Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.

Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)

This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?

Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right.

  • 4
    $\begingroup$ you can also lose (or gain) weight by travelling around the world $\endgroup$
    – Christoph
    Commented Jun 25, 2014 at 14:53
  • 1
    $\begingroup$ @EricLippert No, he's referring to how the force from the sun opposes the force of gravity during the day and adds to it at night. $\endgroup$
    – Señor O
    Commented Jun 25, 2014 at 15:41
  • 1
    $\begingroup$ @SeñorO: Right. I've expanded this into an answer. $\endgroup$ Commented Jun 25, 2014 at 16:27
  • 1
    $\begingroup$ Well I would like to point out that kilograms is the unit used to measure mass. You are measuring weight so you should use newtons. Just a small thing, but still :D $\endgroup$ Commented Jun 25, 2014 at 16:43
  • 1
    $\begingroup$ @Christoph Yes but that depends on (a) how much you eat and (b) how much you walk while doing so. ;-) $\endgroup$ Commented Jun 26, 2014 at 9:27

9 Answers 9


This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked.


To calculate the orbital velocity at the centre of the Earth, $v$, we just note that the centripetal acceleration must be the same as the gravitational acceleration of the Sun so:

$$ \frac{v^2}{r} = \frac{GM}{r^2} $$

which gives:

$$ v^2 = \frac{GM}{r} \tag{1} $$

which is a well known result. Now consider the point on the Earth's surface nearest the Sun i.e. the black dot. The acceleration due to Earth's gravity is the usual $9.81 m/s^2$, but there will be a correction due to the fact the point is $r_e$ metres nearer the Sun. Let's calculate that correction.

The gravitational acceleration due to the Sun at the black dot is:

$$ a_g = \frac{GM}{(r - r_e)^2} $$

The centripetal acceleration due to the motion of the point around the Sun is:

$$ a_c = \frac{v^2}{r - r_e} $$

where because I've assumed the Earth is tidally locked the velocity $v$ is just the Earth's orbital velocity given by equation (1). If we substitute for this we get:

$$ a_c = \frac{GM}{r(r - r_e)} $$

So the correction to the acceleration at the black dot is:

$$\begin{align} \Delta a &= a_g - a_c \\ &= \frac{GM}{(r - r_e)^2} - \frac{GM}{r(r - r_e)} \\ &= GM \left( \frac{r_e}{r(r - r_e)^2} \right) \\ &\approx GM \frac{r_e}{r^3} \end{align}$$

where the last approximation is because $r \gg r_e$ so $r - r_e \approx r$. Putting in the numbers we get:

$$ \Delta a \approx 2.5 \times 10^{-7} m/s^2 $$

So the fractional change in the weight of an object due to the Sun is:

$$ \frac{2.5 \times 10^{-7}}{g} \approx 2.6 \times 10^{-8} $$

and the object is 0.0000026% lighter. Interestingly if you go through the working for the far side of the Earth you get exactly the same result i.e. the object on the far side is also 0.0000026% lighter. In fact this is why the tidal forces of the Sun (and Moon of course) raise a bulge on both the near and far sides of the Earth.

Incidentally, I note that Christoph guesstimated a correction of $10^{-7}$ and he was pretty close :-)

  • 1
    $\begingroup$ I think he may be right - the Earth provides the support force for the gold (thus giving it its weight). If you and the support surface are in free fall, there is no support force. $\endgroup$
    – Señor O
    Commented Jun 25, 2014 at 14:41
  • 2
    $\begingroup$ @JohnRennie let me try my point in another way: point-A on surface at night --- about 6,000 km of Earth --- point-B on surface by day --- 1 AU of space --- Sun -> would an object's measured weight differ between point A and point B, due to the difference in distance from the sun? I believe it would, even if the Earth wasn't there. $\endgroup$ Commented Jun 25, 2014 at 15:09
  • 4
    $\begingroup$ This is a good answer, but I'm going to pick on your final point about tidal bulges. Tidal forces act as you suggest, but tides operate nothing at all like this. There's is no tidal bulge, it's not like a stretched skin of water around Earth bulging out. Britain, for example, has a continuously rotating high tide on one side and low tide at the other. Many other location on Earth also see this effect. Tides are much more like very long wavelength waves that are driven by resonance with tidal forces. If you removed the driving forces, they would take some time to dissipate. $\endgroup$
    – Benjohn
    Commented Jun 25, 2014 at 21:07
  • 4
    $\begingroup$ For more on this, have a look at wikipedia's entry on tides and in particular, Amphidromic points. I'm sorry to labour this point. It's not just pedantry though – it's a huge lie that physics teachers regularly regurgitate that causes significant actual confusion. Eg – how can tides vary around a country, as they certainly do, if the "bulges" are planet wide? $\endgroup$
    – Benjohn
    Commented Jun 25, 2014 at 21:11
  • 2
    $\begingroup$ That additional factor of 0.5 (making for a total factor of 1.5) is because of the tidal acceleration when the Sun is at the horizon. Here the tidal acceleration is inward rather than outward as it is in your situation. $\endgroup$ Commented Jul 30, 2014 at 8:11

Yes your weight will change. The moon will have a bigger impact than the sun, so you need to look at the position of the moon to decide when you will be heaviest (basically - you are lighter when the moon is overhead, or on the opposite side of the earth; and heaviest when it is on the horizon. So a full moon rising makes you fat...)

The effect (the variation in $g$ over the course of a day) has been measured very carefully:

enter image description here

This figure is on page 93 of "Practical Physics" by Gordon Squires (a classical book, and one that I highly recommend). The method used is a beautiful example of careful experimental work, where the speed with which a corner cube drops is measured using an interferometric measurement. The active vibration damping of the reference mirror, clock calibration, etc are a pleasure to read - especially when you think this was done over 30 years ago. They quote a residual error of $60 nm/s^2$ or about 6 ppb. That is stunning.

Note - there is a clear asymmetry here: it is as though the tides don't pull evenly. I believe the reason for this is the relative tilt between the earth's axis and the plane of rotation of the sun and moon. I explained this with a diagram in my answer to another question.

  • $\begingroup$ You'll be heavier in the middle of the day when the Moon is half-full rather than full. A full Moon means the Moon is at nadir at noon, slightly depressing the value of g. $\endgroup$ Commented Jul 29, 2014 at 22:29
  • $\begingroup$ @DavidHammen you are right - thanks for pointing out my mistake. $\endgroup$
    – Floris
    Commented Jul 29, 2014 at 22:43
  • $\begingroup$ What is the origin of the Fourier component with a period of 24 hours? $\endgroup$
    – user4552
    Commented Jul 30, 2014 at 0:13
  • 1
    $\begingroup$ I@BenCrowell - I suspect it is a thermal effect: the earth crust expanding as it is heated by the sun, or the atmosphere heating up. One part in $10^7$ change only requires your observation point to move a mm or so... Do you think that makes sense? $\endgroup$
    – Floris
    Commented Jul 30, 2014 at 0:33
  • $\begingroup$ @Floris: If it was a thermal effect, I would think it would be very hard to calculate so accurately...? I don't know. Does Squires explain it at all? I don't have access to the Zumberge paper. $\endgroup$
    – user4552
    Commented Jul 30, 2014 at 1:37

Let's simplify.

Let's eliminate the Moon.

Let's get rid of the Sun temporarily.

Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way.

We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical springs. Each experiences a 9.8 N force towards the center of Iron Earth. The amount of distortion of each spring is identical.


Now we add in an Iron Sun. Let's connect the Iron Earth and the Iron Sun with a perfectly rigid, massless bar that prevents relative motion between them. Again, no spinning, etc.

Now what are the forces on the test masses? Call the one close to Iron Sun the noon mass.

The noon mass has a force of 9.8 N towards the center of the Iron Earth, and an opposing force of gravity from the Iron Sun towards the Iron Sun.

The midnight mass has a force of 9.8 N towards the center of the Iron Earth and a slightly smaller than before because it is farther from the Iron Sun force of gravity towards the Iron Sun, which adds to the force towards the Iron Earth.

So in this scenario, the springs are stretched by different amounts. The noon spring is stretched less than in our first experiment, and the midnight spring is stretched more than in our first experiment. There is a slight difference in the magnitude of the differences in stretchiness because of the diameter of the Earth.

Now let's eliminate the impossible massless bar connecting Iron Earth and Iron Sun, and replace it with two rockets, one on each planet, that magically pushes the Iron Earth away from the Iron Sun and vice versa exactly enough to counteract the force of gravity from each on the other. So again, they are stationary with respect to each other.

How do the springs change?

They don't. This is the same situation as before. The compression force opposing gravity that previously held Iron Earth and Iron Sun apart despite a huge force of gravity between them has been replaced by a propulsion force; this is a difference that makes no difference.

Now we turn off the rockets, so Iron Earth and Iron Sun start falling straight towards each other. What happens to the springs in the first minute?

This is the question you actually need to answer. If you work it out, you'll see that the accelerating movement of the Iron Earth towards the Iron Sun is exactly enough to compress the noon spring and stretch the midnight spring. There will still be a small difference, but it will be to the difference in the force of gravity from the Sun across the diameter of the Earth; hence my confusion in my original comment to your question. That's the difference I thought you were asking about.

The Earth is of course not falling in a straight line towards the Sun, but that's irrelevant; the acceleration vector is in that direction and that's what matters.

Fun bonus question: If we turn the rockets off at exactly the same time (as observed by an observer at rest with respect to the planets, halfway between them) does the Iron Earth start falling towards the Iron Sun immediately, or do we have to wait eight minutes for the gravity to get from the Iron Sun to the Iron Earth at the speed of light? If it happens instantly, is this a way of communicating faster than the speed of light? Isn't that supposed to be impossible? See if you can work out what happens and why.

  • $\begingroup$ As for the bonus question...the answer has a lot to do with why you needed the rockets in the first place. :) $\endgroup$
    – cHao
    Commented Jun 27, 2014 at 3:15
  • 5
    $\begingroup$ "Let's get rid of the Sun temporarily." AAAAAAAAAAAAAAH! $\endgroup$
    – Adam Davis
    Commented Jun 27, 2014 at 17:49
  • $\begingroup$ I would say the earth would immediately fall towards the sun because the earth is already in the gravitational field of the sun but I have a feeling there might be more to it. Love to hear the answer. $\endgroup$
    – strattonn
    Commented Jun 29, 2014 at 23:36
  • $\begingroup$ @strattonn: That's right. If the Sun changed its position or its mass density or some such thing then the Earth doesn't feel the change until eight minutes later. But the supposition of the problem is that the Earth is already being pulled by the Sun, so the gravity has already "gotten" to the Earth when the rockets turn off. $\endgroup$ Commented Jun 30, 2014 at 2:48

The mistake you're making is that you're looking at the full acceleration when you should look at the relative one.

At distance $R=1\mathrm{au}$ from the sun, the gravitational acceleration is given by $$ a_0 = \frac{GM_\odot}{R^2} $$

Assuming a spherical cow earth (in vacuum), at midday at the equator, we're one earth-radius $r$ closer to the sun, ie $$ a_d = \frac{GM_\odot}{(R-r)^2} = \frac{GM_\odot}{R^2} \frac{1}{(1-\frac rR)^2} \approx \frac{GM_\odot}{R^2}\left( 1 + 2\frac rR \right) $$

At midnight, we're one earth-radius farther away, ie $$ a_n = \frac{GM_\odot}{(R+r)^2} = \frac{GM_\odot}{R^2} \frac{1}{(1+\frac rR)^2} \approx \frac{GM_\odot}{R^2}\left( 1 - 2\frac rR \right) $$

With $$ \Delta a = \frac{2GM_\odot r}{R^3} $$ this reads $$ a_d \approx a_0 + \Delta a \\ a_n \approx a_0 - \Delta a $$

In both cases, we get an additional acceleration away from the center of the earth, effectively reducing the gravity of earth $g$ by $\Delta a$.

Just looking at the powers of ten $$ G=\mathcal O(10^{-10} \mathrm N \mathrm m^2\mathrm{kg}^{-2}) \\M_\odot=\mathcal O(10^{30}\mathrm{kg}) \\R=\mathcal O(10^{11}\mathrm m) \\r=\mathcal O(10^7\mathrm m) $$ shows that this is a miniscule effect of order $-10+30+7-3\cdot11=-6$.

Note that this ignores any fictious forces that need to be accounted for in an earth-based reference frame.

  • $\begingroup$ Can you explain a little more why the relative acceleration is the correct way to look at it? I'm imagining a one-dimensional free-body diagram with the Sun (S), the Earth (E), an observer O1 on the daylight Earth side, and an observer O2 on the night-time Earth side all in a straight line, with the Earth at the origin. O1 experiences a weight of g_E and a negative force of g_S (g_E - g_S); O2 experiences a weight of g_E and a positive force of g_S (g_E + g_S). In this system the only objects are the observers, the Earth, and the Sun. What's incorrect here? $\endgroup$ Commented Jun 25, 2014 at 15:52
  • $\begingroup$ @JohnFeminella Your mistake is pretty subtle here (leading to huge error in numbers). You are forgetting about acceleration. g_S pulls you and everything on Earth towards the Sun, and so we are rotating around it, and our acceleration in any part of the Earth is the same. The difference in weight comes from the fact that g_S is different for O1 and O2. They both experience the same acceleration and the same pull from the Earth's gravity. So, g_S_1 - g_S_2 is the difference. $\endgroup$
    – sashkello
    Commented Jun 27, 2014 at 1:29

Well, at that point it's pretty small, right? And I guess maybe you could take into account gravity from all other "nearby" sources (i.e. the moon and maybe other planets). In fact, the most notable effect from gravity from the sun (and the moon) is in the tides. They both affect tide cycles and strength of the high and low tides which is kind of interesting.

As for your thought about buying gold and selling (it's a tempting idea to be sure), typically I imagine that people "mass" the item instead of weighing it, which shouldn't change (theoretically).

  • 2
    $\begingroup$ Note that the effect on the tides comes more from the gravitational gradient than the actual gravitational strength (the sun's gravity is stronger but the moon affects the tides more). $\endgroup$
    – Señor O
    Commented Jun 25, 2014 at 14:30
  • $\begingroup$ Regarding the last point: When you weigh an object on a scale you are comparing the response of the balance to the test weight with the response to a calibration weight. And the two ought to change in the same way. I wrote an answer you might find illustrative: physics.stackexchange.com/a/138296/26969 $\endgroup$
    – Floris
    Commented Jul 29, 2019 at 12:23

If you are that worried about precision, you should use an inertial balance. You should search for some images, it is pretty neat. It uses a spring mechanism to measure an object's mass. Quoting the wiki:

The object to be measured is placed in the inertial balance, and a spring mechanism starts the oscillation. The time needed to complete a given number of cycles is measured. Knowing the characteristic spring constant and damping coefficient of the spring system, the mass of the object can be computed according to the harmonic oscillator model.

Otherwise, if you are going to be that worried abou the sun's influence on an object's mass, you could just as well take into account the moon's influence. Also take notice that both the orbits of Earth and the Moon are not perfect circles. So an object's weight by night at the perihelion would not be the same as by night at the aphelion, for example.

To add further confusion to it all, you could also take into account that Earth's gravity is not the same across all points in its surface, even at sea level. You could get different measures in the middle of the Atlantic and in the middle of the Pacific even if the sun and the moon didn't have any influence whatsoever.

So either go for the inertial balance, or don't lose your sleep over small variances. For sensitive things like gold, as you mention, I am sure that specialists that deal with such things have their ways to handle quantities of those materials besides working just with weight.


This calculation is wrong. The earth is falling towards the sun and there is also centripetal acceleration, both things that change the weight of an object. The acceleration of the earth, and the centripetal acceleration both act on all objects on earth and (as the equivalence principle is remarkably accurate, if not exact) completely cancel out the effects of the gravitational acceleration of any far-away body. There are remaining effects on the weight that result from the gradient of the acceleration of the Sun and Moon, resulting in tides. These effects are small and (as others have said) are taken into account when great accuracy is needed by measuring the mass, using a balance, rather than the weight. However, actually, these effects are important to our current understanding of measurements of mass / the kilogram. The current definition of kilogram is in terms of a couple of objects (in Paris and Washington DC.) We would like to have there be a better definition of the kilogram, in terms of the force between two wires carrying known currents. Making this better than those artifacts is hard, in part because of tides.

  • $\begingroup$ The current definition of mass is a property of matter that allows bodies to interact with each other through gravitation. Those objects are not the definition of mass - as far as I know, at least the one in France is the kilogram standard. It is just an object for which the mass is, for all purposes, was considered to be one kilogram. But that is not a definition of mass. $\endgroup$ Commented Jun 25, 2014 at 14:51
  • $\begingroup$ Thanks. I agree. Changed it to "kilogram", which is more accurate. $\endgroup$
    – user47505
    Commented Jun 26, 2014 at 15:19

Setting aside tidal force. Distance to sun aprox. 150*10^9m +(-)637*10^3m ,and since earth(masspoint) always in free fall it makes on average 0.006m/s2*(150E9m+637e3m)^2/(150E9m-637E3m)^2 -0.006m/s2 = 1.02E-7 This mean 1*10^-7m/s2/9.8m/s2*70kg/2 => 0.35mg HEAVIERon night and same 0.35 mg LIGHTER days at equator.Days we are pulled 9.8+1*10-7/2 and nights 9,8-1*10^-7/2. Centripetal force ads 0.006*(150E9+6.37E3)/150E9-0.006 => +(-) 0.18 mg This classically , but the fact that sun have 35% greater gravitation (g) on objects at orbit speed 0m/s compared to 30000m/s +460m/s -450m/s , is still under work Yours Timo Moilanen


The simple question asks whether we gain weight as the Earth rotates us away from the Sun (towards evening) and we lose weight as Earth rotates towards the Sun in the morning. Considering hysteresis, it would be maximized slightly after sunset and after sunrise. The effect would be a sideways pull at sunset not directly subtracting from the earthly gravitational direction and mirrored for sunrise. The poles would be null effect. The value would be doubled for 2 effects. Seemingly a 45 degree angle would be reasonable to use. Then at 9:PM and 9:AM. The proximity factors would also be considered because after sunset you would be farther to the sun and after sunrise you would be closer. The Earths spin can not be excluded from the parameters. They add the only interesting point to this whole question. The language would not be other than a real world case. Keep it simple. The speed of acceleration from the sun must be calculated. We reach close to 1670 KM per hour away then the same towards the Sun on return. A body of 100 KG would be pulled by .06 kg at rest using the .0006 times the Earths pull for the suns strength. The weight would be added in the afternoon and lessened in the morning. Like spinning a rock at the end of a string while standing here now. The answer is yes we are heavier and lighter at the respective opposite times of day. .06 KG equals about 1 ounce. Maybe .06 + .06 times .707 = .084 Kg. I submit the gravity would be doubled when accelerating away as any acceleration however provided even if by the rotation of the Earth. That as in a ball thrown up while standing here now. What is the force of us accelerating from the sun at that rate? A 100 KG mass. The overall effect is cancelled out in terms of projectile geometry of the Earth through orbit and outer space, (allowing for tidal effects) but the hysteresis allows the obvious detection at specific times. The question is How would you measure it if all the scales floated likewise? Balance scales would not work to detect. It would have to measure force directly. A word to the wise: only buy gold weighed at 9:am sell gold weighed at 9:pm.

  • 1
    $\begingroup$ The other answers suggest a significantly smaller increase/decrease with the tidal force, so how did you get to your numbers? $\endgroup$
    – Kyle Kanos
    Commented Sep 28, 2015 at 21:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.