Moving body approaching Earth

MOVING BODY APPROACHING EARTH

Consider a body of mass m moving with an initial speed of 10 km/s. This body approaches the Earth and is at a distance of 4R from the surface of earth. As the body enters Earth’s gravitational field, it begins accelerating. This type of acceleration is continuous acceleration as it's continuously being accelerated due to Earth’s gravity.

The final velocity of the moving body when it touches Earth’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 Earth [R = 6371000m]

M - Mass of Earth, M = 6 x 10²⁴ kg

h - height above the surface of Earth, 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 + [(320.16/31855000)*10¹³]

  v² = 10⁸ + 100505415.2

      = 200505415.2

v    = 14159.993 m/s

Thus the final velocity of the body on the surface of Earth is 14159.993 m/s or 50975.97 km/hr if it's moving with an initial velocity of 10km/s. The change [increase] in velocity is,

 Δ = v – u

     = 14159.993 – 10000

     = 4159.993 m/s

The percentage increase in velocity is,

∆% = (4159.993/10000)*100 = 41.59%

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.