Gravitation & Weight
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Gravitation
Gravitation is the force that describes the gravitational attraction between two
bodies with mass. The gravitational attraction force is defined by Newton's universal
law of gravitation (classical physics).
F = G. (m1 . m2 ) / r
^{2}
where:
G: Gravitational constant with the value of 6.67 × 10
^{−11} N m
^{2}
kg
^{-2}
M1: The mass of body 1 in kg
M2: The mass of body 2 in kg
We will use this formula to calculate the force between an object and the earth.
We can make the following observations: the distance r is a practical constant when
considering objects that are close to the surface of the earth. By substituting G, r and
the mass of the earth with their values, we get:
F = 9.8 m
Compare this to the formula F = m.a, and it follows that for objects in the vicinity
of the earth's surface, the acceleration 'a' equals 9.8 m/s
^{2}.
This value is called the earth's gravitational acceleration g.
F = g . m
where:
F : Force in N
g : Gravitational acceleration in m/s
^{2}, with
g = 9.8
m : Mass in kg
Weight
When an object is attracted with a force F = g.m, the weight of the object
resting on a solid surface is defined as the reaction force against the surface.
Explanation
The force that an object encounters when near to the earth's surface can be calculated
from the universal law of gravitational attraction. It appears that a body with
mass, m, encounters an acceleration force, g, of 9.8 m/s^{2}. This results in the earth's
gravitational force, F = g.m. The assumptions we make is that the object's mass
is much less than the mass of the earth. The same must apply to the diameters. The
earth's diameter must be much bigger than the object’s diameter. These assumptions
hold true at the earth’s surface, but when looking, for example, at the gravitational
attraction forces between bodies in space and the earth (e.g. the moon and sun) these assumptions do not apply.
In aviation, these assumptions generally hold true, with the only factor of influence
being the distance to the earth. Let’s look at the influence of the distance to
earth with an example.
An airplane is flying at a height of 10 Km (32,808 feet). By what amount is the
plane attracted to the earth compared to a grounded airplane?
Answer:
The difference in attraction force at a height of 10 km can be found by:
100% . (1/r
^{2} - 1/(r + 10)
^{2})
/ (1/r
^{2})
Diameter earth: = 12,756 Km, r = 6378
Km.
1/r
^{2}=1/6378
^{2}
= 2,4582778622933706834238618738901e-8
1/(r + 10)
^{2} =
1/(6378+10)
^{2} = 2,4505873371682737945168794495314e-8
(2,4583 - 2,4506) / 2,4583 = 0,0031 -> 0,31%
Conclusion, at a height of 10 km (32,808 feet) an aircraft will experience 0.31%
less gravitational attraction force. In most cases, this is of no concern.
Examples
A helicopter with a mass of 500 Kg is flying in a hover at a height of 2 m (6.6
feet). How great is the gravitational force working on the helicopter which
must be overcome by the vertical thrust of the rotor system?
Answer:
F = g.m = 9.8 . 500 = 4900 N.
A helicopter is diving down, and for a brief moment reaches an acceleration of -9.8
m/s
^{2} (the acceleration is directed towards the earth's surface). What is
the weight the pilot will experience sitting in his seat? The pilot has a mass
of 95kg.
Answer:
Weight is a reaction force. There are two acceleration forces acting upon the pilot.
The first is due to the gravitational acceleration, g. The second is due
to the acceleration of the helicopter (relative to the earth's surface):
F1 = m . g
F2 = m . a
g = 9.8
a = -9.8
F
_{total} = F1 + F2 = m . (9.8 - 9.8) = 0 N
The pilot experiences weightlessness.
Now, the helicopter starts decelerating in a vertical direction at a rate of 5 m/s
^{2}. What is the weight
of the pilot?
Answer:
F
_{total} = 95 (9.8 + 5) = 1406N. Compared to the norm (only gravitational
pull), the weight has increased by (9.8 + 5) / 9.8, or 51%. Of course, the
mass remains at 95kg. Note that this increase in force working
on the pilot leads to a reaction force at the seat, holding him in place.