# How to get energy of collision if you know force of gravity of an object($m \rightarrow F=mg$)?

How to get energy of collision if you know force of gravity of an object($m \rightarrow F=mg$)? You get energy of collision by kinetic energy $E_k= \frac{1}{2}mv^2$, but if you use just force of gravity($F=mg$), how you get then the energy of collision? I know that work(energy?) $W$ made by object is $W=F \Delta s$, where $F$ is force exerted to object and $\Delta s$ traveled distance, but this distance makes it hard to calculate any collision energy of a meteorite for example.

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As you pointed out right: $W = F \Delta s$ and the energy is derived form integrating over the path. If we assume that the gravitational field remains constant (so $g$ is independent on the position of the considered object) and that an object falls down form the height $h$ we have to integrate $mg\Delta s$ from $h$ to $0$, which yields $-mgh$. If you are familiar with concept of energy conservation, this(the potential energy) is equivalent to the kinetic energy the mass wins in a free fall from height $h$, which is $E_k = \frac{1}{2}mv^2$ –  dedoco Dec 1 '12 at 11:18
If you're talking about a meteorite, $F\neq mg$. that formula is only valid if your height above the ground is negligible when compared to the radius of the Earth. See Chris's answer. –  Jerry Schirmer Jan 29 '13 at 15:20

Let us assume Earth is the only object in our universe, apart from a meteor way off in the distance (approximately $\infty$ m away). The meteor will gradually begin to approach Earth due to gravity, and we wish to calculate the energy the meteor has when it collides with Earth.

To calculate the energy, we must integrate $W = Fds$ over the path taken by the object.

In this case, $F = \frac{Gm_1m_2}{r^2}$, where $m_1$ is the mass of the Earth, $m_2$ is the mass of the meteor and $r$ is the distance between the Earth and the meteor. The path integral to compute the total energy obtained from travelling to the Earth thus becomes:

$W = \int_R^\infty \frac{Gm_1m_2}{r^2} dr = [-\frac{Gm_1m_2}{r}]_R^\infty = -\frac{Gm_1m_2}{\infty} +\frac{Gm_1m_2}{R} = 0 + \frac{Gm_1m_2}{R} = \frac{Gm_1m_2}{R}$.

where R is the radius of the Earth.

For cases where the meteor is much closer to the Earth, we can assume that the force of gravity is constant, and simply use the equation $W = mgh$.

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