Fields Tutorial 2  Gravity Fields and Energy
The
gravitational potential of a point 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, 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:
[G
= 6.67
´ 10^{11} Nm^{2} kg^{2};
M
= mass of Earth or other planet;
r =
distance from centre.]
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
Question 1 
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:
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:
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. Instead use
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) 
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
Ep
=  6.67
´
10^{11} Nm^{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
Ep
=  6.67
´
10^{11} Nm^{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? 
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 ´
60 ´
60
Þ w = 2 ´ p ´ 1.16 ´ 10^{5} Hz = 7.27 ´ 10^{5} rad/s. 
Now
we need to consider the centripetal acceleration:
a = w^{2}r. We also know that the acceleration is also given by g = GM/r^{2}
Þ w^{2}r
=
GM/r^{2}
[We will ignore the
minus sign.]
Þ
r^{3}
= GM
= 6.67
´
10^{11} N m^{2} kg^{1}
´
5.98 ´
10^{24} kg = 7.55
´
10^{22} m^{3}
(7.27
´
10^{5} Hz)^{2}
Þ r = ^{3}Ö 7.55 ´ 10^{22} m^{3} = 4.24 × 10^{7} m 
Now
we can work out the forward velocity:
v = wr = 7.27 ´ 10^{5} rad/s ´ 4.24 ´ 10^{7} m = 3100 m/s. 
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.
(a)
How long does it take to orbit?
(b)
What is the centripetal acceleration at this speed?
(c)
What is the acceleration due to gravity at this distance? (d) Will it remain in that orbit? Mass of Mars = 6.42 × 10^{23} kg 
Equipotentials in Gravity Fields
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.
A satellite orbiting the Earth along the 40 MJ kg^{1} equipotential does zero work. 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) 