# Dropping Objects of Different Masses

It's a commonly taught principle that dropping two objects of different mass at the same height, given that there is no air resistance, will result in the two objects making contact with the ground at the same time. That's physics 101. However, I'm wondering if this is proven physics, or simply a rounding up.

According to Newton's law of gravitation,

$F=G\frac{m_1 m_2}{r^2}\$

Does that not mean that a difference in the mass of the object being dropped will result in a larger/smaller acceleration? Is the reason we don't bother with this because the change in $g$ is so minute that it's trivial, or is there physics which states that, no matter what, two objects without air resistance dropped at the same height will hit the ground at the same time?

• the $g$ on Earth depends on Earth's mass alone, and on the distance from its centre, as you can see from the formula you have written. What is usually neglected is the dependence of $g$ on $r$ on Earth's surface, since a difference of 1 metre is unnoticeable on Earth's surface. – Phoenix87 Jan 5 '15 at 0:37
• Good question; possible duplicate physics.stackexchange.com/q/3534/d, and physics.stackexchange.com/q/156636 was asked 5 minutes after you. – HDE 226868 Jan 5 '15 at 0:49
• Without resorting to the law of gravitation it's possible to disprove that acceleration depends on the mass of the falling object. "If we assume heavier objects do indeed fall faster than lighter ones (and conversely, lighter objects fall slower), the string will soon pull taut as the lighter object retards the fall of the heavier object. But the system considered as a whole is heavier than the heavy object alone, and therefore should fall faster. This contradiction leads one to conclude the assumption is false." wiki – Rnhmjoj Jan 5 '15 at 13:11

As long as the mass that we aren't dropping is very large and is kept constant, then the mass of the object we are dropping has no considerable effect on its acceleration. This is because of Newton's 2nd Law:

$$F = ma$$

Where $m$ is the mass that is accelerating, i.e. the smaller mass we are dropping. So, if $F = G\frac{Mm}{r^2}$, where $m$ is the mass we dropped, and $M$ is the big mass that the object we dropped is fall to, then:

$$a = \frac{F}{m} = G\frac{M}{r^2}$$

So, while acceleration is dependent in $M$, it does not depend on the mass of the dropped object.

The constant value $g$ is actually only true on the earth's surface, and is appropriately defined as:

$$g_{earth} = G\frac{M}{(R_{earth})^2}$$

Where $R_{earth}$ is the radius of the Earth.

Notice that I said the bigger mass, $M$ (or, the mass that is causing the gravitational field) is, indeed, big. If it were not that big, the object of the mass we dropped (by Newton's 3rd Law) would cause a force on $M$ that results in a significant acceleration of $M$. This means that both masses are significantly accelerating towards each other, so that the effective acceleration of the dropped object (i.e. the acceleration in a reference frame where $M$ is stationary) would be the sum of the magnitudes of the accelerations of each object in a non-accelerating frame of reference:

$$a_{effective} = a_m + a_M = G\frac{M}{r^2} + G\frac{m}{r^2} = G\frac{M+m}{r^2}$$

In which case, the mass of the dropped object does indeed affect the effective acceleration.

(Note that the "effective acceleration" is the acceleration of the dropped object when $M$ appears stationary to the observer. This does not appear to obey Newton's 2nd Law (i.e. $a_{effective} \neq a_m$ because this frame of reference is accelerating, and accelerating reference, otherwise know as non-inertial reference frames, do not obey Newton's 2nd Law! In fact, this question assumes that, in the inertial frame of reference, the two masses are heading towards each other in a straight line...)

Does that not mean that a difference in the mass of the object being dropped will result in a larger/smaller acceleration?

No. It will result in a larger/smaller force, not a larger/smaller acceleration.

As the mass $m$ gets larger, $f$ gets larger, so their ratio $f/m$ (which is acceleration $a$) stays the same.