We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth.
We know that
`F = m xx a`, .......(10.13)
that is,
`F = m xx g`. .......(10.14)
The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same in Eq. (10.14), we have
`W = m xx g` .......(10.15)
As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction.
We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, `W ∝ m`.
It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. The mass of an object remains the same everywhere, that is, on the earth and on any planet whereas its weight depends on its location.
`ulbb"WEIGHT OF AN OBJECT ON THE MOON"`
We have learnt that the weight of an object on the earth is the force with which the earth attracts the object. In the same way, the weight of an object on the moon is the force with which the moon attracts that object.
The mass of the moon is less than that of the earth. Due to this the moon exerts lesser force of attraction on objects.
Let the mass of an object be `m`. Let its weight on the moon be `W_m`. Let the mass of the moon be `M_m` and its radius be `R_m`.
By applying the universal law of gravitation, the weight of the object on the moon will be
` W_m = G (M_m xx m)/R_m^2` ....(10.16)
Let the weight of the same object on the earth be We. The mass of the earth is `M` and its radius is `R`.
From Eqs. (10.9) and (10.15) we have,
` W_e = G (M xx m)/R^2` .....(10.17)
Substituting the values from Table 10.1 in Eqs. (10.16) and (10.17), we get
` W_m = G ( 7.36 xx 10^(22) kg xx m)/( 1.74 xx 10^6 m)^2`
` W_m = 2.431 xx 10^(10) G xx m` ....(10.18a)
and `= 1.474 xx 10^(11) G xx m` .....(10.18b)
Dividing Eq. (10.18a) by Eq. (10.18b), we get
`W_m/W_e = ( 2.431 xx 10^(10))/( 1.474 xx 10^(11))`
or ` W_m/W_e = 0.165 approx 1/6`.......(10.19)
` text( Weight of the object on the moon)/text( Weight of the object on the earth) = 1/6`
Weight of the object on the moon
`= (1//6) xx` its weight on the earth.
We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth.
We know that
`F = m xx a`, .......(10.13)
that is,
`F = m xx g`. .......(10.14)
The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same in Eq. (10.14), we have
`W = m xx g` .......(10.15)
As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction.
We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, `W ∝ m`.
It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. The mass of an object remains the same everywhere, that is, on the earth and on any planet whereas its weight depends on its location.
`ulbb"WEIGHT OF AN OBJECT ON THE MOON"`
We have learnt that the weight of an object on the earth is the force with which the earth attracts the object. In the same way, the weight of an object on the moon is the force with which the moon attracts that object.
The mass of the moon is less than that of the earth. Due to this the moon exerts lesser force of attraction on objects.
Let the mass of an object be `m`. Let its weight on the moon be `W_m`. Let the mass of the moon be `M_m` and its radius be `R_m`.
By applying the universal law of gravitation, the weight of the object on the moon will be
` W_m = G (M_m xx m)/R_m^2` ....(10.16)
Let the weight of the same object on the earth be We. The mass of the earth is `M` and its radius is `R`.
From Eqs. (10.9) and (10.15) we have,
` W_e = G (M xx m)/R^2` .....(10.17)
Substituting the values from Table 10.1 in Eqs. (10.16) and (10.17), we get
` W_m = G ( 7.36 xx 10^(22) kg xx m)/( 1.74 xx 10^6 m)^2`
` W_m = 2.431 xx 10^(10) G xx m` ....(10.18a)
and `= 1.474 xx 10^(11) G xx m` .....(10.18b)
Dividing Eq. (10.18a) by Eq. (10.18b), we get
`W_m/W_e = ( 2.431 xx 10^(10))/( 1.474 xx 10^(11))`
or ` W_m/W_e = 0.165 approx 1/6`.......(10.19)
` text( Weight of the object on the moon)/text( Weight of the object on the earth) = 1/6`
Weight of the object on the moon
`= (1//6) xx` its weight on the earth.