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Can someone good at physics suggest how much lighter I become when the moon is overhead, compared with on the opposite side of the planet? And how might one measure such an effect?Chris
So how would modern scales measure the effect? Wouldn't the scales be pulled towards the moon too and because you are in the way (standing on them), they'd record the same weight?Chris
should we not also take into effect the orbital wobble of the Earth/Moon system as it rotates about the Sun.
In addition to the night-day variation there is a 28 day cycle due to the wobble of the C of G of the earth-moon system in its orbit about the Sun, this C of G is approximately 2565 Km below the Earths surface so there will be a 5130 Km variation in orbital distance each 14 days.Numerous simplifying assumption have been made, ie that the orbital plane of the earth-moon coincides with that of the the orbit around the sun and is circular and no gravitational shielding exists.A precise calculation taking all the factors would be rather challenging!
Are there many 70 Kg men ?, after xmas I am at least 85
Sun's gravitational effect: Lightarrow, in addition to the acceleration to the Sun's gravity - which gives you factor of +/- 40g effective weight, there is the acceleration due to the circular motion of the Earth's orbit around the Sun.
I remember I did some sums a long while ago. The effect of, shall we call it centrifugal force, is in the opposite direction to that of the gravity and more or less cancels out. In the same way that an astronaut feels weightless when in orbit around the Earth, the kg mass and the Earth are both in 'free fall' around the Sun -so there is much less difference than the 40g. I calculated the difference to be about 1/1000 of that, taking both accelerations into account.I was, originally, trying to work out the effect on tides but gave up when I realised that the tides actually consist of a wave which sloshes around the Earth once a day (ish) and the actual Q factor of what is, in effect, a resonator, would be a major factor in determining the actual heights of tides.I calculated the sums of the rotational and gravitational effects due to the Solar orbit and the Lunar 'wobble'. This suggested that the difference due to the Moon's effect would be about 4e-5 N/kg whilst the effect of the Sun would be about 2.6e-6N/kg. It would mean a variation in measured weight of 4e-4% for the moon and 2.6e-5% for the Sun.This isn't far from the ratio which they quote for the relative effect of Sun and Moon on the tides.
And yes - lightarrow must be in very good shape if he reckons on 70kg. I've been 85 kg for years and years.