Fields Tutorial 1  Force and Gravity Fields
Contents 
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’s Inverse Square Law of Gravitation
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.
Often the equation is written with M being the code for the big mass (e.g. a planet) and m for the little mass (e.g. a satellite).
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.
We model planets (and other objects) as point masses, with all the mass in the centre of the sphere.
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:
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^{1} causes an acceleration
of 9.81 m s^{2}.
Gravitational Field Strength between the Earth and the Moon
The gravitational field between the Earth and Moon can be represented using field line like this:
Worked example Calculate the distance from the Earth's surface of the point which the the gravitational field strength is zero, given the following data: Distance between the Earth and the Moon = 384 000 km; Mass of the Earth = 5.98 × 10^{24} kg; Mass of the Moon = 7.35 × 10^{22} kg; Radius of the Earth = 6.37 × 10^{6} m.
Give your answer to an appropriate number of significant figures. 
Answer Using:
we can write:
The minus signs and the G terms cancel out:
This rearranges to:
Then we can substitute for the masses:
Rearranging gives:
Therefore:
We also know that:
So we can substitute:
r_{M} = 3.84 × 10^{8} m ÷ 10.02 = 3.83 × 10^{7} m Therefore: r_{E} = 3.84 × 10^{8} m  3.83 × 10^{7} m = 3.46 × 10^{8} m
We haven't finished yet. We need to subtract the radius of the Earth.
Distance = 3.46 × 10^{8} m  6.37 × 10^{6} m = 3.39 × 10^{8} m (3 s.f.)
3 significant figures are appropriate as the data are to 3 significant figures. 
The distance to this point from the Earth's surface is about 340 000 km, about 90 % of the total distance between the Earth and the Moon. It is sometimes called the Lagrange Point, or the Gravitational Null Point.
Explain what, if any, would be the effect on the result of including the mass of the space probe. 
Derivation of Gravitational Field Strength
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.
Newton’s Law of Gravitation:
Adding the constant of proportionality:
Gravitational field strength is defined as force per unit mass:
This shows consistent with Newton II:
Therefore:
Gravitational Field strength is a vector because it has a direction.
Radius of the Earth
Remember always to take the radius of the Earth into account.
We assume that the Earth is a point mass.
So anything on the surface is R metres from the point mass.
If something is h metres above the surface, we need to add the radius of the Earth.
r = (R + h)
Worked Example
Assuming
the Earth to be a perfect sphere, find its mean density from the following
data:

Answer Work out the mass using
M = gr^{2} = 9.81
N kg^{1}
´
(6.37
´ 10^{6} m)^{2}
= 5.96
´
10^{24} kg G 6.67 ´ 10^{11} N m^{2} kg^{2}

Now work out the volume of the Earth using:
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}

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? 
Variation of Gravitational Field Strength with Distance
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 
In close proximity to the Earth's surface, the gravitational field strength can be considered to be constant at g = 9.81 N kg^{1} and the acceleration due to gravity would be 9.81 m s^{2}.