7.4.1 Potential vs GPE

As a glider soars and dives in the sky, it trades between GPE and KE as it gains or loses height. A spacecraft travelling from the Earth to the Moon experiences similar energy changes. At the beginning when it is leaving Earth, it is gaining GPE (mainly because of Earth’s gravitational field) and losing KE. Towards the end when it is approaching the Moon, it is losing GPE (mainly because of Moon’s gravitational field) and gaining KE. Somewhere between the Earth and the Moon, the spacecraft is at the “summit” of its journey where it has its maximum GPE.

It feels as though when a mass moves in a gravitational field, with or against the gravitational force, it is rolling down or up some gravitational slope. Wouldn’t it be nice to have a number to mark out the “gravitational height” of the gravitational landscape?

This “gravitational height” is called the gravitational potential Φ (pronounced phi). Basically, you can think of Φ (of a location) as the “gravitational potential energy U per unit mass” (of that location).

\displaystyle \phi =\frac{U}{m}

In other words, if a mass m is situated in a gravitational field where the gravitational potential is Φ, then its GPE (at that location) would be

U=m\phi

Example

A 580 kg asteroid is hurtling through outer space. It passes location A at a speed of 1600 m s-1 with GPE of −900 MJ, and arrives at location B which has gravitational potential of -2.1 MJ kg-1.

a) Calculate the gravitational potential at location A.

b) Calculate the KE of the asteroid at location B.

Solution

a)

\phi =U\div m=-900\div 580=-1.55\text{ MJ k}{{\text{g}}^{{-1}}}

b)

\displaystyle \begin{aligned}(KE+GPE)\text{ at A}&=(KE+GPE)\text{ at B}\\\frac{1}{2}(580){{(1600)}^{2}}+(-900\times {{10}^{6}})&=K{{E}_{B}}+(580)(-2.1\times {{10}^{6}})\\742.4\times {{10}^{6}}-900\times {{10}^{6}}&=K{{E}_{B}}-1218\times {{10}^{6}}\\K{{E}_{B}}&=1060\text{ MJ}\end{aligned}

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