(a)      Since g = $G \: \frac{me}{R^{2}}$ …….. (i)

Equation (i) shows that the value of g does not depend upon the mass of the body. This means that light and heavy bodies should fall toward the center of the Earth with the same acceleration.

 

(b)      The value of g varies inversely as the square of the distance i.e. $g \: \alpha \frac{1}{R^{2}}$ if the distance from the center of the earth is increased then the value of g will decrease. That is why the value of g at hills (Murree) is less than its value of the seashore (Karachi).

 

(c)       Earth is not a perfect sphere. It is flattened at the poles, for this reason, the value of g at the pole is more than at the equator. Because the polar radius is less than the equatorial radius.  $g \: \alpha \frac{1}{R^{2}}$

 

 

MINI EXERCISE

1. Does an apple attract the Earth towards it?

Ans:    Yes, according to the law of gravitation an apple attracts Earth towards it but its attraction is very small and cannot be felt.

 

2. With what force does an apple weighing 1N attract the Earth?

Ans:    The force of attraction is equal to the weight of the object. So, an apple weighing 1N attracts the Earth with 1N force.

 

3. Does the weight of an apple increase, decrease, or remain when taken to the top of a mountain?

Ans:    The value of g varies inversely as the square of the distance i.e. g  1/ R²

                Therefore, the weight of an apple decreases when taken to the top of a mountain due to the less gravity on Earth.

 

DO YOU KNOW?

The value of g on the surface of a celestial object depends on its mass and its radius. The radius of g on some of the objects is given below:

Object	g(ms-2)
Sun	274.2
Mercury	3.7
Venus	8.87
Moon	1.62
Mars	3.73
Jupiter	25.94