Gravitation and Weightlessness - Part II

2. Acceleration due to gravity (g)  
The concept of acceleration follows from Newtonís second law of motion. Newtonís second law of motion states that the rate of change of momentum of a body is directly proportional to the force acting on it and takes place in the direction of the force.

That is if  momentum of a body is given by p  =  mv, then according to Newtonís second law of motion

                                                           d (mv)
Force  F = rate of change of (mv)    =   
                                                            dt

where d (mv) is the small change occurring in momentum (mv) in a short time dt.

The above equation can be written as

                      d (mv)
  F     =         
                       dt   

It is safe to assume that mass is constant and does not change with time.

                             d (v)
   F    =   m   x     
                              dt

The rate of change of velocity by definition is the acceleration felt by the mass m.

Thus

                      d (v)
  a     =         
                        dt

and F = m a

or  acceleration a  = Force/mass

Acceleration produced by gravity on objects will be equal to the gravitational force divided by the mass of the object, which is getting attracted, to the gravitational mass. The gravitational acceleration is denoted by g. The gravitational force is given by Newtonís inverse squared law of gravitation.

                                    F_{gravitational}
Thus        g =                       ,    and
                                        mass

                                      m₁ m₂
   F_{gravitational}   =    G     
                                         r²

Example 1 : Calculate the gravitational acceleration produced by earth on a ball that weighs 2 kg. The mass of the earth is 6 x 10²⁴ kg and the radius of the earth is 6.4 x 10³ km. (G = 6.673 x 10^-11   N.m²/kg²).

Gravitational force is given by

                                          m₁ m₂
    F_{gravitational}   =    G         
                                             r²

m_earth = 6 x 10²⁴ kg,      m_ball = 2 kg,      r = 6.4 x 10³ km = 6.4 x 10⁶ m

                       6.673 x 10^-11   (6 x 10²⁴) x (2)
   F_{gravitational}  =         Newton  
                                    6.4 x 10⁶   x  6.4 x 10⁶

            =   19.5 N

To calculate the gravitational acceleration caused by earth on the ball, we have to divide this force F_{gravitational} by the mass of the ball.

          F_{gravitational}                              19.5 N
g =            =            =      9.8 m/s².
        mass of the ball                   2 kg

Thus the ball is accelerated towards the earth by a value of g = 9.8 m/s².

Similarly if we calculate the acceleration caused by the ball on the earth, it will come out to be very small, almost negligible.

Extending the discussions given above, we can say that all bodies are attracted to the earth by gravitational acceleration g.

If a stone or a ball is at a distance R from the center of the earth, then the force can be written as (ball or the stone is at the surface of the earth),

                                    Mass of earth (M) x  mass of the stone (m)
F_{gravitational}   =    G       
                                                                    R²

Acceleration felt by the stone g = F_{gravitational}  x mass of the stone (m)

                         M
Thus g =   G   
                         R²

G = gravitational constant = 6.673 x 10^-11   N.m²/kg²

M = mass of the earth = 6 x 10²⁴ kg

R = radius of the earth = 6.4 x 10³ km = 6.4 x 10⁶ m

g = 9.8 m/s².

This is the acceleration felt by any body on the surface of the earth. This is a constant value as G,M and R are constant values. As one goes above the surface of the earth, g changes; if the distance above the earth is given by r then the g experienced will go as inverse square of (R + r). This change is not significant, hence for all calculations, g is taken as constant. Similar arguments may be applied for g experienced as one penetrates inside the earth.

It has to be borne in mind the g is independent of mass of the body. Hence all bodies on earth experience the same value of g.

Example 2 :  Calculate the value of g on the surface of the moon. Mass of the moon is = 7.4 x 10²² kg, radius of moon = 1740 km and  G = 6.673 x 10^-11   N.m²/kg².

                          M _moon                     6.673 x 10^-11   x (7.3x 10²²)
   g_moon    =   G         =                  m/s².
                          R ² _moon                              1.74 x 10⁶ x 1.74 x 10⁶

   g_moon     =  1.63 m/s².

Thus the acceleration due to moonís gravity is 1/6^th of the gravitational acceleration of the earth on the same body.

Example 3 :  A space ship is flying above the earthís surface at a distance of 300 km. Calculate its gravitational acceleration.

                                       M
   g_space =   G      
                                 (R+ 300 km)²

Substituting values of G,M and R we get

                     6.673 x 10^-11  x 6 x 10²⁴
g_space =         =     8.92 m/ s².
                     (6.4 x 10⁶ + 3 X 10⁵)²

We shall see later on this chapter how and why one feels weightless in space. One of the reasons is that the acceleration due to earthís gravity on the body reduces in space.

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