Transcript Gravity and Projectiles

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?
Variation of g with latitude
RPOLE
Gravity
REQ
REQ = 6378 km
RPOLE = 6357 km
𝜑
Centrifugal
Force
𝑔𝜑 = 𝑔𝑃𝑜𝑙𝑒 - 𝜔2 𝑅 cos 𝜑
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 (𝜑) = 9.78031846 (1+ 0.0053024 sin ² 𝜑
– 0.0000058 sin² 2𝜑)
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
Expected
vertical
path
in the moving body to
the right of its path.
Actual
A particle dropped from a
height of 100 m suffers a
deflection of only 0.0155 cm.
path
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
resistance is taken
into account?
x
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 𝑣 is the velocity of
the ball)
𝐹 = 𝐷𝑣
2
(1)
y
𝑣
𝜃
x
𝐹
−𝑔
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
𝐴𝜌
𝐷=𝐶
(2)
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 𝑣.
The components of the velocity vector are
𝑣𝑥 = 𝑣 cos 𝜃, 𝑣𝑦 = 𝑣 sin 𝜃. (3)
The components of the drag force are
𝐹𝑥 = −𝐷𝑣𝑣𝑥 , 𝐹𝑦 = −𝐷𝑣𝑣𝑦 . (4)
Consequently, the acceleration
experienced by the projectile along the x –
and y – direction is (𝑣 is magnitude of 𝑣)
𝑎𝑥 =
𝐷𝑣𝑣𝑥
−
𝑚
, 𝑎𝑦 = −𝑔 −
𝐷𝑣𝑣𝑦
𝑚
. (5)
We choose a suitably short interval of
time ∆𝑡 during which we can assume the
acceleration to remain constant.
𝑣𝑥 → 𝑣𝑥 + 𝑎𝑥 ∆𝑡 , 𝑣𝑦 → 𝑣𝑦 ∆𝑡.
(6)
The x- and y-coordinates advance to
𝑥
𝑦
1
→ 𝑥 + 𝑣𝑥 ∆𝑡 + 𝑎𝑥
2
1
→ 𝑦 + 𝑣𝑦 ∆𝑡 + 𝑎𝑦
2
∆𝑡
2
,
(7)
∆𝑡 .
(8)
2
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.
50
40
30
y (m)
The values of
the constant
C are
shown along
each curve.
20
C=0.5
C=0.3
C=0.1
C=0
10
0
0
50
100
150
x (m)
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.