Are we heavier at night than at day?

  • #1
The Sun attracts all bodies on the Earth. At midnight, when the Sun is directly below, it pulls on an object in the same direction as the pull of the Earth; at noon, when the Sun is directly above, it pulls on an object in a direction opposite to the pull of the Earth. Hence, all objects should be heavier at night than they are at day.
 

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  • #2
russ_watters
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No. We're on earth and the earth is in orbit, so those forc
es are always cancelled by our acceleration in the orbit.
 
  • #3
D H
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First off, alphabeta1720, are you telling us or asking us?

Secondly, what exactly do you mean by weight? Do you mean the quantity measured by a (very precise) spring scale, or something else?
 
  • #4
nope..we are heavier during day, you'd have been correct provided there was no MOON, the moons gravitational pull is stronger than the sun's ,even when they're on opposite sides of the planet , for the numerical values, they're on google just feed em in the formulae and get your values.
 
  • #5
Borek
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nope..we are heavier during day, you'd have been correct provided there was no MOON
Have you never seen Moon in the middle of the day?
 
  • #6
Have you never seen Moon in the middle of the day?
no, never
 
  • #7
K^2
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Gah!

Yes, you are lighter around Noon because of the Sun, BUT you are also lighter around Midnight. It's called tidal force. Look it up.

Same goes for the Moon. There is a decrease in net gravitational pull when it's right above you OR right bellow you, and either one can be during any time of the day depending on Lunar phase.
 
  • #8
D H
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nope..we are heavier during day
That depends on what you mean by weight. It appears that you are thinking of apparent weight, but you did not specify that.

the moons gravitational pull is stronger than the sun's
And that depends on what you mean by gravitational pull. It appears you are thinking of tidal gravitational forces, but you didn't specify that, either. BTW, that is a rather non-standard meaning of the term "gravitational pull".


Without a clarification of what the OP meant by weight we're just arguing semantics without even having a clear idea of what those semantics are. Why don't we just wait for the OP to say what was meant by the term weight?
 
  • #10
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The Sun attracts all bodies on the Earth. At midnight, when the Sun is directly below, it pulls on an object in the same direction as the pull of the Earth; at noon, when the Sun is directly above, it pulls on an object in a direction opposite to the pull of the Earth. Hence, all objects should be heavier at night than they are at day.
All other things being equal, of course. Taken to the extreme, weight becomes zero at the http://en.wikipedia.org/wiki/Roche_limit" [Broken]
 
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  • #11
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No. We're on earth and the earth is in orbit, so those forces are always canceled by our acceleration in the orbit.
i'm assuming you're referring to the fact that, b/c the earth is in free-fall about the sun, the earth does not feel the sun's gravitational pull? kinda like we don't feel the earth's gravitational pull during sky diving (or any type of free-fall), but we do when we're on the ground and the ground is exerting a force on us? i.e. the force exerted on the earth by the sun is canceled by the earth's free-fall about the sun?

if i understand that much correctly, then i still don't understand how that automatically applies to objects on the earth's surface. after all, your statement doesn't change the fact that the direction of the sun's gravitational influence on earth's surface objects is constantly changing, nor does it change the fact that the distance between the sun and earth's surface objects oscillates once a day (more so for objects near the equator, and less so for objects near the poles). as the earth rotates, objects on its surface more or less move away from the sun and come closer again over a 24-hour rotational period of the earth. in other words, the combination of a surface object's radial and angular velocities with respect to the sun oscillates much more frequently (both every 24 hours and every 365.25 days) than the center of the earth itself (only every 365.25 days). the center of the earth orbits the sun in a near-circular ellipse, but a stationary object on the surface of the earth orbits the sun in a path that resembles a spring or slinky connected end-to-end and stretched out into an ellipse that resembles our planet's orbit about the sun. so while the earth's general radial distance from the sun oscillates in such a way that it has one maximum and one minimum per year, the radial distance of its surface objects to the sun oscillate in such a way that they have one max./min. per day AND one max./min. per year. if the distance between a stationary earth surface object and the sun oscillates every 24 hours, then the sun's gravitational influence on that object must oscillate over the same 24-hour period, since any gravitational force is inversely proportional to the square of that distance.

so if we get constant readings on a household scale for an object of known mass/weight regardless of what time of day it is or what time of year it is, then i have to assume that the sun's gravitational influence over earth's surface objects (as well as the gravitational influence of other massive objects, such as the moon, other planets, etc.) is negligible in comparison to the earth's gravitational influence over its surface objects (due to our proximity to the earth and our vast separation from other massive solar system objects). this is in stark contrast to believing that the many gravitational forces external to the earth are actually canceling each other out exactly such that the only resulting gravitational pull over the earth's surface objects is from the earth alone.

if we were talking about the earth, sun , and moon only, i might be able to stretch my imagination into thinking that, based on your original argument, we don't feel the sun's gravitational force b/c we are in free-fall around it, and likewise, we don't feel the moon's gravitational influence b/c it is in free-fall about the earth, and by convention, we are also in free-fall about the moon. but then how do we explain and cancel out the gravitational influences of bodies that are clearly not in orbit around us (or bodies that we are not orbiting), like the other planets, comets, asteroids and such? again, the only logical explanation i can come up with is that the gravitational influences of other massive solar system bodies are negligible in comparison to earth's gravitational influence, and it is due solely to our proximity to the earth.

this is definitely not the definitive answer, as i'm not super confident in my response. but i would like to know if i'm on the right track...
 
  • #12
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Because the earth is in freefall, only the tidal forces remain, and these give an acceleration of about 0.5*10^(-7) g for the sun. This is 5 mg if you weigh 100 kg. You will be heaviest in the morning/evening and lightest at noon or midnight.
Of course the tidal effect of the moon is more than 2 times larger.
If you're outside, it will likely be warmer during the day, and decreased buoyancy will have a much larger effect than tidal forces.
 
  • #13
pervect
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There will be tidal stretching forces on the Earth due to both the sun and the moon. You'll have "high tides", when the body in question (sun or moon) is either directly overhead OR on the other side of the planet.

This isn't particularly intuitive, but its how tides work. You can't look at the "force of gravity" alone. Consider- what is the "force of gravity" at the center of the Earth? What would you measure there if you put an accelerometer there?

The tidal forces would, in principle, affect the reading of a standard mass on a spring scale, and they are also important in gravimetry - high precision measurements of the Earth's gravitational field - basically a high precision accelerometer. But the cycle would not be a 24 hour cycle as the original poster proposes, you'd have a minimum scale reading when the sun (or moon) was overhead or underfoot. For the case of the sun, the maximum readings would occur about 6am or 6pm, midway between noon and midnight.

For solar tides, the magnitude of the effect is:

(radius of earth) *G * (mass of sun) / (1 au)^3= 2.5*10^-7 m/s^2

For lunar tides:

(radius of earth) *G * (mass of moon) / (384400km)^3= = 5.5*10^-7 m/s^2


It's a bit sketchy, but http://gravmag.ou.edu/reduce/reduce.html [Broken] has some actual raw measurements.
 
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  • #14
D H
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The reason I have asked about weight is because there are at least three distinct meanings. Colloquially and, in some places, legally, "weight" is a synonym for mass. (e.g., "Jack weighs 170 pounds" or "Sue weighs 54 kilograms.") In this sense you are no heavier on Earth than you would be on the surface of the Moon.

A second meaning of weight is mass times the acceleration due to gravity. The original poster is correct if this was the intended meaning. That the Earth is also undergoing gravitational acceleration is irrelevant. Jack, who masses 170 pounds, is about 0.21 pounds force heavier at noon versus midnight per this definition. For Sue, who masses 54 kilos, the difference is about 0.64 newtons.

The third meaning is the quantity measured by your bathroom scale. This is also the cause of the tiredness you feel in your muscles and bones after the end of a long day on your feet, the heaviness you feel when an elevator starts going up, and the tides you see in the oceans. In this case, the original poster has things exactly backwards.

First things first: The Sun makes a person a tiny bit lighter at both noon and midnight compared to the scale weight that would result if the Sun was not present. The effect is a tiny bit larger at noon compared to midnight. You are lighter at noon (by a very, very tiny amount) compared to at midnight. For Jack, the difference is about 2.3×10-17 pounds force. For Sue, it is about 7.1×10-17 newtons.
 
  • #15
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For solar tides, the magnitude of the effect is:

(radius of earth) *G * (mass of sun) / (1 au)^3= 2.5*10^-7 m/s^2

For lunar tides:

(radius of earth) *G * (mass of moon) / (384400km)^3= = 5.5*10^-7 m/s^2

.
Your're forgetting a factor 2 here, that comes from differentiating (1/R^2)
(where R is the distance to the sun or moon)

Acceleration of gravity from sun

[tex] \frac {M_{sun} G } {R^2} [/tex]

tidal force for object at a distance R+r from the sun
(r varies from -(radius of earth) to +(radius of earth)

[tex] \frac {M_{sun} G } {(R+r)^2} \approx \frac {2 M_{sun} G r} {R^3} [/tex]
 
  • #16
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For Jack, the difference is about 2.3×10-17 pounds force. For Sue, it is about 7.1×10-17 newtons.
These are certainly MUCH too low. This is completely unmeasurable, and could never result in measurable tides.
 
  • #17
D H
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These are certainly MUCH too low. This is completely unmeasurable, and could never result in measurable tides.
No, they are not too small, and those differences are not the causes of the tidal forces.

The tidal acceleration at some point on the Earth's surface due to some other body (e.g., the Sun or Moon) is the difference between the gravitational acceleration at the point in question toward that other body and the gravitational acceleration of the Earth as a whole toward that other body.

The tidal acceleration when the Sun is directly overhead (straight up) versus directly underfoot (straight down) toward the center of the Earth are

[tex]\begin{aligned}
a_{\text{noon}} =
\frac{GM_s}{R_s^2}\left(\frac 1 {(1-r_e/R_s)^2} -1\right) \\[10pt]
a_{\text{midnight}} =
\frac{GM_s}{R_s^2}\left(1-\frac 1 {(1+r_e/R_s)^2}\right)
\end{aligned}[/tex]

where G is the universal gravitational constant, Ms is the mass of the Sun, Rs is the distance between the center of the Earth and the center of the Sun, and re is the radius of the Earth. The general equation is a bit uglier. Some items of note:
  • The tidal acceleration is directed away from the center of the Earth when the Sun is directly overhead and when the Sun is directly underfoot.
  • The acceleration is about 5×10-7 m/s2 in both cases.
  • The acceleration at sunrise and sunset is about half the acceleration at noon and midnight in magnitude but is directed inward rather than outward.
  • Small as these forces are, they are the cause of the solar tides. (The lunar tides are about twice as strong as the solar tides.)

That the acceleration at noon versus midnight are not quite equal is a second-order effect. Since the accelerations are already very small, it shouldn't be all that surprising that these second-order effects are exceedingly small.
 
  • #18
atyy
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A second meaning of weight is mass times the acceleration due to gravity. The original poster is correct if this was the intended meaning. That the Earth is also undergoing gravitational acceleration is irrelevant. Jack, who masses 170 pounds, is about 0.21 pounds force heavier at noon versus midnight per this definition. For Sue, who masses 54 kilos, the difference is about 0.64 newtons.

The third meaning is the quantity measured by your bathroom scale. This is also the cause of the tiredness you feel in your muscles and bones after the end of a long day on your feet, the heaviness you feel when an elevator starts going up, and the tides you see in the oceans. In this case, the original poster has things exactly backwards.

First things first: The Sun makes a person a tiny bit lighter at both noon and midnight compared to the scale weight that would result if the Sun was not present. The effect is a tiny bit larger at noon compared to midnight. You are lighter at noon (by a very, very tiny amount) compared to at midnight. For Jack, the difference is about 2.3×10-17 pounds force. For Sue, it is about 7.1×10-17 newtons.
What is the third meaning? I was under the impression that the bathroom scale reads m*g.
 
  • #19
D H
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What is the third meaning? I was under the impression that the bathroom scale reads m*g.
Your bathroom scale measures everything but gravitation. Gravitation is unmeasurable by any local experiment. Your bathroom scale, an accelerometer, the feeling in your gut, ...; those are all local experiments.
 
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  • #20
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Just wondering if anyone was going to bring relativity into it, Like you would have more mass when standing in the sunlight. Like hot water weighs more than cold water.
 
  • #21
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i think nobody has bothered with that b/c you can show that one's weight fluctuates without factoring in the effects of GR. besides, and correct me if i'm wrong, but wouldn't the weight fluctuations due to the effects of GR pale in comparison to the fluctuations caused by actual tidal forces, as well as any general gravitational forces and their ever-changing directions as we both rotate and orbit the sun?

by the way, willem2 said this earlier in the thread:
If you're outside, it will likely be warmer during the day, and decreased buoyancy will have a much larger effect than tidal forces.
...though i don't know if his mentioning being outside was a reference to GR.
 
  • #22
D H
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by the way, willem2 said this earlier in the thread:
If you're outside, it will likely be warmer during the day, and decreased buoyancy will have a much larger effect than tidal forces.
willem2 was correct. A 15°C swing between high and low corresponds to about a 5% change in buoyancy. Since air is 1/800 the density of water, this 5% diurnal variation in buoyancy corresponds to a 1 part in 16 thousand change in scale weight. Compare that to the 1 part in 39 million change in scale weight for solar tidal gravity (sunrise to noon) or the 1 part in 18 million change for lunar tidal gravity (moonrise to zenith).
 

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