Gravitational Potential Energy Calculator

Enter the mass and current height above y=0 to determine the gravitational potential energy of an object.

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Gravitational Potential Energy Formula

Gravitational potential energy is the potential energy an object contains due to it’s mass and the force of gravity with respect to another object, often times a massive object such as the earth.

The formula for gravitational potential energy is the following:

U = – G*M*m/r

Where U is the potential energy, G is the gravitational constant, M is mass 1, m is mass 2, and r is the radius between the two objects.

When looking at the potential energy of an object on earth, this can be simplified into the following:

U= m * g * h

Where U is potential energy, g is the acceleration due to earths gravity, and h is height from surface.

Potential energy is the possible energy an object contains due to gravity. If an object were to obtain all of that energy, it would then be converted into kinetic energy.

How to calculate gravitational potential energy

Unless you are an alien, you spend every day on earth. As a result, in almost any field, the gravitational potential energy is calculated using the simplified equation above. With this in mind, we will take a look at an example of a common physics problem involving gravitational potential energy. We will assume a bowling ball is place on top of a hill, and we want to calculate it’s potential energy.

1. First, we must determine the mass of the bowling ball using either a mass calculator or a scale. For this example we will assume a weight of 200kg.
2. Next, we need to measure the height. Using surveying techniques we determine that the ball sites on a hill that is 200m high.
3. Finally, we plug the information into the simplified equation above and we have our answer. U=200kg * 9.8m/s^2 * 200m = 392,000 Joules
4. Analyze your results and apply to other problems.

As you can see, calculating the gravitational potential energy on earth is easy. It’s a simple plug and calculate. For objects not on earth or at distances far away it becomes much more complicated. The radius and masses of both objects must be taken into account.

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