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What's the weight of the earth?

That is, what's the weight of the earth in the gravitational field of the sun? I imagine it's either

  • $0$, because there's no net fall toward the sun, so the downward component of acceleration is $0$ and $F=ma=0$, or
  • $Gm_1m_2/r^2\approx7\cdot10^{-11}\cdot6\cdot10^{24}\cdot2\cdot10^{30}/(1.5\cdot10^{11})^2\approx4\cdot10^{22}$ Newtons,

but which, and why?

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There's no net fall toward the sun? But then how is the orbit maintained? ;) – MattS Jul 1 '13 at 10:30
Related: – Qmechanic Jul 1 '13 at 23:27

In Newtonian physics, the Earth accelerates towards the Sun. The force between the Earth and the Sun is about 3.4*10^22 Newtons.

The distance between the Earth and Sun is roughly constant, but that isn't enough to say the acceleration is zero. Acceleration is the rate of change of velocity, and velocity is a vector. Velocity changes either when the speed of Earth changes, or when the direction it's moving changes. Because the Earth is going in a circle, the direction of its motion is always changing. That means it's accelerating.

The Earth doesn't move in a circular orbit, but it's pretty close. In a circular orbit of radius $r$ and velocity $v$, the acceleration is

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

which is found using calculus.

Newton's second law, $f =ma$, can be combined with Newton's gravitational law, $f = \frac{Gm_1m_2}{r^2}$ to give

$$\frac{v^2}{r} = \frac{G M}{r^2}$$

with $M$ the mass of the sun. This is called Kepler's third law, and tells you how fast planets go in their orbits at different distances from the sun. It is usually restated to say that the period of the orbit scales as the 3/2 power of $r$.

Your attempt to calculate the force between the Earth and Sun is set up correctly, but executed incorrectly. Always include your units as you go along.

Whether or not the Earth "has weight" is a moot point of semantics. Regardless, in the Newtonian picture, Earth certainly does have a force on it from the Sun and is accelerated by that force.

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+1, and thanks. I've edited the question to read "downward component of acceleration" for clarity. Note that if you're viewing acceleration as including a direction, then its a vector, and you should view $v$ as a vector ($\mathbf v$ I guess) also, in which case you can't square it. I don't understand your formulas, then. – msh210 Jul 1 '13 at 15:44

The answer to your question depends on your definition of weight. A common definition, is that weight is the force on the object resulting from local acceleration/gravity as measured by a scale.

With that definition, an astronaut satellised around the earth is weightless as gravitation is balanced by the acceleration it causes (if you accelerate without gravity, the force due to inertia, measurable as weight, is in the direction opposite to the acceleration). Actually, the same is true if he is not satellised and is just accelerating towards the planet (physically the same situation, though the astronaut might disagree).

Your weight on a planet will depend on the gravity on the surface of that planet (and a little bit on rotation speed), or in any internal part that you can reach.

What is implicit there, is that somehow you need to define the scale used to mesure the weight. On a planet the scale can be anchored to the planet spot where you are measuring weight. In space, you can reasonably anchor it only to something following the same trajectory at the same time, and you will mesure zero in free fall (not quite actually if there are tidal effects), which is what was happening to our astronaut, in both cases.

Hence, I am not sure what weight of the earth would mean. Where is it measured and under what conditions. Essentially I would reach your first conclusion and say that, like the astronaut, Earth is weightless,

Now you could also want to define weight as just the force resulting from gravity produced by a nearby body, by somehow considering that you are at rest with respect to that body. You still have to know exactly where, so as to determine gravitational acceleration. Then you can compute the weight of the earth with respect to the sun, at a distance corresponding to the earth orbit. This gives of course a different result, depending on the mass of the planet as you know and noticed.
You are then ignoring the sun rotation, though rotation is taken into account (however small the effect) if you measure weight at the surface of a planet.

This is the reason why the relevant physical concept is mass rather than weight. The distinction of mass and weight was a major progress in physics.

Note that the first and the second definitions of weight will not give you exactly the same weight for a person at the surface of the planet. What is measured by your bathroom scale is the first, which helps a little bit feeling in shape. Earth rotates.

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+1, and thanks. – msh210 Jul 1 '13 at 15:45

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