# Gravity and mass relation in Conservation of Energy equations

When writing Conservation of Energy equations, if you want to add a gravity component, you write, $$mgh$$. Where $$m$$ is mass, $$g$$ is gravitational constant, and $$h$$ is the height. If a mass has nothing to do with gravity on falling objects, why is it here?

If a mass has nothing to do with gravity on falling objects, why it here?

Its true mass of body has nothing to do with gravitational acceleartion but it has to do with the gravitational force.

We defined the gravitational energy to be the work done by a particle to travel from infinite to that point. For uniform force, We can write the work done by a body when particle travel from point $$a$$ to $$b$$ :

$$W_{ba}=\mathbf{F}_0\cdot(\mathbf{r}_b-\mathbf{r}_a)$$

for gravity $$\mathbf{F_0}=-mg\hat{z}$$, so if the particle moves from $$\mathbf{r}_a$$ to $$\mathbf{r}_b$$, the change in potential energy is

$$U_b-U_a=-\int_{z_a}^{z_b}(-mg)dz=mg(z_b-z_a)$$

So basically the work done depend on the mass and so is the potential energy. It's obvious you need to do more work if you are uplifting a car than to cycle.

• That helped a lot! Nov 1, 2020 at 16:34
• Good to here that. Nov 1, 2020 at 16:36
• "Its true mass of body has nothing to do with gravitational acceleration but it has to do with the gravitational force", what do you mean by this?
– user65081
Nov 1, 2020 at 17:41
• That's means that $g$ doesn't depend on mass of the body (cycle and bus take the same time during the free fall from the same roof). But force exerts by earth $mg$ depends on the mass of the body( the impact made by cycle and bus will be different). Nov 1, 2020 at 17:45

If a mass has nothing to do with gravity on falling objects, why it here?

Not true. Mass has nothing to do with the speed at which an object falls - a small mass, m, and a larger mass, M, when released at the same time from the same height, will hit the ground at the same time (ignoring air resistance). But their kinetic energies when reaching the ground will not be the same: $$\frac{mv^2}{2}$$ < $$\frac{Mv^2}{2}$$ where v is the common final speed.