# How to prove that gravitational potential energy of a body of mass $m$ at a height $h$ is $mgh$?

Many introductory physics books just write that potential energy of a body of mass $m$ at a height $h$ as $U_\text{g}=mgh$. However, they never show how this was derived. I'm interested in knowing this derivation – if possible, avoiding calculus.

• potential energy is defined using calculus. – AccidentalFourierTransform Mar 3 '16 at 17:40
• Vaguely related: physics.stackexchange.com/q/122767/50583 I don't really get what you're asking here - if you don't know calculus, how can you define potential energy, which is just the integral of a conservative force? – ACuriousMind Mar 3 '16 at 17:48
• Hint; if you do not know calculus then divide the height in n-parts and mark those points as y1,y2,...... calculate the work done in elemental paths and try adding it. – drvrm Mar 3 '16 at 19:11
• @ Tweej and Tatan- Don't you think that work done by gravity will be -mgh rather than mgh? – Parth Mar 4 '16 at 0:59

Suppose I let an object fall from rest for a distance $h$. Given that the gravitational acceleration is $g$ the velocity of the object will be given by the SUVAT equation:

$$v^2 = u^2 + 2gh$$

In this case the initial velocity $u=0$ so we just get $v^2 = 2gh$. The kinetic energy of the object is given by:

$$T = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m(2gh) = mgh$$

If energy is conserved the increase in kinetic energy must be equal to the decrease in potential energy, so we get:

$$\Delta U = -\Delta T = -mgh$$

This tells us that if we lower the object by a distance $h$ the potential energy decreases by $mgh$, and conversely that if we raise it by a distance $h$ the potential energy increases by $mgh$.

At the risk of suggesting an overly simple answer, work done on a body is defined as $W = F \cdot d$.

We know that the force of gravity acting on a body is $mg$.

The perpendicular distance it travels is $h$, so $$W=E=F\cdot d= mgh\,.$$

By definition we know that-

Gravitational Potential Energy of a body is the work done against gravity in raising it to a certain height h.

We have$$\text{work}~~=~~\text{force}~\times~\text{displacement} \,.$$

Here force of the body is the weight acting vertically downwards=$mg$ and displacement is $h.$

So,$$\text{work} ~=~mg\cdot h ~=~mgh \,.$$

The form $U=mgh$ is simply an approximation to allow people to quickly calculate small changes in the gravitational potential energy of the system. The actual value of $U_g$ is usually not important in classical mechanics. $\Delta U_g$ is the important concept.

On a planet, small changes in height change $U_g$ in a space where the gravitational force is almost constant, $\vec{F}=m\vec{g}$. A change in potential energy (in classical mechanics) is calculated by the negative work done by a force while the position changes: $$\Delta U_{y'\to y'+h}=-\int_{y'}^{y'+h} mg(-\hat{j})\mathrm\ {d}y$$ The integral is easy: $$\Delta U_{y'\to y'+h}=\left. mgy\right| _y^{y+h}= mg(y'+h-y')=mgh$$

We see that the actual starting point is unimportant as long as the gravitational field can be considered constant. $h$ is positive if the particle is moved opposite the gravitational field (up), and negative if the particle moves down.

## protected by AccidentalFourierTransformJun 16 '18 at 17:51

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