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I am asking what might be the weight (not mass!) of the earth. My weight is the amount of force exerted on me by Earth (say 600N). So, weight of earth should be amount of force exerted on Earth by me (600N)? And it also seems to be changing from person to person. Where am I wrong?

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    $\begingroup$ Weight measured where/how? "Weight" is generally the force directed towards the surface of a significantly more massive object. You might be able to argue that the Earth's weight on you is 600 N; but the total weight of the Earth isn't really something you could determine without more context. $\endgroup$ – JMac Aug 17 '17 at 20:59
  • $\begingroup$ @JMac I think he did provide context in asking if technically the weight of the Earth was also 600N because that's the gravitational force his body is also pulling the Earth with. So probably the answer he's looking for is to just say "weight" is relative to what gravitational field the object is in. $\endgroup$ – ison Aug 17 '17 at 21:23
  • $\begingroup$ @ison Yeah, I get what you mean I guess. I'll try and make my comment into something that fits as an answer. $\endgroup$ – JMac Aug 17 '17 at 21:26
  • $\begingroup$ JMac, what do you mean by 'Earth's weight on you'? And where is my logic flawed in saying, earth's weight should be the amount of force exerted by me on earth? $\endgroup$ – quantised Aug 17 '17 at 21:43
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    $\begingroup$ @Conifold I don't think that's really a good metric to use. The Earth orbits the sun, so it constantly feels weightless relative to it. $\endgroup$ – JMac Aug 18 '17 at 13:06
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It seems you are trying to apply weight inconsistently.

We generally define weight as "force on an object due to gravity."

You cannot really define weight of an individual object. When defining weight we usually are implicitly talking about the objects weight on Earth. We can also find the weight of objects on the moon for example. Objects with the same mass will obviously have different weight in different gravity.

This is because gravity is an attractive force between two objects. The question of "What is the weight of Earth?" has no obvious answer, because generally when we don't specify where, we mean on Earth. The question would then read "What is the weight of Earth on Earth?", which is obviously nonsense.

In your example, you could ask the question "What is the Earth's weight on me?", and then it would be 600 N. The weight of the Earth for someone heavier would be more.

It's not really standard to talk about the Earth's weight on small objects either, which is why it may seem odd to consider that.

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  • $\begingroup$ Good answer except for the comment about the normal force. No! The normal force itself has nothing to do with gravity. In many cases, gravity is needed to manifest the normal force (the normal force of a book on a table) but in other cases, no (the normal force of a hammer struck against an anvil). $\endgroup$ – garyp Aug 18 '17 at 14:47
  • $\begingroup$ @garyp Actually, it is a little vague and potentially confusing, I'm going between multiple frames of reference. I'll remove it. $\endgroup$ – JMac Aug 18 '17 at 15:01
  • $\begingroup$ I don't understand what you are trying to say. Consider the normal force on a book due to the table. It might be numerically equal to the weight of the book, but that's only because Newton's second law tell us that it is: the normal force must balance weight. There is no reciprocity argument, no need to invoke the third law. $\endgroup$ – garyp Aug 18 '17 at 15:01
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Weight is a force, with which one object attracts another via gravity. Therefore weight is relative to the object. If your weight on the Earth is 600N, then your weight on the Moon would be only 100N. And if your weight on the Earth is 600N while mine is 1000N, then the weight of the Earth on you would be 600N, while the weight of Earth on me would be 1000N at the same time. So clearly, as others have pointed out, measuring the weight of Earth has no practical meaning and therefore is not a useful concept.

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I suppose the closest thing would be the pressure at the core, but that's not a weight as such.

For comparison an average human standing on the surface of Earth exerts a pressure of about 16 psi or 110 kPa and the atmospheric pressure is about 100 Kpa.

The pressure at the core is thought to be about 300 GPa. So about 3 million times more at the core that the effect of an average human on the surface.

And to put that in perspective, the average human adult has a mass of around $70 kg$ and the Earth has a mass of around $6\times 10^{24} kg$, so the human is sort of punching well above it's weight, so to speak. :-)

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  • $\begingroup$ If you down-voted I'd prefer you state why because I can't address (or indeed learn) if I don't know your objection. Thanks. $\endgroup$ – StephenG Aug 18 '17 at 19:39
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We can, if we like, work out the "weight" of Earth due to Earth's own gravity, which is far greater than the gravity we individually exert on it. Let's assume constant density (the truth is more complicated, but will give an answer within an order of magnitude of what we get this way).

Say Earth has mass $M$ and radius $R$, so its surface gravity is $g:=GMR^{-2}$. (Obviously, we expect an answer comparable to $Mg$.) However, at a distance $r<R$ from Earth's centre the gravitational field strength is $G\frac{Mr^3/R^3}{r^2}=g\frac{r}{R}$. The spherical shell of radius $r$, thickness $dr$ has volume $4\pi r^2 dr$, which is a fraction $\frac{3r^2}{R^3}dr$ of Earth's volume $\frac{4}{3}\pi R^3$. So this mass-$\frac{3Mr^2}{R^3}dr$ shell feels a gravitational field strength $\frac{gr}{R}$, and has weight $\frac{gr}{R}\cdot\frac{3Mr^2}{R^3}dr=\frac{3Mgr^3}{R^4}dr$. Now we just integrate:$$\int_0^R \frac{3Mgr^3}{R^4}dr=\frac{3Mg}{R^4}\frac{R^4}{4}=\frac{3}{4}Mg.$$As already mentioned, a non-uniform density gets a different coefficient for $Mg$. For what it's worth, $\frac{3}{4}Mg=\frac{3GM^2}{4R^2}$.

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I think that you are right in your analysis.

The weight of an object on the Earth is short for the attractive force on an object due to the object being in the gravitational field of the Earth.
Ignoring the rotation of the Earth it is the reading on the bathroom scales on which the object is placed.

Now look at the situation of the object creating the gravitational field and the Earth being in that gravitational field.
You can then think of the bathroom scales as measuring that force but I do not think that this force is generally called the weight of the Earth.

The reason that you do not usually worry about the force on the Earth due to the object is because that force does not change the motion of the Earth by very much because the Earth is so massive as compared to the object.

Update as a result of a comment made about the following statements made in my answer:

it [the force on the object due to the gravitational attraction of the Earth] is the reading on the bathroom scales on which the object is placed.

You can then think of the bathroom scales as measuring that force [the force on the Earth due to the gravitational attraction of the object].

The reason that bathroom scales show a reading which is equal to the force on the object due to the gravitational attraction of the Earth is as follows.

In static equilibrium on the Earth the object has two equal in magnitude and opposite in direction forces acing on it (Newton's second law) - the force on the object due to the gravitational attraction of the Earth and the force on the object due to the spring in the bathroom scales.

The Earth has two equal in magnitude and opposite in direction forces acing on it (Newton's second law) - the force on the Earth due to the gravitational attraction of the object and the force on the Earth due to the spring in the bathroom scales.

The spring has two equal in magnitude and opposite in direction forces acing on it (Newton's second law) - the force on the spring due to the Earth and the force on the spring due to the object.
These two forces acting on the spring cause a change in the spring which is shown as a reading on the bathroom scales equal to the magnitude of those forces acting on the spring.

The Newton's third law pairs of forces which make them equal in magnitude and opposite in direction are:
The force on the object due to the gravitational attraction of the Earth and the force on the Earth due to the gravitational attraction of the object.
The force at one end of the spring due to the Earth and the force on the Earth due to that end of the spring.
The force at the other end of the spring due to the object and the force on the object due to that other end of the spring.

Since the magnitude of all the forces is the same one can say that the reading on the bathroom scales is equal to the force on the object due to the gravitational attraction of the Earth and also the force on the Earth due to the gravitational attraction of the object.

The force on the object due to the gravitational attraction of the Earth is called the weight of the object but the force on the Earth due to the gravitational attraction of the object is generally not called the weight of the Earth.

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  • $\begingroup$ This kind of argument is often ok, but in a detailed answer to a question like this, more care is needed to avoid creating misunderstanding. A bathroom scale measures gravity only indirectly. It does this by actually measuring the normal force between your foot and the scale, and then invoking Newton's second law to make the connection to gravity. $\endgroup$ – garyp Aug 18 '17 at 14:52
  • $\begingroup$ @garyp The spring in the bathroom scale is subjected to two equal and opposite forces one of which is the downward force on the spring due to the object sitting on the pan. The other force on the spring which acts at the other end of the spring is the force due to the base of the scales pushing up. The spring length changes so that it then exerts equal magnitude forces on the object and the base. Perhaps this is a chicken and egg situation? $\endgroup$ – Farcher Aug 18 '17 at 15:13
  • $\begingroup$ No chicken and egg. Both of those forces, one on either side of the scale, are normal forces. The only gravity force is the force on you due to the earth. (Note no mention of the scale in that last sentence.) That the scale can be used to measure weight is a consequence of Newton's second law. The scale does not directly measure weight. It measures normal force. $\endgroup$ – garyp Aug 18 '17 at 16:14
  • $\begingroup$ @garyp Thank you for your comments which I have acted on by adding an update to my answer. $\endgroup$ – Farcher Aug 19 '17 at 5:48

protected by Qmechanic Aug 18 '17 at 2:36

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