Its embarrassing but I have to say that honestly the centripetal force argument never occurred to me before. Rough calculations seem to indicate that a large man 100Kg should be almost half a pound heavier in the day time as he is at night. Kinda cool.
Now I am dying to get something big and stable enough to see if my home scale can pick it up.
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.
And the Earth is slowly curving in its orbit, generating an apparent centrifugal force that decreases your weight at midnight, and increases your weight at noon. Except for a very tiny tidal correction, these two forces exactly cancel which is why the Earth stays in orbit in the first place. This argument would only be valid if the Earth were suspended motionless on two giant poles running through the axis or something.
The Earth is a (fairly) rigid body held together by its internal structure, and is not required to be moving at orbital velocity at every point on its surface. That is, the effect you mention exists, but it is not clear that it exactly cancels the gravitational effect. (Or equivalently, it’s not obvious that the tidal effect is small.) Don’t forget that the Earth’s rotation is reducing your effective orbital velocity on the day-side, and increasing it on the night-side.
Now, if you have some numbers showing that the cancellation is close to exact for the specific case of the Earth, that’s fine. An argument showing that it’s always going to be close to exact for planet-sized bodies in orbit around stars would also be convincing.
According to Wikipedia, solar tides are about 0.52*10^-7 g, as opposed to lunar tides of about 1.1*10^-7 g. One part in twenty million and one part in ten million, respectively.
The non-cracking of the Earth demonstrates only that the tidal force is small relative to that required to crack the Earth apart, which may not be a particularly strong upper bound on human scales. :) However, RobinZ’s numbers show that it’s also small relative to human weights, so there we go.
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.
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.)
I got 0.6N (=6.7e-11 2e30 100/(1.5e11)^2). Still small, but potentially measurable. (Er, except for the whole frame-of-reference thing mentioned above.)
Don’t forget to adjust your calculations for not being on the equator, and to take into account that ‘nighttime’ is not equivalent to ‘the Sun pulls you directly towards the center of the Earth’. Both tend to make the effect smaller.
I just read their website.
Its embarrassing but I have to say that honestly the centripetal force argument never occurred to me before. Rough calculations seem to indicate that a large man 100Kg should be almost half a pound heavier in the day time as he is at night. Kinda cool.
Now I am dying to get something big and stable enough to see if my home scale can pick it up.
Quick look didn’t find it, but I don’t see why this follows (and at a wild guess, I’m guessing it doesn’t). Can you link?
It doesn’t. My though process was too silly to even bother explaining.
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.
And the Earth is slowly curving in its orbit, generating an apparent centrifugal force that decreases your weight at midnight, and increases your weight at noon. Except for a very tiny tidal correction, these two forces exactly cancel which is why the Earth stays in orbit in the first place. This argument would only be valid if the Earth were suspended motionless on two giant poles running through the axis or something.
The Earth is a (fairly) rigid body held together by its internal structure, and is not required to be moving at orbital velocity at every point on its surface. That is, the effect you mention exists, but it is not clear that it exactly cancels the gravitational effect. (Or equivalently, it’s not obvious that the tidal effect is small.) Don’t forget that the Earth’s rotation is reducing your effective orbital velocity on the day-side, and increasing it on the night-side.
Now, if you have some numbers showing that the cancellation is close to exact for the specific case of the Earth, that’s fine. An argument showing that it’s always going to be close to exact for planet-sized bodies in orbit around stars would also be convincing.
According to Wikipedia, solar tides are about 0.52*10^-7 g, as opposed to lunar tides of about 1.1*10^-7 g. One part in twenty million and one part in ten million, respectively.
This was my original thought until I realized that of course it cancels or else the earth would crack into pieces.
The non-cracking of the Earth demonstrates only that the tidal force is small relative to that required to crack the Earth apart, which may not be a particularly strong upper bound on human scales. :) However, RobinZ’s numbers show that it’s also small relative to human weights, so there we go.
No, because it pulls you, your scale and the Earth all (very close to) equally.
I feel like an idiot for not seeing this earlier: you’re right; this is the tidal force problem.
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.
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.)
I got 0.6N (=6.7e-11 2e30 100/(1.5e11)^2). Still small, but potentially measurable. (Er, except for the whole frame-of-reference thing mentioned above.)
Oops, added a zero typing the numbers into my calculator. :oo
Don’t forget to adjust your calculations for not being on the equator, and to take into account that ‘nighttime’ is not equivalent to ‘the Sun pulls you directly towards the center of the Earth’. Both tend to make the effect smaller.