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Gravity and Projectiles. News Item on 21/07/2011. This morning at a missile test range the DRDO launched its first ever Prahaar missile that can strike within 10 m of a target 150 km away. What are the factors which can change the trajectory of a projectile?. Gravity. Centrifugal Force.
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News Item on 21/07/2011 This morning at a missile test range the DRDO launched its first ever Prahaar missile that can strike within 10 m of a target 150 km away.
What are the factors which can change the trajectory of a projectile? Gravity Centrifugal Force Variation of g with latitude RPOLE REQ - REQ = 6378 km RPOLE =6357 km
Combining variation of g with latitude and variation due to mass distribution inside the earth, scientists have derived the Geodetic Reference System for 1967 g () = 9.78031846 (1+ 0.0053024 sin ² –0.0000058 sin² 2)
Coriolis Force Earth is a rotating frame of reference. A body moving in this frame experiences a force called Coriolis force. This force causes a drift in the moving body to the right of its path. Expected vertical path Actual path A particle dropped from a height of 100 m suffers a deflection of only 0.0155 cm.
A missile travelling with a velocity of 5000 km/hr, during its flight of 1000 s, will drify and miss its target by 100 km. Intended Path Actual Path
Projectile with air resistance We all know that in the absence of any air resistance the path of a projectile is a parabola. y What happens to the path when air x resistance is taken into account?
Suppose the object is a cricket ball bowled by a fast bowler at 144 km/hr (40 m/s). Resistance due to air is called air drag. For the assumed speed, the magnitude of the force of friction due to drag may be taken as (D is the drag coefficient and is the velocity of the ball) (1)
y x
The drag coefficient D depends on the area of the projectile facing the air in the direction of its motion and the density of air through which it moves. If A is the surface area of the projectile and ρ is the density of air, then drag coefficient can be written as (2) Constant C (0.1 – 1) takes care of the shape and nature of the surface of the projectile.
The direction of the resistance or the drag force is always opposite to that of the velocity . The components of the velocity vector are , . (3) The components of the drag force are . (4)
Consequently, the acceleration experienced by the projectile along the x – and y – direction is (is magnitude of ) , . (5) We choose a suitably short interval of time during which we can assume the acceleration to remain constant. , . (6)
The x- and y-coordinates advance to , (7) . (8) The new velocity found in Equations 6 is fed back into Equations 5 to get the new acceleration. Steps 5,6,7 and 8 are then repeated. The process is repeated a number of times and y is plotted against x to get the trajectory.
The trajectories shown in the next slide have been plotted for a cricket ball travelling at a velocity of 40 m/s(144 km/hr) at an angle of 45º to the horizontal. The data was generated by a computer program. The mass of the cricket ball is taken as 0.160 kg and its radius as 0.0360 m (these are typical values). The value for the density of air has been adopted as 1.22 kg/m3, appropriate for sea level at 15 ºC.
The values of the constant C are shown along each curve. We see that the resistance due to air cannot be ignored. Not only does the projectile not reach its expected height, its range is also considerably reduced.