Path followed by projectile : An analysis.

We had studied a lot about projectile motion. First let us revise the classical projectile motion, Before starting the derivation for path of projectile, Time of flight and range we had assumed that

1. Air resistance is neglected.
2. Variation of g neglected.
3. Curvature of earth neglected.

Hence by using classical equations of kinematics (v=u+at , s=ut+1/2at² and v²=u² + 2as) we got
Time of flight T =(2u sinθ)/g
Range = (u² sin2θ)/g
Height = (u² sin² θ)/g

here ux=ucosθ, uy=usinθ

x=u cosθ t --1
t=x/ucosθ --2
and we know that y=uy t + 1/2 ay t²
 therefore y= usinθ t - 1/2 gt² --3
from equation 2 and 3 we get
y= (u sinθ x/u cos θ)- (1/2g x² /u² cos²θ)

and finally we got y= tanθ x - (gx²/2u²cos²θ)
  this can be written in the form of y=ax + bx²
         This implies path followed by a projectile is a PARABOLA

  Now let us think about our assumptions and variation due to our assumptions. 
But under the laws of gravity, a parabola is an impossible shape for an object that's gravitationally bound to the Earth. For flat earth thinkers the acceleration due to gravity always points towards downwards hence without any variation we get a parabola. But even though due to the size shape of earth is neglected it will change your value on a small scale. The g always points towards the centre of earth parabola will always not true. 
  If you think earth is flat rather than curved and g points downwards in your experimental region 
ny time you throw and release an object, therefore, it enters a situation known as free-fall. In the directions that are parallel to the Earth's surface (horizontal), the speed of any projectile will remain constant. In the directions that are perpendicular to Earth's surface (vertical), however, your projectile will accelerate downwards at 9.8 m/s²: the acceleration due to gravity at Earth's surface. If you make these assumptions, then the trajectory you calculate will always be a parabola, exactly what we're taught in physics classes around the globe.

  According to keplers first law the law of elliptical orbit which says every planet revolve around the sun in an elliptical orbit with sun at one of the foci. like wise all satellite which projected from earth and our moon revolve around the earth in an elliptical orbit. From this we can assume that really the path of a projectile is a ellipse with centre of earth at on of the foci.   

When thinking as flat earth in a projectile motion.

The Earth were perfectly flat and the acceleration, everywhere, were straight down all projectiles would make a parabola.

When considering the shape of earth in a projectile motion.

The acceleration due to gravity always point towards the centre of earth.
     
       The practical problem to think path of projectile is a ellipse is if we through a stone or a ball it raise to a little height and stops its trajectory at the surface of earth. I mentioned centre of earth is at one of the focus then the maximum height acquired by our stone/ball may its aphelion.
We can get around this problem, however, by imagining that we had something that didn't interact with normal matter as our projectile. Perhaps it could be a low-energy neutrino; perhaps it could be a clump of dark matter. In either case, this projectile, once we released it, would only experience the gravitational force, and would pass through the surface and interior of the Earth itself under only the force of gravity.

If you expected that this particle would make a closed ellipse, however, and would return to its original location some ~90 minutes later back above the surface of the Earth from where it was first thrown.

     By considering earth as a single point we could found trajectories for our satellites but we cant determine it for an object passing through earth surface easily because in variation in g is calculated by assuming earth has uniform density.

But actually the graph when considering density we get like this

 In Preliminary reference Earth model maximum value for g is at 2890 km below earth surface. As long as you're outside of a mass that's shaped like a sphere (or spheroid), all of that mass gravitationally attracts you towards the centre of the object. But if you're only outside part of that mass (and only part of it is closer to the centre of the world than you are), then all the parts of that mass that are outside of your present location cancel out.

PART TWO WILL PUBLISH SOON

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Composed by

Abhinav P Pradeep, wayanad, Kerala, India

Email:Abhinavppindia@gmail.com

Twitter: Abhinav P Pradeep

(@Abhinavppindia)