# How much lighter do I weigh at noon than I do at dawn?

1. May 7, 2009

### thenewmans

Maybe I should ask about the moon too but once I get the sun down, maybe I can do the moon myself. I think the best indicator of my weight change is the tide. So the moon has a greater effect. Still the sun has some effect as we get closer and farther. I know that the centrifugal force of our orbit is in balance with the sun’s pull. So there should be 2 solar high tides a day. And that’s why I’m comparing noon to dawn.

I’m sure I’m doing this wrong but here it goes. I started with F = G * Mme * Msun / R^2. Since I’m comparing noon to dawn, most of that cancels out to give me Fdelta = Rdawn^2 / Rnoon^2. Sticking to meters (SI units), I get:

149,600,000,000^2 / (149,600,000,000 - 6,378,000)^2 = 1.00008527

I assume this change in force matches my change in weight. So if I way 200lbs, my weight changes by 3oz. That sounds like too much. Where did I go wrong? Is it that assumption?

2. May 7, 2009

### cesiumfrog

chroot, that's off by 3 orders of magnitude. I'm not sure why you were looking at ratios; in fact that equation didn't even depend on the sun's mass, and gave kilograms for the lunar case (rather than just the factor of two, explaining king tides).

Last edited: May 7, 2009
3. May 7, 2009

### chroot

Staff Emeritus
Er, yeah. I'll delete mine. Ha!

- Warren

4. May 7, 2009

### D H

Staff Emeritus
Not so fast, chroot.

That equation doesn't even make sense. It has units of mass. Why are you dividing by "gravity on earth"? Without that division, you get 35 micronewtons, compare this to 686 newtons nominal weight for a 70 kg person.

5. May 7, 2009

### cesiumfrog

How perchance could you make that comparison, except by employing a factor of "gravity on earth"? I made that division because I wanted to convert to the mass units in which people ordinarily recognise weight.

6. May 8, 2009

### D H

Staff Emeritus
No one has answered the original post.

The answer depends on what one means by "weight". IF you define weight as mass times the total gravitational acceleration of an object, a 200 pound mass person will weigh nominally weigh 200 lbf and will weigh
• 3.66×105 lbf more than nominal at dawn (or sunset),
• 0.121 lbf less than nominal at noon,
• 0.121 lbf more than nominal at midnight.

This the definition of weight (mass times gravitational acceleration) is the definition used in many elementary physics texts (up to and including freshman physics). Except for computing the behavior of aircraft, this is not a particularly useful definition. The reason: Weight defined in this sense is not measurable.

If you define weight as that quantity measured by an ideal spring scale you will get a very different picture. The Earth is accelerating toward the Sun, and you need to take this into account when computing the measured weight. The Sun will make the measured weight of a 200 pound mass person be
• 1.03×10-5 lbf less than nominal at noon and at midnight,
• 5.16×10-6 lbf more than nominal at dawn (and at sunset).

7. May 8, 2009

### thenewmans

D H,
Thanks! I see my flaw! 200lbs is my mass. But what I need is my weight from the sun. To the sun, we weigh only 2 ounces.

2 * 1.00008527 = 2.00017054

So I weigh 0.00017 ounces less at noon than at dawn. My answer came out a little different than yours. But at least I got a similar order of magnitude.

8. May 8, 2009

### cesiumfrog

It hadn't occurred to me that, by "weight", the OP meant anything other than what is read when standing on a scale. I understand how you calculated the first number (I gave the formula for it), but where are you getting the second number and how are you trying to define "nominal"?

9. May 9, 2009

### Chronos

That seems a bit high to me. Clocks suggest a smaller difference.

C

10. May 9, 2009

### D H

Staff Emeritus
The common definition of weight (in physics) is mass times gravitational acceleration. To most people, weight is what a scale measures, which is something quite different than this standard definition of weight. (Legally, weight is a synonym for mass, but that's a different issue).

So what do scales measure?
• Newtonian POV: Scales measure the net non-gravitational force acting on a body. The gravitational component of the net force is not measurable because there is no way to shield the gravitational force.
• General relativistic POV: Scales measure the net force acting on a body. There is no gravitational component of the net force as gravitation is a pseudo-force in general relativity.

The total apparent force acting on an object in some arbitrary (Newtonian) reference frame is Ftot=Ff+S+W where Ff is the total fictitious force in this frame, S is the object's scale weight (aka apparent weight), and W is the object's true weight (mass times gravitational acceleration). Suppose the object is stationary in some reference frame. The total apparent force is necessarily zero in this frame. The object's scale weight (a vector) is thus S=-(Ff+W), where Ff is the fictitious force acting on the object in a reference frame in which the object is stationary. Any other means of computing the scale weight must necessarily yield the same result as that obtained by computing forces in a frame in which the object is stationary.

The topic of discussion is the weight (scale weight) of someone standing still on the surface of the Earth. This person is stationary in the acceleration and rotating Earth-fixed frame (e.g., a frame with origin at the center of the Earth and rotating with the Earth). As the desired quantity is the Sun's contribution to scale weight, I'll ignore rotational effects. Both the gravitational acceleration of a person and the Earth toward the Sun contributes to the person's scale weight. The net result:

$$\boldsymbol W_s= GM_sm_p \left( \frac{\boldsymbol r_s - \boldsymbol r_p}{||\boldsymbol r_s - \boldsymbol r_p||^3} - \frac{\boldsymbol r_s}{||\boldsymbol r_s||^3} \right)$$

where $\boldsymbol W_s$ is the Sun's contribution to a person's scale weight, $GM_s$ is the Sun's standard gravitational parameter, $m_p$ is the person's mass, $\boldsymbol r_s$ is the vector from the center of the Earth to the center of the Sun, and $\boldsymbol r_p$ is the vector from the center of the Earth to the person.

Let $theta$ be the angle between $\boldsymbol r_s$ and $\boldsymbol r_p$. Denote two orthogonal unit vectors, $\hat x$ and $\hat y$ such that $\boldsymbol r_s = r_s \hat x$ and $\boldsymbol r_p = r_e (\cos\theta \hat x+\sin\theta \hat y)$. Here r_s (scalar) is the distance between the Earth and Sun, nominally 1AU and r_e is the radius of the Earth (I'm assuming a spherical Earth for simplicity). After a bit of grinding, the Sun's contribution to a person's scale weight is, to first order,

$$\boldsymbol W_s\approx m_p\,\frac{GM_s}{r_s^2}\,\frac{r_e}{r_s}(2\cos\theta \hat x - \sin\theta \hat y)$$

At theta=0 (Sun directly overhead), this is 2mpGMsre/rs^3 directed sunward (i.e., away from the center of the Earth: the person is a bit lighter). At theta=180o (midnight at the equator on one of the equinoxes), this first order approximation has the same magnitude as the theta=0 value, but is directed anti-sunward. This is also directed away from the center of the Earth, once again making the person a bit lighter. The magnitude of the Sun's contribution is at a minimum at theta=90o (sunrise and sunset). However, the direction at this minimum magnitude is inward: The person is a bit heavier.

Gravitational time dilation due to the Sun's gravity is fairly small because the distance between the Earth and Sun is much, much greater than the Sun's Schwarzschild radius (~3 km). The difference in gravitational time dilation at sunrise and noon is extremely small because this involves the difference between two very small and very similar numbers.