Fields Tutorial 2  Gravity Fields and Energy
Gravitational Potential Energy
The
gravitational potential energy of a point is defined as:
The
work done on a mass in moving it to that point from a point remote from all
other masses.
In
other words, this means the work done to move a unit mass from infinity to the
point under consideration.
The
zero point for gravitational potential is at infinity, so as we are moving towards the Earth, we are getting work
out of the system.
Therefore gravitational potential is negative. If we moved the object away to infinity, we would have to do
a job of work on the object. We
work out gravitational potential with the formula:
E_{p} = potential energy (J);
G = 6.67 ´ 10^{11} Nm^{2} kg^{2};
M = mass of Earth or other planet (kg);
r =
distance from centre (m).]
Another
bear trap
Ep = mgDh is only true when we are very close to the Earth’s surface. So do NOT use it for objects out in space. 
Gravitational Potential
The gravitational potential is defined as:
The
work done on a unit mass in moving it to that point from a point remote from all
other masses.
In other words:
The energy per unit mass
So we can summarise this as:
The physics code for gravitational potential is V_{g} and the units are Joules per kilogram (J kg^{1}). We can write a second equation by combining this with the equation for potential energy to give us:
If we go back to our graph of gravitational field strength against distance we get:
We
see that the gravitational potential is the area
under the graph:
since
work = force
´
distance moved.
work done per unit mass = force per unit mass ´ distance moved
The
asteroid Ceres has a diameter of 785 km and a mass of 1.0 × 10^{20} kg.
Calculate:
(a)
the strength of the gravitational field; (b) the gravitational potential at the surface. 
If we plot V against r, we get the following graph:
From
this graph we can see that the gradient gives us the gravitational field
strength,
g.
So we can say that:
Notice the apparent contradiction of the gradient being a positive slope and the equation being negative. The DV term will be negative, while the gradient is positive. Therefore we need to make the gradient negative for the positive slope to be consistent with the negative DV term.
We
can go on to use the idea of gravitational potential to find an expression for
the potential energy.
We have defined potential as energy per unit mass, so we can work out the
total energy for any mass by multiplying the potential by the mass:
Worked Example
A satellite of mass 200 kg is
to be moved from an orbit of 200 km above the Earth's surface to a an
orbit of 400 km. What work
needs to be done? (Radius of Earth = 6.37 ´ 10^{6} m; mass of Earth = 6.0 × 10^{24} kg) 
Answer
We need to calculate the gravitational potential energy
at 200 km:
200
km = 0.2 × 10^{6} m
Þ
r
= 6.57 × 10^{6 }m
E_{p}
=  6.67
´
10^{11} N m^{2} kg^{2} × 6.0 × 10^{24}
kg ×
200 kg
= 1.22 × 10^{10} J 6.57 × 10^{6 }m

Now
do the same for the 400 km:
400
km = 0.4 × 10^{6} m
Þ
r = 7.01 × 10^{6 }m
E_{p} =  6.67
´
10^{11} N m^{2} kg^{2} × 6.0 × 10^{24}
kg ×
200 kg
= 1.18 × 10^{10} J 6.77 × 10^{6 }m

Now we can work out the job of work we have to do to
shift the satellite:
Work
done =
1.18
× 10^{10} J  1.22 × 10^{10} J = +0.04 × 10^{10}
J = +4.0 × 10^{8} J
The
plus sign tells us that a job of work has to be done.

Some deep space
cosmonauts land on a small planet. They
know that the mass of their craft is 500 × 10^{3} kg and that to stop, they had to
use 6.0 × 10^{11} J of energy. They
also worked out that the radius of the planet is 1500 km.
(a)
Explain why the potential energy is  6.0 × 10^{11} J.
Why is the sign negative?
(b)
What is the mass of the planet?
(c) What is its density? 
What a wretchedly bad piece of bad contrived science fiction!
Motion of Masses in Gravitational Fields
Newton’s Laws of Gravitation can be used to explain the motion of planets and stars. Much of modern space exploration uses the three hundredyearold model. Orbiting satellites are NOT doing gravity defiance acts; instead they are actually falling in a curved path towards the Earth all the time. However they have a sufficient forwards velocity to miss the Earth all the time. Since the gravity field is radial, the force acts at 90^{o} to the direction of travel all the time. Therefore the path is circular. If we stopped gravity, the satellite would fly off tangentially into space in a straight line. If we stopped the satellite, it would fall straight back to Earth.
For a satellite to be in a particular orbit, a particular velocity is required or a given distance. Some satellites are placed so that they go in an easterly direction, completing one orbit each day. They remain above one given point on the Earth’s surface, so are called geostationary. This kind of satellite orbit is used in telecommunications. Other satellites move in a polar orbit so that they can perform sweeps of the surface. Spy satellites use a polar orbit.
When considering the motion of satellites in orbit, you have to know the rules of simple circular motion. Click HERE to go back to Further Mechanics Tutorial 3 to revise these.
Useful formulae include:
w = 2pf;
a = v^{2} /r;
a
= rw^{2}
.
Worked
Example A communications satellite is to be placed in a circular geostationary orbit. What must its height and speed be? 
Answer
This
question seems to be remarkably lacking in information, but there is an
answer.
Use Newton’s Laws of Gravitation to solve this. There is a single force acting on the satellite, gravitational attraction, so the satellite is acceleration all the time towards the centre of the Earth. 
We need the satellite to be travelling at a sufficient forwards velocity so that it completes ONE orbit every 24 hours. We need to work out the angular velocity before we can work out the linear speed. 
w = 2pf. We need to work out f.
f
= ____1_____ = 1.16 ´
10^{5} Hz
24 h ´
60 min ´
60
Þ w = 2 ´ p ´ 1.16 ´ 10^{5} Hz = 7.27 ´ 10^{5} rad s^{1}. 
Now
we need to consider the centripetal acceleration:
a = w^{2}r. (Note the minus sign indicating that the direction of the acceleration is to the centre of the circle)
We also know that the acceleration is also given by g = GM/r^{2}
Þ w^{2}r = GM/r^{2} [The minus signs then cancel out.]
Þ
r^{3}
= GM/w^{2}
= 6.67
´
10^{11} N m^{2} kg^{1}
´
5.98 ´
10^{24} kg = 7.55
´
10^{22} m^{3}
(7.27
´
10^{5} rad s^{1})^{2}
Þ r = (7.55 ´ 10^{22} m^{3})^{1/3} = 4.24 × 10^{7} m 
Now
we can work out the forward velocity:
v = wr = 7.27 ´ 10^{5} rad s^{1} ´ 4.24 ´ 10^{7} m = 3100 m s^{1}. 
The planet Mars has a diameter of 6800 km. A satellite is in orbit 5000 km above the planet's surface travelling at a speed of 7100 m s^{1}.
(a) How long does it take to orbit?
(b)
What is the centripetal acceleration at this speed?
(c) Calculate the acceleration due
to gravity at this distance. Give your answer to an appropriate number of
significant figures. (d) Will it remain in that orbit? Mass of Mars = 6.42 × 10^{23} kg 
Equipotentials in Gravity Fields
So far we have looked at radial fields. Close to the surface of a planet, the field can be considered to be uniform. The idea is summarised in the diagram below:
In a uniform gravity field, we are familiar with the equation for potential energy:
We can rewrite the equation in terms of gravitational potential (energy per unit mass):
At a height of 10 m, the potential is 98 J kg^{1}; at 20 m, it’s 196 J kg^{1}, and so on. We can show these potentials as lines. Every point along these lines has the same potential, so they are called lines of equipotential.
If we move along a line, we do no work at all, because we have not moved any distance against the force of gravity.
Contours on a map are lines of equipotential. They are marked in 10 metre intervals rather than 98 J kg^{1} intervals, because the actual potentials would mean nothing to the map user.
In a radial field, the lines of equipotential are not equally spaced, as the energy per unit mass varies inversely with the distance.
A satellite orbiting the Earth along the 40 MJ kg^{1} equipotential does zero work, because the force is always at 90^{o}. This is why it does not lose kinetic energy. If it did, it would fall towards the Earth.
What is the radius of the orbit of the satellite above? How high is the orbit above the Earth's surface? (mass of the Earth = 5.98 × 10^{24} kg; radius of Earth = 6.37 × 10^{6} m) 
Models of Gravity (Extension)
Gravity is quite a difficult concept to conceive.
In the Grand Unified Theory model that particle physicists use, there is a gauge boson for gravity called the graviton. Since gravity is fundamental to our understanding of the universe, it is reasonable to suppose that the Universe is crawling with the little brutes, each acting like a boomerang thrown out by the first object to snare a second object. So far no such particles have been observed. That is hardly surprising as they have zero mass like photons.
Photons are considered to be packets of electromagnetic wave. Now that gravity waves have been detected (with a wave speed of 3.0 × 10^{8} m s^{1}), it might be reasonable to suggest that a graviton is a packet of a gravity wave.
Other physicists model gravity as a kind of rubber sheet. This is easier to visualise. Consider the frame of a trampoline (like the one you may have in your garden), onto which is stretched a uniform rubber sheet. The idea is shown below:
If we place a large heavy ball in the middle of the rubber sheet, we see this side view:
The rubber sheet forms a shape that is the same as a 1/r relationship. The ball is in a potential well. Work has to be done to extract it. The diagram below shows a potential well with its equipotentials. This was drawn with Mathematica software.
Image by AllenMcC WikiMedia Commons
In 1915, Albert Einstein introduced the ideas of SpaceTime to as part of his theory of General Relativity. It considers gravity as having properties of both space and time. It is a complex theory, including concepts of quantum gravity. However it is supported by astrophysics observations. Superficially the diagrams associated with spacetime resemble those of energy wells, but it is wrong to say this.
The diagram below shows how space time is distorted by the presence of a large object like a planet.
Image by Mysid WikiMedia Commons
Using Calculus to link Gravitational Force and Gravitational Energy (Extension)
The graph below was produced to model the force acting on a satellite of a certain mass at a certain height above the surface of the Earth. The mass of the Earth is 5.98 × 10^{24} kg and has a radius of 6.37 × 10^{6} m. The satellite has a mass of 3500 kg.
We know that if we were to increase the height from 0.4 × 10^{6} m to 1.2 × 10^{6} m, we would have to do a job of work, as we are applying a force against the gravity field. On the graph below, the red lines show this and the work done is the area under the graph between the two red lines.
The counting of squares would only give an approximation. Calculus integration would give us a more precise result. So we can write:
The minus sign tell us that we are going in the opposite direction to the force.
We know that:
So we can change the equation to:
So we can write this as:
In this case G, m_{1}, and m_{2} are constants. Using the power rule for integration, we can write:
Maths Note Power rule for integration
For differentiation the converse is true:
Yes, I have missed out the constant, C. In this case, C = 0. 
Worked example A satellite of mass 3500 kg is to be raised from an orbit of 400 km above the surface of the Earth to an orbit of 1200 km above the surface of the Earth. Calculate the work that needs to be done. (Mass of the Earth = 5.98 × 10^{24} kg; radius of Earth = 6.37 × 10^{6} m) 
Answer Use:
Remember to add the radius of the Earth: r_{2} = 1.20 × 10^{6} m + 6.37 × 10^{6} m = 7.57 × 10^{6} m r_{1} = 0.400 × 10^{6} m + 6.37 × 10^{6} m = 6.77 × 10^{6} m
W = [(6.67 × 10^{11} N m^{2} kg^{2} × 5.98 × 10^{24} kg × 3500 kg) ÷ 7.57 × 10^{6} m]  [(6.67 × 10^{11} N m^{2} kg^{2} × 5.98 × 10^{24} kg × 3500 kg) ÷ 6.77 × 10^{6} m]
W = 1.844 × 10^{11} J   2.062 × 10^{11} J = +2.18 × 10^{10} J
The positive sign means that work has to be put in. Watch the signs! The work done results in an increase in potential energy of +2.18 × 10^{10} J.

We can apply the same rule to the relationship between gravitational field strength and gravitational potential. Gravitational field strength is force per unit mass and gravitational potential is energy per unit mass. This is shown on the graph:
We can therefore write:
The converse is also true. The gravitational field strength is the gradient of the graph of gravitational potential against distance. So we can write:
Similarly the force is product of the gravitational field strength and the mass m_{2}, so we can write: