Gravitational potential and gravitational potential energy

​​In a nutshell

Gravitational potential is defined as the work done per unit mass, moving a mass from infinity to its position in the gravitational field. Gravitational potential energy depends on the mass and where the mass is in the gravitational field. Escape velocity is the minimum velocity needed to escape a gravitational field. 

Equations

Constants

Variable definitions

Gravitational potential

Gravitational potential is defined as the work done per unit mass, moving a mass from infinity to its position in the gravitational field. 

In a radial field, like most celestial objects including Earth, is give as:

​$$V_g=-\frac {GM}r$$​​

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Curiosity: The equation for gravitational potential is negative as work has to be done against the gravitational field to move a mass out of it. The maximum value gravitational potential can have is $$0 \, J$$ at a distance of infinity from a gravitational field.

The graph for gravitational potential against distance looks like this:

For a tangent of the curve, the gradient would be given as:

​$$gradient=\frac {\Delta V_g}{\Delta r}$$​​

Substituting in the equation for $$V_g$$:

$$gradient=\frac {\frac {GM}{r}}{r} \newline \\[0.1in]gradient=\frac {GM}{r^2} \newline \\[0.1in] gradient=-g \newline \\[0.1in]$$​​

The gradient of a gravitational potential graph against distance gives the gravitational field strength $$g$$. So gravitational field strength is the rate of change of gravitational potential with respect to the distance. 

Consider two masses at different distances in a gravitational field. The masses will have different gravitational potentials due to having different distances. This means there is a potential difference between the two masses. 

Work done against gravity

When you move a mass further away in a gravitational field, work is being done against gravity and the amount of work depends on where in the field the mass is.

Astronauts on the International Space Station can move "heavy" objects around easily as the gravitational field of Earth is very weak, so the work done is much less than moving the same object on the surface of Earth. 

This can be shown as: 

​$$\Delta W=m\Delta V_g$$​​

The work done is the product of the mass and change in gravitational potential. 

Another graphical relationship is how the size of a force on an object changes due to the distance away from a gravitational field. 

For two points on the curve, the area underneath represents the work done moving the object from one point to the other. 

Gravitational potential energy

As gravitational potential is defined as the work done per unit mass, the work done, or energy transferred, can be given as:

​$$E=mV_g$$​​

Substituting in the equation for $$V_g$$:

$$E=-\frac {GMm}r$$​​

​​Example

The International Space Station has a mass of $$420\, 000\, kg$$ and orbits the Earth at a distance of $$408\,km$$ . Calculate the gravitational potential energy of the International Space Station..

$$M_{Earth}=5.97\times 10^{24} \, kg \newline \\[0.1in]R_{Earth}=6.37\times 10^6 \, m$$​​

Firstly, write down the known values: 

$$m=420\,000 \, kg \newline \\[0.1in]r_{orbit}=408\,km=408\,000 \, m \newline \\[0.1in]M_{Earth}=5.97\times 10^{24} \, kg \newline \\[0.1in]R_{Earth}=6.37\times 10^6 \, m$$

​Next, write down the equations needed and rearrange if necessary:

$$E=-\frac {GMm}r$$

Then, substitute the values into the equation:

$$E=-\frac {6.67\times ^{-11}\times 5.97\times 10^{24}\times 420\,000}{6.37\times 10^6+408\,000} \newline \\[0.1in]E=2.46745...\times 10^{13}$$

Note: The orbital distance needs to be added to the radius of the Earth as the distance is from the centre of the Earth and not the surface.

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$gravitational \ potential\ energy, \ E=2.47\times 10^{13} \, J$$​​​​​

The gravitational potential energy of the International Space Station is $$\underline{2.47\times 10^{13} \, J}.$$​

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Escape Velocity

In order to escape a gravitational field, the object must have enough kinetic energy to leave the field. 

The escape velocity is given as the minimum velocity required, so when calculating the energy, the minimum amount of kinetic energy is used. 

To obtain an equation for the escape velocity, consider when the kinetic energy is just enough to escape the gravitational field, so that the object has a net energy of zero:

$$kinetic \ energy +gravitational \ potential \ energy=0 \newline \\[0.1in]\frac 12 mv^2+(-\frac {GMm}{r})=0 \newline \\[0.1in]\frac 12 mv^2=\frac {GMm}{r} \newline \\[0.1in]\frac 12 v^2=\frac {GM}{r} \newline \\[0.1in]v_{escape}=\sqrt {\frac {2GM}{r} }$$​​

​​Example

Calculate the escape velocity of Mars.

$$M_{Mars}=6.39\times 10^{23} \,kg \newline \\[0.1in]R_{Mars}=3.39 \times 10^{6} \, m$$​​

Firstly, write down the known values: 

$$M_{Mars}=6.39\times 10^{23} \,kg \newline \\[0.1in]R_{Mars}=3.39 \times 10^{6} \, m$$

​Next, write down the equations needed and rearrange if necessary:

$$v_{escape}=\sqrt {\frac {2GM}{r} }$$

Then, substitute the values into the equation:

$$v_{escape}=\sqrt {\frac {2\times 6.67 \times 10^{-11}\times 6.39\times 10^{23}}{3.39\times 10^6} } \newline \\[0.1in]v_{escape}=5014.51...$$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$escape \ velocity, \ v_{escape}=5010 \, m\,s^{-1}$$​​​​​

The escape velocity of Mars is $$\underline{5010 \, m\,s^{-1}}.$$​

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