Free falling body on Alpha Arae

FREE FALLING BODY ON ALPHA ARAE

INTRODUCTION

Alpha Arae is a distant star in our Solar System with a distance of 270 light years from the Earth. Consider a body of mass m to be dropped at a height of 1kilometer above the surface of Alpha arae. The body will experience a force of gravity equal to the acceleration due to gravity of Alpha arae which is g = 85.74 m/s². This value can be considered to be approximately same 1kilometer above Alpha arae’s surface. As the body falls toward Alpha arae, the displacement is considered negative. 

ASSUMPTION

The body approaches equator of the star.

CALCULATION

Initial Displacement y­’ = 0 m

Final Displacement y = -1000 m

Applying first kinematic equation we get,

y – y’ = u*t – (1/2)*g*t²

u – Initial velocity m/s

-1000 = 0*t – (1/2)*85.74*t²

-1000 = -42.87t²

t² = 23.3263

t = 4.8297 s

This is the time taken by the body to cover a distance of 1km.

The final velocity v when the body hits the surface is given by 3^rd Kinematic equation,

v² = u² – 2*g*[y – y’]

v² = 0 – 2*85.74*[-1000-0]

v   = √2*85.74*1000

v    = 414.1 m/s

CONCLUSION

Thus the final velocity of the body when it reaches surface of Alpha arae is 414.1 m/s or 1490.76 kph or 931.725 mph.