# How has the definition of gravitational potential energy been derived?

The definition of gravitational potential energy is - The gravitational potential energy of an object at a point above the ground is defined as the work done is raising it from the ground to that point against the gravity . How has this definition been derived?I think this definition has been derived by using the formula Work Done=Force*Displacement . But I don't know how to do this . can someone explain this derivation? Also, Why can't we define gravitational potential energy like this - The gravitational potential energy can be defined as the amount of work done to lift an object of mass,m to a height,h with acceleration A such that A>g in terms of magnitude? (g=gravitational acceleration)why is this definition wrong?

Technically, the gravitational potential energy is defined as the work done to bring an object from a point at infinity to a point in the gravitational field. Since $W=\int F \cdot dl$, we have, from Newtons law of gravitation :

$E=\int_{\infty }^{R} \frac{GMm}{r^2} dr \\=-\frac{GMm}{R}$

What you're saying is the change of gravitational potential energy is mgh when h is negligible compared to the radius of the Earth. In this case, we have $W=\int_{0}^{h} mg dx\\=mg \int_{0}^{h}dx\\=mgh$

When you're lifting the object with an acceleration, the work you do is not just converted into gravitational potential energy. Due to the acceleration, your body is gaining kinetic energy as well.

• Note that your first equation is the (negative) work done by gravity, not the work done to bring the object in from infinity. You can get your definition to work, but you have to specify how the object is brought it. For example, constant speed plus invoking Newton's second law plus the absence of other forces. $\Delta U = - W_\mathrm{internal}$ avoids that, and other complications. – garyp May 24 at 14:11

$GPE = mgh$ and $Work = Force*distance$. The force required to oppose gravity while lifting an object is equal and opposite to the force exerted on it by gravity, namely mg. To get the work, you can multiply it by how high you lifted it, which is h. I don't know if you were looking for something more with that derivation.

As for the second part of your question, you can't derive potential energy using an acceleration other than g because, although you expend more energy than the potential energy you give it, that goes into the kinetic energy the object possesses as it is lifted. The potential energy only takes into account the work you did against gravity, which has the potential to be exerted on the object later.