Question

You are given the following data: \[g = 9.81\;m{s^{ - 2}}\], the radius of the 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\[\; = 27.3{\text{days}}\]. 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 the first method of finding the mass of Earth, we will use the value of gravitational acceleration on the surface of the Earth to determine its mass. In the second method, we will balance the gravitational force between the Earth and the moon to determine the mass of the Earth.

Formula used: In this solution, we will use the following formula:
- Gravitational force between two objects: $F = \dfrac{{GMm}}{{{r^2}}}$ between two objects of mass $M,m$ and at a distance $r$
- Gravitational acceleration on the surface of Earth $g = \dfrac{{GM}}{{R_E^2}}$ where \[M\] is the mass of the Earth and ${R_E}$ is the radius of the Earth.
- Centripetal acceleration: $F = \dfrac{{m{v^2}}}{r}$ where $m$ is the mass of the revolving object

Complete step by step answer:
We know that the gravitational acceleration on the surface of the Earth is \[g = 9.81\;m{s^{ - 2}}\] and the radius of the Earth is \[ = 6.37 \times {10^6}m\].
We also know that the gravitational acceleration at the surface of the Earth is determined as:
$g = \dfrac{{GM}}{{R_E^2}}$
Then we can shift the above formula and determine the mass of the Earth as
$M = \dfrac{{gR_E^2}}{G}$
Substituting the value of \[g = 9.81\;m{s^{ - 2}}\], ${R_E} = 6.37 \times {10^6}m$ and \[G = 6.67 \times {10^{ - 11}}N\;{m^2}\;k{g^2}\], we get
$M = \dfrac{{9.81 \times {{\left( {6.37 \times {{10}^6}} \right)}^2}}}{{6.67 \times {{10}^{ - 11}}}}$
Which gives us
$M = 5.97 \times {10^{24}}kg$
Second method
When the moon is revolving around the Earth, the centripetal force experienced by the moon is balanced by the gravitational force between it and the Earth. This can be represented mathematically as
$\dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$
In this case, $r$ is the distance between the Earth and the moon
Now the velocity of the moon in its orbit can be determined by the relation of linear velocity with angular velocity as
$v = r\omega $ so we can write the above equation as
$\dfrac{{GMm}}{{{r^2}}} = m{\omega ^2}r$
Now the angular velocity of the moon will be $\omega = 2\pi /T$ where $T$ is the time period of rotation of Earth.
Solving for $M$, we can write
$M = \dfrac{{4{\pi ^2}{r^3}}}{{G{T^2}}}$
Substituting the appropriate values, we get
$M = 6.02 \times {10^{24}}kg$
Which is the mass of the Earth

Note: The value of the gravitational acceleration on Earth depends on the distance of the surface from the centre of mass of the Earth. Also, the moon has an elliptical orbit which means its distance from the Earth is not constant but for the sake of this problem, we can consider the orbit to be circular.