Moving body approaching Neutron Star

MOVING BODY APPROACHING NEUTRON STAR 

INTRODUCTION

The Neutron Star is a highly compact star primarily composed of neutrons. It has density the same as that of an atom. A Neutron star is born from a supernova explosion. Consider a body of mass m moving with an initial speed of 10 km/s approaching the star at a distance of 4R from its surface. As the body enters Neutron Star’s gravitational field, it begins accelerating. This type of acceleration is continuous acceleration as it's continuously being accelerated due to Neutron Star’s gravity.

CALCULATION

The final velocity of the moving body when it touches Neutron Star’s surface is,

v² = u² + 2GM [(1/R) – (1/R+h)] 

Where, G = 6.67*10^-11 Nm²/kg² [Universal Gravitation Constant]

R – Radius of Neutron Star [R = 11000m]

M - Mass of Neutron Star, M = 2*2*10³⁰ kg

h - Height above the surface of Neutron Star, h = 4R

u - Initial velocity of the mass, u = 10km/s

v - Final velocity of the mass

v² = u² + 2GM [(1/R) – (1/R+4R)]

v² = u² + 2GM [(1/R) – (1/5R)]

v² = u² + 2GM [4R/5R²]

v² = u² + GM [8/5R]

Substituting all values we get,

v² = 10⁸ + [3.88*10¹⁶] 

v² = 3.88*10¹⁶

v    = 196,977,156 m/s

CONCLUSION

Thus the final velocity of the body on the surface of Neutron Star is 196,977,156 m/s or if it's moving with an initial velocity of 10km/s. It depends on the initial velocity of the body, mass and radius of planet but independent of mass of object.

Moving Body approaching Alpha Arae

MOVING BODY APPROACHING ALPHA ARAE 

INTRODUCTION

The 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 moving with an initial speed of 10 km/s approaching Alpha arae at a distance of 4R from its surface. As the body enters Alpha arae’s gravitational field, it begins accelerating. This type of acceleration is continuous acceleration as it's continuously being accelerated due to Alpha arae’s gravity.

CALCULATION

The final velocity of the moving body when it touches Alpha arae’s surface is,

v² = u² + 2GM [(1/R) – (1/R+h)] 

Where, G = 6.67*10^-11 Nm²/kg² [Universal Gravitation Constant]

R – Radius of Alpha arae [R = 3,130,650,000m]

M - Mass of Alpha arae, M = 1.92 x 10³¹ kg

h - Height above the surface of Alpha arae, h = 4R

u - Initial velocity of the mass, u = 10km/s

v - Final velocity of the mass

v² = u² + 2GM [(1/R) – (1/R+4R)]

v² = u² + 2GM [(1/R) – (1/5R)]

v² = u² + 2GM [4R/5R²]

v² = u² + GM [8/5R]

Substituting all values we get,

v² = 10⁸ + [6.545*10¹¹]

v² = 6.546*10¹¹

v    = 809,073.54 m/s

CONCLUSION

Thus the final velocity of the body on the surface of Alpha arae is 809,073.54 m/s if it's moving with an initial velocity of 10km/s.

The change [increase] in velocity is,

 Δ = v – u

 Δ = 809,073.54 – 10000

 Δ = 799,073.54 m/s

The percentage increase in velocity is,

∆% = (799,073/10000)*100 = 7990.73%

Thus the final velocity of object depends on the initial velocity of the body, mass and radius of planet but independent of mass of object.

Moving body approaching Sun

MOVING BODY APPROACHING SUN

 INTRODUCTION

Sun is a star in the center of our Solar System. Consider a body of mass 'm' moving with an initial speed of 10 km/s approaching Sun at a distance of 5R from its surface where R is the radius of Sun. As the body enters Sun’s gravitational field, it begins accelerating. This type of acceleration is continuous acceleration as it's continuously being accelerated due to Sun’s gravity.

ASSUMPTIONS

• The body can withstand Sun's gravity and will not collapse before touching Sun's surface.
• The body can withstand the temperature gradients close to Sun. 

CALCULATION

 

The final velocity of the moving body when it touches Sun’s surface is,

v² = u² + 2GM [(1/R) – (1/R+h)]  

Where, G = 6.67*10^-11 Nm²/kg² [Universal Gravitation Constant]

R – Radius of Sun [R = 695700000m]

M - Mass of Sun, M = 2*10³⁰ kg

h - Height above the surface of Sun, h = 5R

u - Initial velocity of the mass, u = 10km/s

v - Final velocity of the mass

v² = u² + 2GM [(1/R) – (1/R+5R)]

v² = u² + 2GM [(1/R) – (1/6R)]

v² = u² + 2GM [5R/6R²]

v² = u² + GM [5/3R]

Substituting all values we get,

v² = 10⁸ + [3.1958*10¹¹]

v² = 3.1968*10¹¹

v    = 565404.45 m/s

CONCLUSION 

Thus the final velocity of the body on the surface of Sun is 565404.45 m/s if it's moving with an initial velocity of 10km/s. The change [increase] in velocity is,

 Δ = v – u

     = 565404.45 – 10000

     = 555404.45 m/s

The percentage increase in velocity is,

∆% = (555404.45/10000)*100 = 5554.04%

Thus the final velocity of object depends on the initial velocity of the body, mass and radius of star but independent of mass of object.