Question

Show that the weight of an object on-
A. Moon is ${(1/6)^{th}}$ its weight on earth.
B. What is the mass of an object whose weight is $49N$ on earth?

Answer 1

Hint: Weight of an object is defined as the force of gravity between a planet (or a celestial body) and the object. The weight of an object is the product of its mass with the acceleration due to the gravity of that planet.

Complete step by step answer:
Moon is a satellite of earth, although the moon is smaller than the earth, it still has a gravitational force that it can exert a significant gravitational pull on the objects that are on its surface. This pull is called weight. The weight of an object is defined as-
$W = mg$
Where, $m$ is the mass of the object
And $g$ is the acceleration due to gravity on that planet.
A. The acceleration due to gravity is calculated as-
$g = \dfrac{{GM}}{{{R^2}}}$
Where $G$is the universal Gravitational Constant.
$M$is the mass of the planet and
$R$ is the radius of the planet (or the distance between the object and the centre of mass of the planet, at surface this distance is equal to the radius of the planet).
For earth,
The radius of the earth is- ${R_e} = 6378km$
In SI units, ${R_e} = 6.378 \times {10^6}m$
And the mass is, ${M_e} = 5.972 \times {10^{24}}kg$
The universal gravitational constant is $G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
Therefore, acceleration due to gravity for earth,
${g_e} = \dfrac{{6.67 \times {{10}^{ - 11}} \times 5.972 \times {{10}^{24}}}}{{{{\left( {6.378 \times {{10}^6}} \right)}^2}}}$
$ \Rightarrow {g_e} = \dfrac{{39.833}}{{40.678}} \times {10^{ - 11 + 24 - 12}}$
$ \Rightarrow {g_e} = 0.979 \times 10$
This is approximately equal to,
${g_e} = 9.8m{s^{ - 2}}$
For moon,
The radius of the moon is- ${R_m} = 1.737 \times {10^6}m$
And the mass of the moon is, ${M_m} = 7.347 \times {10^{22}}kg$
Therefore, the acceleration due to gravity of the moon is,
${g_m} = \dfrac{{6.67 \times {{10}^{ - 11}} \times 7.347 \times {{10}^{22}}}}{{{{\left( {1.737 \times {{10}^6}} \right)}^2}}}$
$ \Rightarrow {g_m} = \dfrac{{49.004}}{{3.017}} \times {10^{ - 11 + 22 - 12}}$
$ \Rightarrow {g_m} = 16.242 \times {10^{ - 1}}$
This is approximately equal to,
${g_m} = 1.6m{s^{ - 2}}$
Comparing the acceleration due to gravity on earth with that of moon we have,
$\dfrac{{{g_m}}}{{{g_e}}} = \dfrac{{1.6}}{{9.8}}$
$ \Rightarrow {g_m} = 0.163{g_e}$
Writing this approximately as a fraction,
$ \Rightarrow {g_m} = \dfrac{{16}}{{100}}{g_e}$
$ \Rightarrow {g_m} = \dfrac{4}{{25}}{g_e} = \dfrac{1}{{6.25}}{g_e}$
Therefore the acceleration due to gravity on moon is roughly ${\left( {1/6} \right)^{th}}$ that of earth.
B. We know that the acceleration due to gravity on earth is, $g = 9.8m/{s^2}$
Given that the weight of the object is $W = 49N$
Then the mass of the object is given by,
$m = \dfrac{W}{g}$
$m = \dfrac{{49}}{{9.8}} = 5$
The mass of the object is $5\;kg$.

Note: The mass of the moon is approximately ${\left( {1/100} \right)^{th}}$ times the mass of the earth and its radius is approximately ${\left( {1/4} \right)^{th}}$ that of earth. This can be used to simply calculate that the acceleration due to gravity on the moon is ${\left( {1/6} \right)^{th}}$ of the earth’s.