# Simple Missile Ballistics, Orbits and Aerodynamics

## Lift and Drag

### The Artilleryman’s Range Equations

The ballistic trajectory of a projectile follows a parabolic path defined by the following equations:

At any time t, the horizontal and vertical displacement: x and y from the origin are given by:

x = V0.t.cos(Θ)

y = V0.t.sin(Θ) - ½.g.t2

The total flight time is given by:

t = 2.V0.sin(Θ)/g

where

t is the time of flight

g is the acceleration due to the Earth's gravity at sea level = 9.8 m/s2

V0 is the Intial velocity

Θ is the initial angle of elevation

x is the horizontal displacement

y is the verical displacement

Assumptions:

No aerodynamic drag

Constant g force

Launch and target both on same level

Range = V02sin2Θ/g

Maximum Range = V02/g occurs when the angle of elevation of the initial launch = 45°

Maximum Altitude During Trajectory = V02sin2Θ/2g

Maximum Possible Altitude = V02/2g which occurs when the angle of elevation of the initial launch Θ = 90° (A missile lauched straight up would fall back on the head of the launcher)

Note that the 70 and 20 degree trajectories have the same range, as do any pair of launches at complementary angles.

### Drag Force (Newtons) = 0.5 x P x V2 x Cd x A

where:

P = Density of Air (kg/m3)

The density of air decreases linearly with altitude decreasing the drag by the same ratio

P ≈ 1.29 kg/m3 @ sea level

P ≈ 0.232 kg/m3 @ 12,000 m (7.5 Miles) = 18% of the air density at sea level

P ≈ 0.001 kg/m3 @ 50 kilometers (The stratopause)

V = Velocity (m/s) or Air speed

The drag increases as the square of the velocity

Mach 1 (the speed of sound) = 340 m/s @ sea level

Mach 1 ≈ 295 m/s @ 12,000 m altitude

Cd = Co-efficient of Drag

The drag coefficient is constant for a given missile.

It depends mostly on the shape of the missile and its appendages

Cd ≈ 0.6 to 0.95 for rockets

A = Sectional Area (m2)

The effective cross sectional area is constant for a given missile

A ≈ 2.76 m2 for a 1.65 m diameter missile. (The V-2)

See more about the Components of Drag

### Aerodynamic Lift

For a wing or a lifting body at a given angle of attack the lift force is given by:

### Aerodynamic Lift Force (Newtons) = 0.5 x P x V2 x CL x A

where

CL = Co-efficient of Lift

The lift coefficient is constant for a given angle of attack and a given wing/body shape.

It depends on the cross sectional profile of the wing or lifting body and the angle of attack.

CL ≈ 0 to 1.6 for a light aicraft for angles of attack betwen 0° and 17°

CL ≈ 0 to 1.3 for a fighter jet for angles of attack betwen 0° and 30°

A = The effective area of the wing or lifting surface (m2)

As with drag, the lift increases with the square of the velocity but decreases with altitude as the density of the air decreases.

### Aerodynamic Efficiency

The Lift over Drag Ratio:  L / D  or  CL/ Cd at a given speed and angle of attack gives a measure of the Aerodynamic Efficiency of the aerofoil at that speed and angle of attack.

Note:

At very low speeds both the drag and the lift are also very small.

Because lift and drag forces are both proportional to the square of the velocity, the lift and drag at Mach 3 will be 9 times greater than the lift and drag at Mach 1.

Maximum range can be achieved by flying at very high altitudes where the density of the air is very low and the drag will consequently also be very low.

But

Aerodynamic flight is not possible above the atmosphere in the vacuum of space since the density of the ambient environment will be zero and so both the lift and drag will also be zero.

## Guided Missiles

### Free Flight

The ballistic trajectory of projectile is set by its speed and direction at the instant when the initial propulsion has stopped. After this point no further control of the projectile's trajectory is possible. For an artillery shell this point is when the projectile emerges from the gun barrel. For rocket powered ballistic missiles the balistic trajectory starts when the rocket motor is turned off.

As noted above, the maximum range on level ground occurs when launch angle of elecation of the projectile is 45°. But a rocket begins its ballistic trajectory at high altitude, equivalent to starting somewhere along the trajectory of a projectile launched from the ground. In this case the maximum range will occur at a angle of elevation offset from 45°. Similarly the maximum range of an artillery shell lauched from an elevated site to a target below will also be offset from 45°

The actual trajectory followed by a projectile will be influenced by the prevailing atmospheric wind conditions encountered during the flight as well as the density of the air through which it travels which affects the drag (See opposite top). For very high trajectories the affect of reduced gravity at high altitudes must be also taken into account. (See below)

### Controlled Flight

Controlling the direction and speed of a projectile such as a cruise missile or a spacecraft in flight requires the addition of a guidance control system to provide a reference direction and a means of keeping it on track. Thus it needs an on board power source and a means of changing the vehicle's speed and direction

Aerodynamic controls which depend on the air pressure on control surfaces such as wings, rudders and fins can only work in the Earth's atmosphere. About 99% of the total air mass of the atmosphere can be found in the bottom two layers, the troposphere and The stratosphere. Beyond the Earth's atmosphere there is no aerodynamic drag (or lift) at all. At the altitude of the stratopause which is the boundary between the stratosphere and the mesosphere, typically at an altitude 50 to 55 km (160,000 to 180,000 ft) the air density and pressure are only 1/1000 of that at sea level and insufficent to support practical aerodynamic lift or controls.

Aerodynamic controls are also ineffective at very low vehicle speeds.

Beyond the Earth's atmosphere, controlling a space vehicle's attitude and speed needs an active method to deflect the direction of the thrust of the main rocket or jet exhaust such as mounting the rocket engine on a steerable gimbals or using rudders within the exhaust stream, Alternatively, a series of small auxilliary jet thrusters could be used to alter the attitude of the missile.

## Entering Space

### Gravitational Force at Different Altitudes

The Inverse Square Law

The gravitational force between 2 bodies is proportional to their masses and inversely proportional to the distance between their centres.

Thus the gravitational force “F” between two bodies is given by:

F = mg = GMm

d2

where

G = The gravitational constant = 6.67 x 10-11 m3/kg. s2

M = Mass of body 1

m = Mass of body 2

d = Distance between the centres of the 2 bodies

In the case of the Earth's gravitational pull

M is the mass of the Earth = 5.976x1024 kg

R is the radius of the Earth = 6380 kms (at the equator)

g is the acceleration due to the Earth's gravity at sea level = 9.8 m/s2

The radius of the second body is negligibly small compared with the radius of the Earth

Then the acceleration due to gravity at sea level is given by:

g = GM/R2

The acceleration due to gravity ga at any height "h" above the surface of the Earth is given by:

ga = GM/(R+h)2

So that

ga /g= [R/(R+h)]2

Thus at an altitude of 100 kms (62 miles) above sea level, known as the Kármán line, which is commonly used to define the boundary between the Earth's atmosphere and outer space, the acceleration due to gravity is given by;

ga= g(6380/6480)2 = 96.94% of g at sea level.

and the force of gravity is 3.06% less than gravitational force at sea level.

The effect of this would be to increase slightly the range of a high altitude projectile.

Similarly at 100,000 kms above the surface of the Earth

ga = g(6380/106,380)2 = 0.0036g or 0.36% of g on Earth

so that spacecraft at this altitude are still under the weak influence of the Earth’s gravity

### Escape Velocity

For a spacecraft to escape from the Earth's gravity its kinetic energy must be equal to or greater than the potential energy of gravitation.

Thus

½m Ve2 = Fd = GMmd/d2

so that

Ve= √2GM/d

where

F = The gravitational force

Ve = The escape velocity

d = The distance between the centre of the Earth and the spacecraft

G = The gravitational constant = 6.67 x 10-11 m3/kg. s2

M = The mass of the Earth = 6.976x1024 kg

m = The mass of the spacecraft

On the surface of the Earth, d = 6380 km so that:

The Escape Velocity Ve is about 11.2 kilometers per second (≈25,000 miles per hour) which is approximately 33 times the speed of sound (Mach 33).

At the Kármán line (100 kilometers above sea level) the escape velocity Ve is reduced to 11.1 kilometers per second

At 10,000 kilometers altitude in "space", Ve reduces to 7.0 km/s.

and at 100,000 kiliometers Ve reduces to 2.7 km/s

### Orbital Velocity

For a circular orbit, the centripetal force on the spacecraft should exactly balance the gravitational force. Thus:

GMm/d2 = mVo2/d

Where

Vo = The orbital velocity of the spacecraft

So that

Vo = √GM/d ≈ 17,800 miles per hour

and

Ve = Vo 2

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