Question

You are given the following data: $g = 9.81m{s^{ - 2}}$ , radius of earth $= 6.37 \times {10^6}m$, the distance the Moon from the earth $= 3.84 \times {10^8}m$ and the time period of the Moon's revolution =27.3 days. Obtain the mass of the earth in two different ways.
$G = 6.67 \times {10^{ - 11}}N{m^2}k{g^2}$

Answer 1

Hint: In order to find the mass of the earth first we'll take earth alone and by the use of the formula for the gravitational constant of a single body we will find the mass. In the second case we will take earth and moon and then we will consider the gravitational force between both these bodies same as the centripetal force of the moon.

Formula used- $G = \dfrac{{g{R^2}}}{M},{F_G} = \dfrac{{G{M_1}{M_2}}}{{{r^2}}},{F_c} = mr{\left( {\dfrac{{2\pi }}{T}} \right)^2}$

Complete step-by-step solution:
Given that
Radius of earth ${R_e} = 6.37 \times {10^6}m$
Distance the moon from the earth $r = 3.84 \times {10^8}m$
And the time period of the Moon's revolution $T = 27.3days = 27.3 \times 24 \times 60 \times 60s$
We know that the gravitational constant is given as:
$G = \dfrac{{g{R^2}}}{M}$
Where G is the gravitational constant, M is the mass of the object, g is acceleration due to gravity and R is the radius of the body.
Let this body be Earth so the formula becomes.

$$\because G = \dfrac{{g{R^2}}}{M} \\\ \Rightarrow G = \dfrac{{gR_e^2}}{{{M_e}}} \\\ \Rightarrow {M_e} = \dfrac{{gR_e^2}}{G} \\\$$

Let us substitute all the values in this equation to find the mass of earth

$$\because {M_e} = \dfrac{{gR_e^2}}{G} \\\ \Rightarrow {M_e} = \dfrac{{9.81 \times {{\left( {6.37 \times {{10}^6}} \right)}^2}}}{{6.67 \times {{10}^{ - 11}}}} \\\ \Rightarrow {M_e} = 5.97 \times {10^{24}}kg \\\$$

Alternate method.
As we know that the moon moves around the earth only due to the gravitational pull of the earth. So the centripetal force of the moon is provided by the gravitational force between earth and moon. So let us equate these two forces for finding the mass of earth.
Gravitational force between two objects is given by:
${F_G} = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}$
Here ${F_G}$ is the gravitational force, G is the gravitational constant, ${M_1}$ is mass of first object, ${M_2}$ is the mass of second object and r is the distance between them.
Considering the given situation of earth and moon this formula becomes:
${F_G} = \dfrac{{G{M_e}{M_m}}}{{{{\left( {{r_{e - m}}} \right)}^2}}}$---- (1)
Here ${M_e}$ is the mass of earth, ${M_m}$ is the mass of the moon and r is the distance between them.
Also the centripetal force is given as:
${F_c} = mr{\left( {\dfrac{{2\pi }}{T}} \right)^2}$
Here m is the mass of the moon or revolving object which is moon in this case, r is the distance and T is the time of revolution.
So the formula becomes.
${F_c} = {M_m}r{\left( {\dfrac{{2\pi }}{T}} \right)^2}$----- (2)
As the centripetal force is same as gravitational force, so let us compare equation (1) and equation (2)
So, we get:

$$\because {F_G} = {F_c} \\\ \Rightarrow \dfrac{{G{M_e}{M_m}}}{{{{\left( r \right)}^2}}} = {M_m}r{\left( {\dfrac{{2\pi }}{T}} \right)^2} \\\$$

Let us derive the equation for the mass of earth from the above equation by cancelling some terms and bringing others to RHS.

$$\Rightarrow \dfrac{G}{{{{\left( r \right)}^2}}}{M_e} = r{\left( {\dfrac{{2\pi }}{T}} \right)^2} \\\ \Rightarrow {M_e} = \dfrac{{4{\pi ^2}{r^3}}}{{G{T^2}}} \\\$$

Let us now substitute the given values in the above formula to find the mass of earth.
So we have:

$$\because {M_e} = \dfrac{{4{\pi ^2}{r^3}}}{{G{T^2}}} \\\ \Rightarrow {M_e} = \dfrac{{4 \times {{\left( {\dfrac{{22}}{7}} \right)}^2} \times {{\left( {3.84 \times {{10}^8}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}} \times {{\left( {27.3 \times 24 \times 60 \times 60} \right)}^2}}} \\\$$

After solving the above term we get:
$\Rightarrow {M_e} = 6.02 \times {10^{24}}kg$
Hence the mass of earth is $6.02 \times {10^{24}}kg$ .

Note- The gravitational constant is the constant of proportionality used in the Theory of Gravitation by Newton. G denoted universal gravitational constant is the force of attraction between any two unit masses separated by a unit distance. This is an observable, physical constant found in the science of gravity. It is also defined as Constant by Newton. The gravitational constant 's value is similar in the universe. The value of G varies from g and corresponds to the acceleration due to gravity.