Fields Tutorial 1  Force and Gravity Fields
Gravity
Fields
A
field is a region in which a force is felt.
Gravity
is a very mysterious force. Nobody
knows why objects have this attractive force between them, even if they are far
apart. This attraction occurs for
any object with mass, however small. The
force is very small, and always attractive. We never
get a repulsive gravitational force.
Newton
found that all objects accelerate towards the Earth at 9.81 m/s^{2}. He also worked out the moon has an acceleration towards the
Earth of 2.72 m/s^{2}. Therefore:
Acceleration
of Moon = 2.72 × 10^{3} m/s^{2} =
1 =
1
Acceleration of apple
9.81 m/s^{2}
3610 (60.1)^{2}
He
also worked out the distances between the moon and the centre of the Earth and
the distance between the apple and the centre of the Earth.
Their ratio also turned out to be 60.1:1. This formed the basis of
Newton’s
Inverse Square Law of Gravitation.
Every particle of matter in the
Universe attracts every other particle with a gravitational force that is
proportional to the products of the masses and inversely proportional to the
square of the distance between them.
We
can write this in code form:
We
can now add the constant of proportionality:
The Gravitational Constant, G, has a value of 6.67 ´ 10 ^{–11} N m^{2} kg^{2}. The minus sign tells us that gravity is an attractive force. In gravity problems, we always assume that the masses are point masses.
Question 1 
What is the force of attraction between two giant ships each of mass 100 000 tonnes, 50 m apart? (1 tonne = 1000 kg) 
Gravity is a very small force, but it exists between all objects. There is a tiny gravitational attraction between you and the student sitting next to you in your physics lesson. You can't feel it because it's negligible. The only reason we feel gravity is that the Earth is a very large object. Remember that as well as the Earth attracting us towards it, we are attracting the Earth towards us. The tiny attraction has important implications for dust particles coming together in space to form stars and planets.
We are all familiar with the magnetic field of a bar magnet. The field lines show up as areas of attraction and repulsion at the poles and the concentration of field lines at any point. The closer we are to the magnet, the stronger the lines of force.
We
can do a similar exercise with a gravity field; only this time there are
attractive forces involved and no repulsive forces.
The Earth, in common with many other planets, is very nearly a perfect sphere (relatively smoother than a billiards ball). Its gravity field is radial with the pull being directly towards the centre. The closer in we get, the stronger the pull.
Question 2 
How can you tell from the diagram that the gravity field gets stronger? 
The
concentration of gravitational field
lines is an indication of the gravitational
field strength at any point, which is formally defined as:
The
gravitational force per unit mass at that point.
So
we can write that statement in code as:
g = gravitational force = F [Units – newtons per kilogram (N/kg)]
mass m
You
will have met the expression
F/m
in the context of
a = F/m,
so it
doesn’t take a genius to see that gravitational field strength is the same
thing as acceleration. A gravitational
field strength of 9.81 N/kg causes an acceleration
of 9.81 m/s^{2}.
From our derivation of gravitational field strength
and Newton’s Law of Gravitation, we can
derive an equation to tell us the value of the field strength at a distance
r
from the centre of the Earth, mass
M.
F = GmM Þ g = F/m Þ g = GmM = GM
r^{2} m r^{2} r^{2}
Gravitational Field strength is a vector because it has a direction.
Remember always to take the radius of the Earth into account.
Worked Example
Assuming
the Earth to be a perfect sphere, find its mean density from the following
data:

Work
out the mass using g = 
GM
r^{2}
M = gr^{2} = 9.81
N/kg
´
(6.37
´ 10^{6} m)^{2}
= 5.96
´
10^{24} kg G 6.67 ´ 10^{11} Nm^{2} kg^{2} 
Now
work out the volume of the Earth using
V
= 4/3
pr^{3}:
V
= 4
´
p
´
(6.37
´
10^{6} m)^{3}
= 1.083
´
10^{21} m^{3} 3 
Now
use density = mass
¸
volume to get the answer:
Density
= 5.96
´
10^{24} kg = 5500 kg m^{3} 1.083 ´ 10^{21} m^{3} 
Question 3 
Jupiter has a mass of 1.91 × 10^{27} kg and its radius is 7.14 × 10^{7} m. What is its gravitational field strength? What is the acceleration due to gravity? 
If
we draw a graph of the gravity field strength going to the centre of the Earth,
we would not find a simple inverse square relationship, as if the Earth were a
point mass. Instead we would see a more complex variation. The Earth
is not a uniform sphere, but has a dense core and a less dense crust.
The value of g can be calculated from
seismic surveys and the graph above shows its variation with distance.
Note that the graph is drawn as a negative function as the field is
attractive. You may see it drawn as
a positive function.
At the centre of the Earth, g = 0, because matter will be pulled equally in all directions so that the overall force is zero.
A very common beartrap
When we work out g for distances away from the Earth's surface, we MUST add on the radius of the Earth. Suppose a satellite were flying 1.0 × 10^{6} m above the Earth's surface the distance r would NOT be 1.0 × 10^{6} m but (1.0 × 10^{6} m + 6.37 ´ 10^{6} m) = 7.37 × 10^{6} m.
A satellite is 4000 km above the Earth's surface. What is the acceleration due to gravity at this point? Mass of Earth = 6.0 × 10^{24} kg 