A2 Physics 


Gravity Fields and Potentials
Apollo 8 Mission  24 December 1968
Contents:
Field lines
Newtons law of gravitation
Gravitational field strenght, g
Gravitational potential
Equipotentials and potential gradients
Kepler's law
The mass of an object creates a gravitational field around it. A mass placed in that field generates a force of attraction on that mass. 2 masses always attract each other with an equal and opposite force (but some forces are too small to measure, e.g. a person pulling the Earth).
Field LinesIf you place a small ‘test’ mass near a massive object it will follow a path called a field line.
There are 2 types of fields… 1. Radial – field lines are like spokes on a wheel. Close field lines = stronger force i.e. closer to the object. Field lines always point towards centre of mass. 'g' is bigger nearer centre. 
2. Uniform field  gravitational field strength is the same in magnitude and direction throughout. This is the approximated field at the Earth's surface. 
Newton's Law of Gravitation
This law describes the gravitational force, F, that one mass, m1, exerts on another mass, m2, at a certain distance apart, r.
or just
G = Gravitational constant = 6.67x1011m3kg^1s^2. This constant is very small ans is testimony to how weak the force of gravity really is. Compare this with the equivalent electrostatic force equation between two charges (Coulomb's law) whos constant is 10^20 times larger.
This equation follows an inverse square relation i.e. when the distance is doubled the force decreases by a factor of 4 and when the distance is trebled the force decreases by a factor of 9.
Tip: the value for r is from the centre of the planet. When doing satellite calculations you are often given only the height from the Earth's surface. You must remember to add on the Earth's radius to calculate the force exerted on it (RE=6400km).
This equation follows an inverse square relation i.e. when the distance is doubled the force decreases by a factor of 4 and when the distance is trebled the force decreases by a factor of 9.
Tip: the value for r is from the centre of the planet. When doing satellite calculations you are often given only the height from the Earth's surface. You must remember to add on the Earth's radius to calculate the force exerted on it (RE=6400km).
Example
Question = Calculate the force exerted on a 75kg man standing on the surface of the Earth.
Answer =
Answer =
Gravitational Field Strength, g
The strength of a gravitational field, g, is the force per unit mass on a small test mass placed in that field. The test mass needs to be small because otherwise it will have an effect on the larger mass itself and would therefore alter the value for g.
g is also known as the ‘acceleration due to gravity’ because (when there are no external forces) any object will accelerate towards the centre of a large object with this speed. On Earth g ≈ 9.81ms^2. On the Moon g ≈ 1.6ms^2. Gravitational field strength, g, is measured in Nkg1.
By equating Newton's gravity equation with F=ma we can arrive at a useful equation for g.
g is also known as the ‘acceleration due to gravity’ because (when there are no external forces) any object will accelerate towards the centre of a large object with this speed. On Earth g ≈ 9.81ms^2. On the Moon g ≈ 1.6ms^2. Gravitational field strength, g, is measured in Nkg1.
By equating Newton's gravity equation with F=ma we can arrive at a useful equation for g.
This equation is only valid when the test mass is very small; otherwise the test mass will affect the gravity field of the larger object, altering the effect of its own pull of gravity i.e. M>>m (i.e. M is a lot larger than m). Again, this equation is only valid beyond the surface of the Earth; once you tunnel into the Earth the parts of the Earth that you have just tunneled past are contributing to a pull away from the centre of the Earth. The further in you go towards the centre the less mass in the centre there is to pull you towards the centre.
The graph below shows how g varies with distance from the centre of a large spherical object (e.g. a planet or star) or uniform density.
The graph below shows how g varies with distance from the centre of a large spherical object (e.g. a planet or star) or uniform density.
As you can see, g varies with the inverse proportional to the distance, r, from the centre of the planet. However, this is only from the surface of that object i.e. 'R' is the radius of the planet. If you were to tunnel through the Earth towards the core the gravitational field strength would decrease linearly, until it reached zero when you were in the centre. Essentially you would hover in the centre of the Earth if you reached it and it was hollow. Conceptually, this is because all the mass will be pulling you equally in every direction, making the resultant force zero, meaning that you wouldn't accelerate in any direction.
The reason why this is so is given below:
Volume of a sphere (i.e. a planet):
The reason why this is so is given below:
Volume of a sphere (i.e. a planet):
Density, p (Greek letter 'rho'), of a uniform sphere:
Using both equations and rearranging to get m the subject of the forumla:
Use the equation for g and this equation for mass to get the following:
Finally:
And so, within the Earth's surface (i.e. r<R) the gravitational field strength, g, is directly proportional to the distance, r, from the centre of the planet.
Gravitational Potential
“The gravitational potential, V, (Jkg^1), at a point is the work done, W, per unit mass, m, to move a small object from a place far away from gravity fields”. It can also be defined as the energy required per unit mass to remove that mass from the gravity field (i.e. place it far away). This is a useful variable to define since it gives us an idea of how much energy is involved in moving objects within a gravity field.
If a small object of mass, m, is moved from a gravitational potential of V1 to V2 work is done = m(V1V2) or just mΔV. This work is equal to the objects gain in gravitational potential energy, Epot, measured in Joules.
Close to the Earth's surface the work done in lifting an object is equal to the force x height raised. This work done is essentially the amount of gained gravitational potential energy, Epot. Since the force we are working against is the weight (=mg) and the height raised is just Δh, then:
This equation on works for small changes in h. This is because the field strength depends on how far away from the object you are. Therefore, if Δh is too large then g will vary and the equation will not hold true. It is important to note that Epot is not an absolute value, it is a relative value. it makes no difference where we choose our zero point of potential energy, it is arbitrary, as long as we are able to define how much that energy increase has been. We often define our zero energy as being at the floor.
The gravitational potential, V, at an infinite distance from a mass is zero. When the object moves closer to that mass we are getting work out of the system. When we try to push a mass away from an object we are doing work on the system. The general form for finding the absolute value for the in gravitational potential, V, is:
The gravitational potential, V, at an infinite distance from a mass is zero. When the object moves closer to that mass we are getting work out of the system. When we try to push a mass away from an object we are doing work on the system. The general form for finding the absolute value for the in gravitational potential, V, is:
The negative nature of V implies that gravity is an attractive force and that the system binds mass together and that you need energy to escape the system.
To find the change in gravitational potential, ΔV , its the potential at one distance minus the potential at the other distance:
To find the change in gravitational potential, ΔV , its the potential at one distance minus the potential at the other distance:
To calculate the energy (or work done) required to go from potential V1 to V2 you must multiply by the mass of the object. Since V is the energy per unit mass. The total energy to complete this movement is just:
This can be represented on a graph. The area under a forcedistance graph is the work done.
The gravitational potential near the Earth is depicted in the following graph:
Note: this is not an inverse square law. Rather, when the distance doubles the potential halves. When the distance trebles the potential decreases by a third. Also notice the potential is negative. This type of graph is often spoken about in terms of a 'potential well'; you need energy input into the system to escape this well.
Equipotentials and Potential Gradients
Equipotentials are 3D surface of constant gravitational potential (on the diagram opposite they appear like contours on a 2D OS map). At an increased distance from the surface, the gravitational field becomes weaker so the gain of gravitational potential energy per meter of height gained becomes less and less i.e. for the same difference in potential these equipotentials get more spread out the further from the object. Near the surface of the object the equipotentials are horizontal (parallel to the ground) i.e. they are uniform over small regions. The potential gradient (Jkg^1m^1) in a gravity field is the change of potential per metre at that point. Near the Earth's surface a 1kg mass gains 9.81J of energy per meter it is raised. Further from the Each this becomes less and less. 
In general, the potential gradient is:
Another useful form of this equation is: