Is the acceleration due to gravity the same no matter the mass of the object? [duplicate]

This question already has an answer here:

The University of Illinois Department of Physics article Q & A: Heavy and Light - Both Fall the Same states that the acceleration due to gravity is the same no matter the mass of the object.

However if an object, say for example the moon, were falling to the Earth, gravity would have to pull harder to speed it up the same. Would the acceleration of gravity be the same?

My question is not the same as Don't heavier objects actually fall faster … because I'm asking about acceleration of gravity, I'm asking if heavier objects would fall slower, not faster. So, wouldn't they fall slower because gravity has to pull harder?

Side note: Let's just say we also ignore the effects the other question mentions.

marked as duplicate by caverac, sammy gerbil, Chris♦, Jon Custer, valerioFeb 24 '18 at 9:45

• OK, so let's believe that the moon falls slower because it's larger. Cut it in half. How fast does each half fall? – Hot Licks Feb 22 '18 at 0:02

Although the "gravity has to pull harder" to accelerate a more massive object, the gravity also has "more to pull on."

Consider Newton's second law $F = ma$. As there is more mass, there is more force. Therefore the acceleration stays the same.

No, the acceleration is the same, and so masses of any weight fall at the same speed after a given time. To see this consider two masses, $m$ and $M$, and assume that $M\gg m$ or, equivalently, that $M$ is fixed in space somehow (this is to avoid the problems associated with $M$ moving which causes heavier objects to effectively fall faster as discussed in the other question.

So, what is the (magnitude of the) force due to gravity between $m$ and $M$? Well, Newton talls us that it is

$$F = \frac{GmM}{r^2}$$

So, now, what is the (magnitude of the) acceleration of $m$? Well, Newton also tells us that $F = ma$, so $a = F/m$, so

\begin{align} a &= \frac{GmM}{mr^2}\\ &= \frac{GM}{r^2} \end{align}

And look: there is no dependence on $m$ at all, because it cancels out.

That this is true is actually a surprising fact: there are, really, two definitions of mass here:

• inertial mass is the mass which occurs in $F = ma$, Newton's second law;
• gravitational mass is the mass which occurs in $F = Gm_1 m_2/r^2$, Newton's law of gravitation.

And it turns out that these two masses are always the same for a given object (or, strictly, are always in the same proportion, with the constant of proportionality being absorbed into $G$). The above result, that all things fall with the same acceleration, is because of this equality between inertial and gravitational mass.

This is not something that needs to be the case: experimentally it is the case to the best of our knowledge (ie no experiments we have been able to do have ever shown it not to be the case), but nothing says it has to be the case. If it turned out not the case then General Relativity, for which this equality is a rather basic assumption, would fail.

(Just to be clear: although I've made the point that the equality of inertial and gravitational mass is an experimental fact, which I think is important to undertand, I strongly believe it to be true. I'm not trying to push some alternative theory of gravity!)

• Equivalently, all experiments to measure G find the same value, whatever test masses they use. OTOH, it's notoriously hard to measure G to more than a handful of decimal places, so our empirical verification of this isn't as strong as, for example, the constancy of the speed of light. – PM 2Ring Sep 17 '18 at 9:14

All masses will experience the same acceleration in a gravitational field due to the fortuitous equivalence of inertial and gravitational mass. This equivalence was experimentally verified to high accuracy more than 100 years ago by the Hungarian physicist Lorand Eötvös. The equivalence principle was one of the starting points of Einstein's General Theory of Relativity.