## Energy and Satellite Motion

### A satellite in orbit will have both Kinetic Energy and Gravitational Potential Energy. The Kinetic Energy is easy to calculate as it simply follows the standard form of E_{k} = ^{1}/_{2}mv^{2}, but calculating Gravitational Potential Energy is more complicated.

### This is because at National 5 Level, the Gravitational Potential Energy could be calculated by E_{p} = mgh. This simplified formula can only be used if the value of "g" is constant, which only works over relatively small values of "h". In order to calculate the Gravitational Potential Energy over larger ranges, the more complication version is required :-

### Gravitational Potential Energy can be found using:-

### E_{p} = mg h

### However, by substituting the Gravitational Force ( mg ) formula from earlier within the Advanced Higher course, the Gravitational Potential Energy can be shown as :-

### mg = ( GMm ) / r^{2}

### E_{p} = ( ( GMm ) / r^{2} ) x h

### If the Value of h is taken as the height above a point mass then h = r , and the Gravitational Potential Energy can be found as :-

### E_{p} = ( GMm ) / r

### However, by convention, we state that Gravitational Potential Energy is equal to 0 J at infinity, and as such , the formula is given as :-

### Where :-

### E_{p} = Gravitational Potential Energy ( J )

### G = Gravitational Constant = 6.67x10^{-11} m^{3}kg^{-1}s^{-2 }

### M = Large Object Mass ( kg )

### m = Small Object Mass ( kg )

### r = The separation between Masses ( center to center ) ( m )

### Note - The negative sign in the above formula means that as the value of "r" increases, the value of E_{p} becomes less negative, which is means the value is increasing, which is what you would expect for an increased radius. All that has been done is that the scale has been re calibrated to a new zero point. This was done similarly in the Higher course when discussing Electron Energy levels within an atom.

## Example 1 -

### A spacecraft of Mass 5 x10^{3} kg burns its engines to move away from the Moon. At the start of the maneuver, the spacecraft is 2 x10^{6} m from the centre of the Moon and moves to a point 2.5 x10^{6} m from the centre of the Moon. If the Mass of the Moon is 7.3 x10^{22} kg, calculate the Work Done that is required to move the spacecraft.

### The spacecraft has an Initial Gravitational Potential Energy of :-

### E_{p( start )} = - ( 6.67x10^{-11} x 7.3x10^{22} x 5x10^{3} ) / 2x10^{6}

### E_{p( start )} = - 1.217275x10^{10} J

### The Spacecraft has a Final Gravitational Potential Energy of :-

### E_{p( end )} = - ( 6.67x10^{-11} x 7.3x10^{22} x 5x10^{3} ) / 2.5x10^{6}

### E_{p( end )} = - 0.973820x10^{10} J

### The Change in Gravitational Potential Energy is given by :-

### ΔEp = E_{p( end )} - E_{p( start )}

### ΔEp = - 0.973820x10^{10} - - 1.217275x10^{10}

### ΔEp = 2.43455x10^{9} J

### The Work that must be Done is equal to the Change in Potential Energy :-

### Work Done = 2.4 GJ

## Gravitational Potential

### The Gravitational Potential ( V ) at a point in a Gravitational field is defined as the Work Done by an external Force to move a unit Mass from infinity to that point.

### By applying the above Gravitational Potential Energy formula to a unit mass, the Gravitational Potential can be found :-

### V = - ^{GM }/ _{r }

### Where :-

### V = Gravitational Potential ( J kg^{-1} )

### G = Gravitational Constant = 6.67x10^{-11} m^{3}kg^{-1}s^{-2 }

### M = Large Object Mass ( kg )

### r = The separation between Masses ( center to center ) ( m )

### The diagram below shows how the Gravitational Potential varies as a spacecraft moves away from the Earth.

### This means that a 1 kg mass dropped from infinity will lose 62.8 MJ of Gravitational Potential Energy by the time it hits the surface of the Earth.

### If air resistance is ignored, this lost Gravitational Potential Energy is converted to Kinetic Energy.

### Note - The formula for Gravitational Potential look very similar to the formula for Electrical Potential (as seen in Unit 2). Both are identical in concept, each simply referring to a different type of field.

## Example 2 -

### If the Earth has a mass of 6x10^{24} kg and a mean radius of 6.4x10^{6} m, calculate the Gravitational Potential at the Earth's surface .

### V = ?

### G = 6.67x10^{-11} m^{3}kg^{-1}s^{-2}

### m = 6x10^{24} kg

### r = 6.4x10^{6} m

### V = - ( 6.67x10^{-11 }x 6x10^{24 }) / 6.4x10^{6}

### V = - 63x10^{6} Jkg^{-1}

## Kinetic Energy of a Satellite

### As previously stated, the Kinetic Energy of a satellite can be found be deriving a form of E_{k}=^{1}/_{2}mv^{2}, based upon a Centripetal Force calculation :-

## Combined Energy of a Satellite

### The combined energy of a satellite can be found simple by the addition of the Kinetic Energy and the Gravitational Potential Energy :-

## Movement within a Gravitational Field

### Gravitational fields are conservative fields. In a conservative field, any round trip (A trip that starts and ends at the same point) requires no Work to be Done. If no Work is Done, then Energy is conserved. The diagram below shows a route that a spacecraft could take within a Gravitational field :-

### If the only Force is the Gravitational Force acting downwards, Work would only be Done moving the spacecraft vertically.

### As the spacecraft has overall no vertical displacement, the Work Done is zero.

### Note - The above only applies in the absence of Friction. If Friction is taken into account, Energy is not conserved, and as such Friction is an example of a non - conservative Force.