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How do we calculate weight of an object which is orbiting another bigger object in the space?

For example, comparing the pair of Sun-Earth, Earth is orbiting Sun due to the force applied by Sun on Earth. Earth is undergoing non-uniform acceleration.

Does Earth has weight? I think it has weight, but no sensation of it since it is in free fall. Is the weight of Earth changing every moment? Why it is believed popularly that Earth doesn't have weight?

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    $\begingroup$ Do you know the difference between weight and mass? $\endgroup$ – Steeven Nov 10 '18 at 11:45
  • $\begingroup$ Yes, I know. That is reflected in the question. no where, I mentioned mass. $\endgroup$ – KawaiKx Nov 10 '18 at 15:38
  • $\begingroup$ Okay, then let me ask: what is your weight if you were floating around in space? $\endgroup$ – Steeven Nov 10 '18 at 16:28
  • $\begingroup$ There is no floating around in space exactly. You would be pulled to the nearest big object. Either you go straight toward it or orbit around it if you other velocities too. So called 'free fall' is due to weight only. $\endgroup$ – KawaiKx Nov 10 '18 at 18:18
  • $\begingroup$ And if no other big objects existed in any significant nearby location? Then you would free float, right? What would the weight then be? $\endgroup$ – Steeven Nov 10 '18 at 18:24
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Weight and mass are two different things:

  • The mass of an object is an intrinsic property of the object itself. Mass does not have a direction. Mass is often expressed in kilograms. A person's body has the same mass whether that person is standing on the earth, standing on the moon, or falling off a roof.

  • The weight of an object depends on how it is moving, and weight has a direction. An object in free-fall is weightless. An object that is not in free-fall has weight. The amount of weight is has depends on how much the object is deviating from free-fall, and the direction of its weight depends on the direction in which it is deviating from free-fall. Weight is often expressed in pounds or Newtons. The same person's body has one weight when standing on the earth, a different weight when standing on the moon, and no weight at all when falling off a roof. (Please don't try this at home.) We can even say that the same person's body has different weights when standing on opposite sides of the earth — same magnitude, but different direction.

So, what about the earth?

  • The earth has an enormous mass. Many, many, many kilograms.

  • The earth, regarded as a single object, is in free-fall around the sun. Therefore, regarded as a single object, the earth is weightless.

For an explanation (with a nice picture) of why an orbiting object qualifies as a falling object, see this post.

If we used a giant rocket engine to push the earth out of its orbit, then the earth would have weight while it is being pushed by the rocket, because it is deviating from free-fall. The amount of weight it would have would depend on how hard the rocket is pushing, and the direction of the weight would depend on the direction in which the rocket is pushing.

We can also regard the earth as a being composed of lots of parts — mountains, oceans, magma, and so on. If we consider just one part of the earth, say just one mountain, then that mountain has a large weight. It has a large weight because the substrate underneath the mountain is pushing up on it with just enough force to prevent the mountain from freely falling toward the center of the earth. The earth has a large mass, and mass creates gravity, so all the parts of the earth would fall toward the center if they were not prevented from doing so by pushing on each other. In this sense, each part of the earth has weight, because those individual parts are not in free-fall; the other parts underneath them (underneath = closer to the center of the earth) are pushing up on them (up = away from the center of the earth) just enough to prevent them from freely falling toward the center.

But when the earth is regarded as a single object in orbit (freely falling) around the sun, the earth is weightless overall.

The question mentioned the word "acceleration." This is worth clarifying. The word acceleration is used with two different meanings:

  • It is used to mean that the distance between the object and some other reference object (like the distance of a falling rock from the surface of the earth) is changing at a non-constant rate. This is sometimes called "relative" acceleration. It is undefined unless a reference is specified. When the word is used this way, the same object may be accelerating with respect to one reference and not accelerating with respect to a different reference. This has nothing to do with the object's weight.

  • It is used to mean that the object not in free-fall. (More technically: it is used to mean that an object is not following a geodesic in spacetime.) This does not depend on any external reference. If an object is accelerating in this sense, then it has weight.

If we carelessly use the same word with both meanings, we end up with crazy-sounding statements, kind of like this famous example: "Feathers are always light (opposite of heavy) even though sometimes they're not light (opposite of dark)."

In summary:

  • If the earth is regarded as a single object in orbit (freely falling) around the sun, the earth is weightless. We can rightly say that it weighs zero pounds (or zero Newtons). Individual parts of the earth (like individual mountains) have weight because they are not in free-fall, and their weights are in different directions depending on where they are located on the earth. But overall, the earth is weightless, because it is in free-fall overall when regarded as a single object.

  • Mass is simpler. The earth's overall mass is just the sum of the masses of all of its parts. Mass does not have a direction and does not depend on how the object is moving. The earth has a very, very large mass — lots and lots of kilograms.

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  • $\begingroup$ isn't free fall due to 'weight' ? if there was no weight, how can something fall freely towards other object? $\endgroup$ – KawaiKx Nov 10 '18 at 15:42
  • $\begingroup$ @KawaiKx My answer was about language. It tried to clarify what physicists usually mean when they use the word "weight". But when the goal is to really understand nature, words are not enough, because words are too vague. (That's why physicists like equations.) If you can ask your question without using the word "weight", that might help us understand what you're trying to ask. $\endgroup$ – Chiral Anomaly Nov 10 '18 at 21:23
  • $\begingroup$ Thanks! It's clear now. I was a victim of definition. Term 'weight' is to be used wrt Earth only. $\endgroup$ – KawaiKx Nov 11 '18 at 5:32
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Weight is just another name for gravitational force. It has the general formula:

$$F_g=G\frac {Mm}{r^2}=mg$$

In the latter version, we have collected a lot of parameters in the symbol $g$: $g=G\frac {M}{r^2}$.

At any point in space, this can be calculated between all attractive objects. For the example of earth orbiting the sun, this can be calculated as well by inputting the distance $r$. The force is the same if the object is orbiting or just falling straight down (the difference there is just the tangential speed).

Usually, we won't call it weight in astronomical cases, though. The term "weight" is typically used when referring to the force on a very tiny object being pulled in by a very large object. The earth pulls in you with a force which you call your weight, but remember that by Newton's 3rd law, the earth is being pulled in with that same force towards you. It is correct to say that the gravitational force by the earth on you equals the gravitational force by you on the earth. But would you say that the earth's weight equals your weight? No, because that is not how we use the word "weight", although it would strictly be correct.

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    $\begingroup$ Thanks! It's clear now. I was a victim of definition. Term 'weight' is to be used wrt Earth only. $\endgroup$ – KawaiKx Nov 11 '18 at 5:31
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If you use the ISO definition of weight ($F_g=mg$), all you need is the mass $m$ of the object (e.g. the Earth) and the local acceleration of free fall, named $g$ in this definition (not to be confused with $g$ the standard gravity at Earth's surface).

Hint: If we assume that the Earth is kept in a stable circular orbit around the Sun, and affected by no additional forces, the local acceleration is the centripetal force for this circular motion, i.e. $g = r\omega^2$ where $r$ is the Sun-Earth distance, and $\omega$ is the angular frequency of the orbit (2$\pi\; \mathrm{radians}/1\; \mathrm{year}$). Plug in the appropriate numbers to calculate the weight $F_g$.

EDIT: Note that in a reference frame that is co-moving with the orbital motion, the Earth's weight disappears since $F_g$ now also includes the centrifugal force that appears in the rotating frame. This cancels the centripetal force exactly.

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