Question

Estimate the mass of the earth, given radius of the earth =$6.4 \times {10^6}$m, acceleration due to gravity= $9.8m/{s^2}$ and gravitational constant =$6.67 \times {10^ - }11$ S.I units.

Answer 1

Hint: Gravitational acceleration is referred to as the acceleration of an object.
This is the constant gain in speed because exclusively by the force of gravitational attraction
The measurement and analysis of the gravitational rates are known as gravimetry

Complete answer:
The value of gravity ‘\[g\] ’ is not the same at all places of the earth. The gravity value depends on two factors:
1.The distance of the earth's center to the place
2.The place latitude is the angle of the place with horizontal
 The value of gravity is found to be \[9.8{\text{ }}m/s2\] at the equator and more at the poles as the radii of the earth are less than that of the equator about \[21{\text{ }}km.\]
It comes from Newton’s law of gravitation.
The formula for the Gravitational force is
\[{F_G} = \dfrac{{GMm}}{{{r^2}}}\]
 the acceleration due to gravity on an object with mass m is,
\[{a_G} = \dfrac{{{F_G}}}{m} = \dfrac{{GM}}{{{r^2}}}\]
Take \[G\] to be Newton’s Gravitational constant, \[M\] to be the mass of the Earth, and \[r\] to be the Radius of the Earth and you will find
\[{a_G} = 9.8m/{s^2}\]
 Now,
\[F = \dfrac{{GM}}{{{R^2}}}\]
Given,
\[G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\]
\[R = 6.4 \times {10^6}m\]
\[g = 9.8m{s^{ - 2}}\]
We know that,
\[M = \dfrac{{g{R^2}}}{G}\]
Put the values in the above equation,
\[M = \dfrac{{g{R^2}}}{G} = \dfrac{{9.8 \times 6.4 \times 6.4 \times {{10}^{12}}}}{{6.67 \times {{10}^{ - 11}}}}\]
Now simplify the above equation we get,
$ \approx 6 \times {10^{24}}kg$.
Here the mass of the earth is $ \approx 6 \times {10^{24}}kg$

Note:
The gravity denoted by g is the acceleration the Earth imparts to objects on or near its surface.
The acceleration is measured in meters per second squared or equivalently in newtons per kilogram \[\left( {N/kg} \right)\] in SI units.
The approximate value of 9.81 m/s2, it means that, ignoring the effects of air resistance