Question

Estimate the mass of earth when it is given that radius of earth is $6.4{\text{ }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}$, acceleration due to gravity is ${\text{9}}{\text{.8 m }}{{\text{s}}^{ - 2}}$ and the gravitational constant is $6.67{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ N k}}{{\text{g}}^{ - 2}}{\text{ }}{{\text{m}}^2}$

Answer 1

We have to calculate the mass of heavenly objects like earth whose radius is given. We will use the concept of law of attraction for any two objects in the universe. We will calculate the estimated mass of earth by using the relation between radius of object, acceleration due to gravity and gravitational constant.

Formula Used:
$g{\text{ = }}\dfrac{{GM}}{{{R^2}}}$
Here, $g$= acceleration due to gravity, $G$= gravitational constant, $M$= mass of the earth and $R$= radius of the earth.

Complete step by step answer:
Let us assume $g$ be the acceleration due to gravity experienced by earth, $R$ be the radius of earth, $G$ be the gravitational constant and we will find $M$ which is the mass of earth. Since we know that mass of heavenly objects like earth can be calculated by using the formula,
$g{\text{ = }}\dfrac{{GM}}{{{R^2}}}$
We are given with, ${\text{R = }}6.4{\text{ }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}$, ${\text{g = 9}}{\text{.8 m }}{{\text{s}}^{ - 2}}$ and ${\text{G = }}6.67{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ N k}}{{\text{g}}^{ - 2}}{\text{ }}{{\text{m}}^2}$. On substituting the given values we get the mass of earth as,
$g{\text{ = }}\dfrac{{GM}}{{{R^2}}}$
${\text{9}}{\text{.8 m }}{{\text{s}}^{ - 2}}{\text{ = }}\dfrac{{6.67{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ N k}}{{\text{g}}^{ - 2}}{\text{ }}{{\text{m}}^2}{\text{ }} \times M}}{{{{\left( {6.4{\text{ }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}} \right)}^2}}}$
${\text{M = }}\dfrac{{{{\left( {6.4{\text{ }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}} \right)}^2} \times {\text{ }}9.8{\text{ m }}{{\text{s}}^{ - 2}}}}{{6.67{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ N k}}{{\text{g}}^{ - 2}}{\text{ }}{{\text{m}}^2}}}$
On solving the above we get the mass of earth as,
$\therefore {\text{M }} \approx {\text{ 6 }} \times {\text{ 1}}{{\text{0}}^{24}}{\text{ Kg}}$

Hence we the approximate mass of earth as ${\text{6 }} \times {\text{ 1}}{{\text{0}}^{24}}{\text{ Kg}}$.

Note: The weight of the body is different from the mass of the body. When we multiply the mass of the body with the acceleration of gravity then we get the weight of the body. Mass of the body is always less than the weight of the body. While calculating the mass of such a heavenly body we take values of approximation. We cannot calculate the exact mass of such a heavenly body therefore we can only calculate the estimated mass of earth. We have to take value of $g$ which is given in the question, if it is not mention about $g$ then take it as ${\text{9}}{\text{.8 m }}{{\text{s}}^{ - 2}}$.