Why do 2 bodies of different masses reach the ground at the same time?

Question

Since inertia is the quantity that resists change in motion, and is represented by mass, and if two different objects are dropped from the same height, shouldn't the heavier mass accelerate less than the one with the lesser mass, since it has a greater inertia?

Answer 1

The usual answer is that the Newtonian gravitational force is proportional to the gravitational mass of the object and so, if inertial mass and gravitational mass are equivalent, a hammer and a falcon feather fall at same rate (in a vacuum).

$$F_g = Km_g = m_ia$$

$$m_g\equiv m_i \Rightarrow a = K$$

But Einstein's theory of gravity, General Relativity, explains the observed result differently; the freely falling hammer and falcon feather are not accelerated at all!

We observe them to be accelerating together, not because there is a gravitational force on each that just happens to make them accelerate at the same rate, but because we are in the accelerated reference frame of the surface of a planet (or moon). From this accelerated reference frame, inertial (non-accelerated) objects appear to accelerate and this naturally explains why they accelerate together.

Answer 2

According to Newton's law of gravity, two bodies with different masses do not reach the ground at the same time. The body with greater mass reaches the surface of the Earth faster than the body with the lesser mass.

Indeed, let $M$ be Earth's mass, let $m_1$ be the mass of body $1$ and let $m_2$ be the mass of body $2$. Then, let us look at the two body system: Earth - body $1$. According to Newton's principles combined with his law of gravity, Earth attracts body $1$ as well as body $1$ attracts Earth. Then, after deriving the equations of motion for Earth and body $1$ as a two-body system, one obtains the following equations of motion for the vector $\vec{r}=\vec{r}(t)$ connecting Earth's center of mass with body $1$'s center of mass:

$$\frac{d^2\,\vec{r}}{dt^2} \, = \, - \, \left(\, \frac{(M+m_1) \, G}{|\vec{r}|^3} \, \right)\, \vec{r}$$

Denote by $r(t) = |\vec{r}(t)|$ the length of vector $\vec{r}(t)$. Then $r=r(t)$ is the distance between Earth and body $1$ at time $t$. When we say body $1$ falls towards Earth's surface, technically we should think of body $1$ falling towards Earth and at the same time Earth falling towards body $1$. Thus, to find the time when body $1$ reaches Earth's surface, we need to calculate how the distance $r(t)$ goes from the initial height $r(0)$ of body $1$ plus Earth's radius, to the time $t_1$ when $r(t_1)$ is equal to Earth's radius (assuming body $1$ is a point-mass). Assuming that body $1$ falls along a straight-line trajectory, which is a valid solution of the equation I wrote above, then distance $r=r(t)$ satisfies the differential equation

$$\frac{d^2\,{r}}{dt^2} \, = \, - \, \frac{(M+m_1) \, G}{{r}^2}$$ which yields the law of conservation of energy equation

$$\frac{1}{2} \, \left(\frac{dr}{dt}\right)^2 \, -\, \frac{(M+m_1) \, G}{{r}} = -\, \frac{(M+m_1) \, G}{{r(0)}}$$ so one gets $$\frac{dr}{dt} = \, \pm\, \sqrt{2\,(M+m_1) \, G\, \left( \frac{1}{r}\,-\, \frac{1}{r(0)} \right)\,}$$ $$\frac{dr}{dt} = \, \pm\, \sqrt{2\,(M+m_1) \, G\, \left( \frac{\, r(0) - r \,}{r(0) \, r} \right)\,}$$

After choosing the minus because $r(t)$ should be decreasing as body $1$ is falling, the latter equation can be integrated as follows, yielding the time of fall for body $1$

$$t_1 = \sqrt{\frac{r(0)}{2\, (M+m_1)\, G}} \,\int_{r(t_1)}^{r(t_0)} \, \sqrt{\frac{r}{\, r(0) - r \, }} \, dr$$ or equivalently, $$\big(\sqrt{\, M+m_1 \,}\big) \, t_1 = \sqrt{\frac{r(0)}{2\,G}} \, \int_{r(t_1)}^{r(t_0)} \, \sqrt{\frac{r}{\, r(0) - r \, }} \, dr$$ Analogously, the same calculation for body $2$ yields $$\big(\sqrt{\, M+m_2 \,}\big) \, t_2 = \sqrt{\frac{r(0)}{2\,G}} \, \int_{r(t_2)}^{r(t_0)} \, \sqrt{\frac{r}{\, r(0) - r \, }} \, dr$$ Since both bodies are dropped from the same height $r(0) = r_0$ and the motion ends when $r(t_1) = r(t_2) = r_{\text{earth}}$ is the end distance (Earth's radius $r_{\text{earth}}$) one obtains the identities

$$\big(\sqrt{\, M+m_2 \,}\big) \, t_2 \, = \, \sqrt{\frac{r(0)}{2\,G}} \, \int_{r_{\text{earth}}}^{r_0} \, \sqrt{\frac{r}{\, r(0) - r \, }} \, dr \, = \, \big(\sqrt{\, M+m_1 \,}\big) \, t_1 $$ that is, $$\big(\sqrt{\, M+m_2 \,}\big) \, t_2 \, = \, \big(\sqrt{\, M+m_1 \,}\big) \, t_1 $$ which is equivalent to $$t_2 = \sqrt{\frac{M+m_1}{M+m_2}}\, t_1$$ Now, if $m_1 > m_2$ then $\sqrt{\frac{M+m_1}{M+m_2}} > 1$ so $$t_2 = \sqrt{\frac{M+m_1}{M+m_2}}\, t_1 \, > t_1$$ which means that the time of fall $t_2$ of the lighter body $2$ is longer than the time of fall $t_2$ of the lighter body $t_1$, i.e. it takes longer for the lighter body to fall on the ground than for the heavier.

However, in practice, Earth's mass $M$ is so overwhelmingly greater than either of the masses $m_1$ and $m_2$, that $M+m_1 \thickapprox M \thickapprox M + m_2 $ and so consequently $t_1 \thickapprox t_2$.