Gravitational Fields
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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/s2. He also worked out the moon has an acceleration towards the
Earth of 2.72 m/s2. Therefore:
Acceleration
of Moon = 2.72 × 10-3 m/s2 =
1 =
1
Acceleration of apple
9.81 m/s2
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:
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The Gravitational Constant, G, has a value of 6.67 ´ 10 –11 N m2 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.
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.

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/s2.
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
r2 m r2 r2

Gravitational Field strength is a vector because it has a direction. Remember always to take the radius of the Earth into account.
Worked Example
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Assuming
the Earth to be a perfect sphere, find its mean density from the following
data: |
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Work
out the mass using g = - GM
r2
M = gr2 = 9.81
N/kg
´
(6.37
´ 106 m)2
= 5.96
´
1024 kg G 6.67 ´ 10-11 Nm2 kg-2 |
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Now
work out the volume of the Earth using V
= 4/3
pr3: V
= 4
´
p
´
(6.37
´
106 m)3
= 1.083
´
1021 m3 3 |
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Now
use density = mass ¸
volume to get the answer: Density
= 5.96
´
1024 kg = 5500 kg m-3 1.083 ´ 1021 m3 |
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 bear-trap

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 × 106 m above the Earth's surface the distance r would NOT be 1.0 × 106 m but (1.0 × 106 m + 6.37 ´ 106 m) = 7.37 × 106 m.
Gravitational
Potential
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 Nm2 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.
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
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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 ´ 106 m; mass of Earth = 6.0 × 1024 kg) |
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We need to calculate the gravitational potential energy
at 200 km: 200
km = 0.2 × 106 m Þ r = 6.57 × 106 m Ep
= - 6.67
´
10-11 Nm2 kg-2 × 6.0 × 1024
kg ×
200 kg
= -1.22 × 1010 J 6.57 × 106 m
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Now
do the same for the 400 km: 400
km = 0.4 × 106 m
Þ r = 7.01 × 106 m Ep
= - 6.67
´
10-11 Nm2 kg-2 × 6.0 × 1024
kg ×
200 kg
= -1.18 × 1010 J 6.77 × 106 m
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Now we can work out the job of work we have to do to
shift the satellite: Work
done = -1.18
× 1010 J - -1.22 × 1010 J = +0.04 × 1010
J = +4.0 × 108 J
The
plus sign tells us that a job of work has to be done.
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