Free falling body on Sun

FREE FALLING BODY ON SUN

Consider a body of mass m to be dropped at a height of 1kilometer above the surface of Sun. The body will experience a force of gravity equal to the acceleration due to gravity of Sun g = 275.54 m/s². This value can be considered to be approximately same 1kilometer above Sun’s surface. As the body falls toward Sun, the displacement is considered negative.

ASSUMPTIONS

·         1. Effect of air resistance is negligible

·         2. Effect of wind is negligible

·         3. Effect of rotation of Sun is negligible

Initial Displacement y­’ = 0 m

Final Displacement y = -1000 m

Applying first kinematic equation we get,

y - y' = u*t - (1/2)*g*t^2

u – Initial velocity m/s

-1000 = 0*t - (1/2)*275.54*t^2

-1000 = -137.77*t^2

t^2 = 7.2584

t = 2.6941 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,

v^2 = u^2 - 2g(y - y')

v^2 = 0 - 2*275.54*(-1000-0)

v^2 = 2*275.54*1000

v = 742.34 m/s                             

                                                                                                                                              

Thus the final velocity of the body when it reaches surface of Sun is 742.34 m/s or 2672.45kmph or 1670.28mph

Free falling body on Jupiter

FREE FALLING BODY ON JUPITER

Consider a body of mass m to be dropped at a height of 1kilometer above the surface of Jupiter. The body will experience a force of gravity equal to the acceleration due to gravity of Jupiter g = 25.9 m/s². This value can be considered to be approximately same 1kilometer above Jupiter’s surface. As the body falls toward Jupiter, the displacement is considered negative.

ASSUMPTIONS

·         1. Effect of air resistance on is negligible

·         2. Effect of wind is negligible

·         3. Effect of rotation of Jupiter is negligible

Initial Displacement y­’ = 0 m

Final Displacement y = -1000 m

Applying first kinematic equation we get,

y - y' = u*t - (1/2)g*t^2

u – Initial velocity m/s

-1000 = 0*t - (1/2)*25.9*t^2 

-1000 = -12.95*t^2

t^2 = 77.22

t = 8.7874 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 3rd Kinematic equation,

v^2 = u^2 - 2g(y - y')

v^2 = 0 - 2*25.9*(-1000-0)

v^2 = 2*25.9*1000

                     

 v   = √2*25.9*1000

      

 v  = 227.59 m/s

Thus the final velocity of the body when it reaches surface of Jupiter is 227.59 m/s or 819.34kmph or 512.09mph

Free Falling body on Earth

FREE FALLING BODY ON EARTH

Consider a body of mass m to be dropped at a height of 1kilometer above the surface of Earth. The body will experience a force of gravity equal to the acceleration due to gravity of Earth g = 9.8 m/s² This value can be considered to be approximately same 1kilometer above Earth’s surface. As the body falls toward Earth, the displacement is considered negative.

ASSUMPTIONS

·         1. Effect of air resistance is negligible

·         2. Effect of wind is negligible

·         3. Effect of rotation of earth is negligible

Initial Displacement y­’ = 0 m

Final Displacement y = -1000 m

Applying first kinematic equation we get,

y - y' = u*t - (1/2)g*t^2

u – Initial velocity  (m/s)

-1000 = 0*t - (1/2)*9.8*t^2

-1000 = -4.9t^2

 t^2 = 204.08

 t = 14.2857 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,

v^2 = u^2 - 2g(y - y')

v^2 = 0 - 2*9.8*(-1000 - 0)

                                                             

v^2 = 2*9.8*1000

v = 140 m/s

Thus the final velocity of the body when it reaches surface of Earth is 140 m/s or 504kmph or 315mph