Question

The mass of $1kg$ on earth weighs $\dfrac{1}{6}kg$ on the moon. The radius of the moon is $1.738\times {{10}^{6}}\text{m}$. The mass of the moon is $\left( {{g}_{e}}=9.8m{{s}^{-2}},G=6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}} \right)$.
$\begin{align}
  & \text{A}\text{. }7.4\times {{10}^{22}}kg \\
 & \text{B}\text{. }4.35\times {{10}^{16}}kg \\
 & \text{C}\text{. }4.4\times {{10}^{23}}kg \\
 & \text{D}\text{. }7.4\times {{10}^{20}}kg \\
\end{align}$

Answer 1

Hint: The weight of an object is the product of the mass of the object and the acceleration due to gravity (g). Acceleration due to gravity varies with the mass and size of the astronomical body. So, the weight of an object varies at the surface of different astronomical objects.

Formula used:
Acceleration due to gravity of a planet of mass M and radius R is given by,
$g=\dfrac{GM}{{{R}^{2}}}$
Weight of an object on the surface of any planet is given by,
$W=mg$

Complete answer:
The acceleration due to the gravity of a planet or satellite of mass M and radius R is given by,
$g=\dfrac{GM}{{{R}^{2}}}$
Where G is the gravitational constant.
So, the acceleration due to the gravity of the earth is denoted by ${{g}_{e}}$ , and the acceleration due to the gravity of the moon is denoted by ${{g}_{m}}$.
Let ${{M}_{m}}\text{ and }{{\text{R}}_{m}}$ be the mass and radius of the moon, respectively. So, the acceleration due to gravity on the moon’s surface can be written as,
${{g}_{m}}=\dfrac{G{{M}_{m}}}{{{R}_{m}}^{2}}$ … equation(1)
In the problem, an object which weighs 1 kg on the earth’s surface, will weigh about $\dfrac{1}{6}kg$ on the moon’s surface. So, we can write the equation for both these cases.

In the case of the earth, we can write,
$1kg=m{{g}_{e}}$ …. equation (2)
In the case of the moon, we can write,
$\dfrac{1}{6}kg=m{{g}_{m}}$… equation (3)
Dividing equation (3) by equation (2), we get,
$\dfrac{1}{6}=\dfrac{{{g}_{m}}}{{{g}_{e}}}$
$\Rightarrow {{g}_{m}}=\dfrac{{{g}_{e}}}{6}$
$\therefore {{g}_{m}}=1.64\text{ m/}{{\text{s}}^{2}}$
From equation (1) we can write ${{g}_{m}}$ as,
$\dfrac{G{{M}_{m}}}{{{R}_{m}}^{2}}=1.64m/{{s}^{2}}$
Substituting the values of radius of the moon and G, we get,
$\dfrac{\left( 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}} \right)\times {{M}_{m}}}{{{\left( 1.738\times {{10}^{6}}\text{m} \right)}^{2}}}=1.64m{{s}^{-2}}$
$\Rightarrow {{M}_{m}}=\dfrac{1.64m{{s}^{-2}}\times {{\left( 1.738\times {{10}^{6}}\text{m} \right)}^{2}}}{\left( 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}} \right)}$
$\therefore {{M}_{m}}=4.35\times {{10}^{16}}kg$

So, the correct answer is “Option B”.

Note:
The escape velocity of a planet of mass M and radius R is given by,
${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$
This equation can also be written in the form,
${{v}_{e}}=\sqrt{2gR}$
Where g is the acceleration due to gravity, the escape velocity of the earth is $11.19km{{s}^{-1}}$ and of the moon is $2.38km{{s}^{-1}}$.
The value of acceleration due to gravity decreases as we go from the surface of the earth to higher altitudes or deeper depths.