Question

The weight of a body on the surface of the moon is $\dfrac{1}{6}th$ of that on the earth’s surface. It is because acceleration due to gravity on the moon is six times that of the earth.
(A) True
(B) False

Answer 1

Hint : The weight of a body at the surface of a celestial body is directly proportional to the acceleration due to gravity on the surface of the body. A decrease in acceleration due to gravity leads to a decrease in the weight of a body.

Formula used: In this solution we will be using the following formula;
$\Rightarrow W = mg$ where $W$ is the weight of a body on the surface of a celestial body, $m$ is the mass of the body, and $g$ is the acceleration due to gravity of the celestial body.

Complete step by step answer
It is understood based on Newton’s law of universal gravitation, that any two objects which have masses must attract each other with a force which is proportional to the product of the masses and inversely related to the square of the distance between their centres. This force of attraction is what allows a person on the surface of a planet to not fall out of the planet. An effect of this force of attraction is the weight of the body on that planet. The weight of a body on a planet or any celestial body is given by
$\Rightarrow W = mg$ , where $m$ is the mass of the body, and $g$ is the acceleration due to gravity of the celestial body. This implies that an increase in the acceleration due to gravity must in turn increase the weight of the body on the celestial body. Hence, the statement is false.
The correct option is B.

Note
Alternatively, by mathematical reasoning, we have that, $W = mg$
The weight on the moon is given by
$\Rightarrow {W_m} = m{g_m}$ .
Now since the weight on the surface of the moon is one-sixth of the earth, then,
$\Rightarrow {W_m} = \dfrac{{{W_e}}}{6} = m{g_m}$
Thus
$\Rightarrow {W_e} = 6m{g_m}$
Substituting into ${W_e} = m{g_e}$ . We have
$\Rightarrow 6m{g_m} = m{g_e}$
$\Rightarrow {g_m} = \dfrac{1}{6}{g_e}$
This implies that if the weight on the moon is one-sixth that of the earth, then it must be that the acceleration due to gravity is also one-sixth, and not time six.
Hence, the statement is false.