It should be a very fundamental thing, a very simple question. But there's something I want to understand.

We know that when we throw an apple vertically upwards, it experiences a force of gravity due to the Earth, and in turn, the Earth also experiences a force acting on it, equal in magnitude. That's what Newton's 3rd law says would happen. Although the acceleration of apple towards the Earth is much largerer (because of its must smaller mass compared to Earth's) than the rate at which Earth accelerates towards the Apple. Earth's acceleration is negligible, but it is not zero.

Here's what I want to understand. Since motion is always relative, what if I am asked this question : Apple's and Earth's accelerations are with respect to which observer (or which reference frame)?

I could say that the Apple accelerates with respect to the Earth's reference frame. Because when we consider the acceleration of the Apple relative to Earth, we assume that Earth is at rest even if it is moving (relative motion). Similar to

$a_{AB}$ $=$ $a_A$ $-$ $a_B$

Here $B$ is observing $A$ and $B$ is treated to be at rest (relative to Earth) even though $B$ has its own acceleration, its acceleration is added to $A$ with a negative sign.

So I could say that the Apple is accelerating at whatever rate it is accelerating at, $relative$ to $Earth's$ reference frame. What about Earth's acceleration? I can't say that the earth is accelerating relative to the Apple because in that case its acceleration would be equal to Apple's acceleration with a minus sign. So the Earth accelerates relative to which frame? Also, is it correct to say the Apple is accelerating relative to Earth's frame (just want to confirm).

  • $\begingroup$ The amount of acceleration of the Earth due to its attraction to the apple is very tiny, as mentioned in this question. $\endgroup$ – PM 2Ring Nov 24 '19 at 14:47

Apple's and Earth's accelerations are with respect to which observer (or which reference frame)?

In any inertial frame, you will measure the same acceleration of the Earth and the same acceleration of the apple.

One example is the one in which the Earth is initially at rest (neglecting the Earth's motion about the sun, etc).

Another could be the frame in which the sun is at rest (neglecting its motion about the galactic center and due to whatever other gravitational forces it experiences)

  • $\begingroup$ So can I say that Apple’s acceleration and Earth’s acceleration, both are relative to any inertial frame? Also, when I say any inertial frame, they are inertial relative to which frame? Because Earth itself is accelerating towards the Apple so those “inertial frames” that I am talking about can’t be inertial relative to Earth’s frame? $\endgroup$ – π times e Nov 24 '19 at 5:00
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    $\begingroup$ An inertial frame is an inertial frame without reference to any other frame. Notice I gave an example of "the frame in which the Earth is initially at rest", not "the frame in which the Earth is at rest at all times". They are different, precisely because of the acceleration caused by throwing the apple, and only one is an inertial frame. $\endgroup$ – The Photon Nov 24 '19 at 5:02
  • $\begingroup$ Looks like you added to your answer when I was typing my question. I understand it better now, thanks. When you say “the frame in which the Earth is initially at rest”, you mean, a reference frame that is not experiencing any gravitational force? Due to the sun, the moon, the centre of the Milky Way..... It is not experiencing any force at all? $\endgroup$ – π times e Nov 24 '19 at 5:08
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    $\begingroup$ Reference frames don't experience forces. They aren't objects and don't have mass, so they aren't affected by gravity. If you measure in the frame where the Earth is initially at rest, the Earth will be accellerating in this frame due to those gravitational sources you mention. But AFAIK (I haven't done any calculation to check) that would not be significant in the time it takes to throw an apple in the air and have it fall back down again. (Maybe more significant than the reaction force from that action on the Earth, again I've done no calculation to check). $\endgroup$ – The Photon Nov 24 '19 at 5:20

The third law asserts that whenever two objects are exerting a force upon each other, such as gravity, the two objects are both accelerating towards the common center of mass.

(And this of course generalizes to any number of mutually force exerting objects.)

In other words, the common center of mass of the objects involved will remain in inertial motion.

Therefore with any collection of objects the natural choice of reference frame to express their acceleration is the common center of mass.

In astronomy the choice of reference frame is guided by the desired level of accuracy. The higher the required level of accuracy, the wider the required scope.

In the case of the Earth-Moon system the Moon and Earth are both revolving around their common center of mass. Actually, the Earth is so much heavier than the Moon that the common center of mass is not somewhere in between, the common center of mass is inside the Earth. Anyway, the Earth center of mass is not in inertial motion. Therefore: for a minimal level of accuracy the common center of mass of the Earth and Moon must be used.

If higher accuracy than that is needed you have to move to the common center of mass of the Solar system as a whole. Jupiter is so heavy that the common center of mass of the Solar system is a bit outside the Sun. To calculate the motion of the Moon you need to express all of the motions of the celestical bodies of the Solar system with respect to the Solar system's center of mass. From that you can evaluate all the gravitational pulls that have a noticable effect, and then proceed to evaluate the future motions of the celestial bodies.

Moving to larger and larger scales:

The Solar system is in orbit around the center of mass of our Galaxy.

Our Galaxy and the Andromeda Galaxy are exerting a gravitational pull on each other, so they are being accelerated towards their common center of mass. Given the distance between the two galaxies, and their masses, it can be predicted after how many billions of years a process of galaxy merging will commence

  • $\begingroup$ We have not studies centre of mass at school yet, but I kind of got the gist of your explanation. I am going to save this answer and will come back to it after I have studied center of mass. I know there's quite a lot for me to gain from this answer. Thanks! $\endgroup$ – π times e Nov 24 '19 at 10:29

One way to think of it is that Earth and Apple are accelerated in reference to their common center of gravity.


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