave their crowns to their sons. Shmuel HaKatan took all the treasures in the world, and went away.”
at first, I would think you would weight less when the moon is directly overhead but there are many other factors to concider including the earth moon system. The system is spining at a point between the earth and the moon and therefor the part away from the moon is moving faster in a circle (really some orbit, but for simplecity, lets say a circle) due to centrifical acceleration, you might weigh less at the far end where the centri. force os ‘pulling’ you off the earth then when you are twards the moon and the cent. f. is pushing you down, increasing your weight. also rmeber that the moon gravity is apx 1/6 that of earth on the surface of the moon, you can’t expect it to decrease your weight significanly this far away. If you want to lose some weight and live in the LI NY area, I know an excellent cardio instructor.
Also I assume the tides are as big as they are due to steady buildup over time, much like rocking a big , well rock, till it gets moving. the moon if if was a stray rock caught by earths grav. probally took 100’s to 1000’s of yrs to get a halfway decent tide going. just a guess, surf’s up.
“If the gravitational attraction of the moon is strong enough to draw countless tons of seawater to one side of the earth, then why don’t people weigh significantly less during high tide?”
Comment 1: Why is this under this Forum? Did Cecil recently comment on tides? Reference to the column would help.
Comment 2: The question contains a false assumption, namely, that the high tide occurs when the moon is overhead (thus causing a decreased net gravitaional pull ‘down’ (see the recent column in the mailbag on the meaning of ‘down’)). Actually, high tides occur twice per day, so even if they occurred at the moment the moon was straight ‘up,’ there would also be a high tide when the moon was straight ‘down,’ with a net pull greater than ‘normal.’ Fortunatly, the tides lag with relation to the relative position of the moon with respect to any location on the globe, so the assumption isn’t even accurate once per day.
Comment 3: Even if we disassociate the ‘tide’ issue from the ‘weight’ issue, and address the ‘weight’ issue with relation to the lunar position, the question still can’t easily be answered. Simplifying the force diagram involved to three bodies (Earth, Moon, Person), one would have to clarify several things. One would have to determine such things as the position on the globe of the person, the relative position of the moon, the ‘norm’ against which the measurement is to be taken, etc. People with too much time on their hands can do the math for us if they want (my Halliday and Resnick stays firmly put on the shelf whereon it was deposited after freshman Physics).
Comment 4: Mostly, of course, you wouldn’t weigh ‘significantly’ less because the difference in the net force acting on you wouldn’t be significant. As pointed out already, the tides aren’t caused by water being pulled to one side of the earth by the attraction of the moon, they are caused by difference in gravitational attraction over the distance of roughly 8000 miles on opposite sides of the globe, a relatively minute difference that builds up by acting on the liquid molecules in the water over time. If you want to feel the human equivalent, try a mosh pit.