F = G. (m1 . m2 ) / r

where:

G: Gravitational constant with the value of 6.67 × 10

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

This value is called the earth's gravitational acceleration g.

F = g . m

where:

F : Force in N

g : Gravitational acceleration in m/s

m : Mass in kg

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?100% . (1/r

Diameter earth: = 12,756 Km, r = 6378 Km.

1/r

1/(r + 10)

(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.

F = g.m = 9.8 . 500 = 4900 N.

F1 = m . g

F2 = m . a

g = 9.8

a = -9.8

F

The pilot experiences weightlessness.

Now, the helicopter starts decelerating in a vertical direction at a rate of 5 m/s

- Advertisements -

Comments are disabled.

2: (Book) Principles of Helicopter Flight

3: Microsoft FSX Steam Edition

4: Logitech Extreme 3D Pro Joystick

5: Saitek Pro Flight Rudder Pedals

[ Log In ]