Why is it that, in basic physics, it is taught that the acceleration of gravity is the same for objects of any mass (at a constant distance from earth)? Why then is the acceleration of gravity different on planets with different mass? If acceleration is always $9.8\,\mathrm{m/s^2}$ for gravity, since gravity exists between all objects, won't all objects be accelerating towards each other at $9.8\,\mathrm{m/s^2}$? Or is the difference in acceleration so negligible that it is thought to be non-existent?
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/6074/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Sep 18, 2015 at 6:30
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$\begingroup$ en.wikipedia.org/wiki/Surface_gravity $\endgroup$– Alfred CentauriCommented Sep 18, 2015 at 11:04
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If we have an object of mass $m$ at a distance $r$ from the earth then Newton's law of gravitation tells us that the force between the object and the earth is given by:
$$ F = \frac{GM_em}{r^2} \tag{1} $$
where $M_e$ is the mass of the earth and $G$ is the gravitational constant. Since the Earth is much heavier than (most) objects near it we can take the Earth to be fixed and the object to accelerate towards it. Newtons second law tells us that $F = ma$, so the acceleration of the object $m$ is given by:
$$ a = \frac{F}{m} = \frac{GM_e}{r^2} \tag{2} $$
Note that the mass of the object $m$ does not appear in this equation, and that's why the acceleration doesn't depend on the mass of the falling object.
But the acceleration does depend on the mass of the Earth $M_e$ and the distance $r$. So if you go to a different planet, say with a mass $M_x$, the acceleration would now be:
$$ a = \frac{GM_x}{r^2} $$
That's why the acceleration at some fixed distance $r$ is different on different planets.
If we don't make the approximation that the Earth is fixed then the motions of the two bodies become more complcated because both accelerate towards the centre of mass. For example in the Earth-Moon system both bodies accelerate towards a point about two thirds of the way from the Earth's centre to its surface.
However the force on the two bodies is still given by my first equation, so the acceleration (that is the quantity you'd measure using an accelerometer on the object) is still given by my second equation. But if you define the acceleration by the change in the distance between the objects:
$$ a_r = \frac{d^2r}{dt^2} $$
then this would be greater than equation (2) calculates.
answered Sep 18, 2015 at 5:09
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$\begingroup$ But supposing we did not assume earth to be fixed, wouldn't the acceleration then depend on the mass of the objec as well, since the earth would be accelerating (at an extremely small rate) towards the object, thus making the apparent acceleration of the object (towards the earth) seem greater? Understanding that it is negligible in practice of course. $\endgroup$ Commented Sep 18, 2015 at 5:13
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$\begingroup$ @sneelhorses: I've extended my answer to respond to your comment $\endgroup$ Commented Sep 18, 2015 at 5:35
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$\begingroup$ Thank you. I understand it more clearly now. I suppose instead of acceleration, I was looking for the 2nd derivative of the distance between the two objects. Is there a term for that that is less cumbersome? $\endgroup$ Commented Sep 18, 2015 at 5:38
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$\begingroup$ @sneelhorses: you have to calculate the position of the centre of mass and the effective mass of the system, then calculate the acceleration of the two objects towards the cebtre of mass. It's actually not that hard but it's messy enough that I feel no great urgency to do it here! $\endgroup$ Commented Sep 18, 2015 at 5:42