Gravitational Potential Energy : Educating Physics

Objectives:

You should be able to calculate the amount of energy associated with lifting an object
The gravitational potential energy of an object can be calculated using the equation; $E_{p} = m \times g \times h$

Anything that is up in the air and has the ability to fall has what we call gravitational potential energy. What this means is that if you took away the supports of something up in the air then it would fall and ‘lose’ this form of energy.

The gravitational potential energy is governed by three factors:

how heavy the object is that you are lifting (the heavier it is the more energy it will require to lift it up)
the height you want to lift it up by (the higher, the more energy will be required)
where the object is and therefore the gravitational pull due to Earth (if you were in space where gravity is much smaller, then it would be easier to lift the object up)

The equation linking these three quantities together is:

$E_{p} = mgh$

where;
$E_{p}$ is the gravitational potential energy, measured in joules, $J$
$m$ is the mass, measured in kilograms, $kg$
$g$ is the acceleration due to gravity, measured in metres per second squared, $m/s^{2}$
$h$ is the heigh, measured in metres, $m$

The acceleration due to gravity on Earth is $g = 9.8 \ m/s^{2}$ , remember this value and always use it unless a different value is given to you in an examination. Sometimes exams can have questions about astronauts lifting objects up on the moon in which case the acceleration due to gravity is different, its approximately $g_{moon} = 1.6 \ m/s^{2}$

Examples of using this equation

How much gravitational potential energy is required to lift a television, or mass $2.4 \ kg$ up three flights of stairs of $late 5.2 / m $ each?
A runner is hoping to run up mount Snowdon which has a vertical prominent height of $1038 \ m$ . The runner has a mass of $72 \ kg$
An astronaut of mass $75 \ kg$ on the international space station (ISS) returns home after one of their 6 month voyages. Upon return they lose $285 \ MJ$ of gravitational potential energy. How high up is the ISS from the surface of Earth?

Solutions

$m = 2.4 \ kg$
$g = 9.8 \ m/s^{2}$
$h = 5.2 \ m$
$E_{p} = mgh = 2.4 \times 9.8 \times 5.2$
$E_{p} = 122.3 \ J$
$m = 72 \ kg$
$g = 9.8 \ m/s^{2}$
$h = 1038 \ m$
$E_{p} = mgh = 72 \times 9.8 \times 1038$
$E_{p} = 732,413 \ J = 732 \ kJ$
$E_{p} = -285 \ MJ = -285,000,000 \ J$
$m = 75 \ kg$
$g = 9.8 \ m/s^{2}$
$E_{p} = mgh \rightarrow -283,000,000 = 75 \times 9.8 \times h$
$-285,000,000 = 735 \times h$
$h = -387,755 \ m$
This means the astronaut has fallen through a distance of $388 \ km$ and therefore the ISS is $388 \ km$ up.