Question

Write the units of g and G.

Answer 1

Hint: For finding the units of g and G consider the gravitational force of attraction. Thus by rearranging the equation we will get the value of G. Hence, by substituting the units of force, mass of earth, mass of planet and radius of circular orbit we will get the unit of gravitational constant. The acceleration can also be explained as the rate of change of velocity with the time.

Complete step by step answer:
When a body is falling freely downwards, there will be an acceleration due to gravity towards the earth. The value of acceleration due to gravity is generally equal to $9.8m/s^2$. This can be denoted by the letter ‘g’. The value of acceleration due to gravity may vary with the altitude. Its value generally depends on its gravitational environments. On the earth surface also there will be some variations in the value of acceleration due to gravity. The acceleration can also be explained as the rate of change of velocity with the time.
 Due to the gravitational force of attraction, an object acquires acceleration due to gravity.
That is,
$F=mg$
where, F is the force
m is the mass
g is the acceleration due to gravity.
The value of acceleration due to gravity has maximum value at poles and minimum value at equator.
The acceleration due to gravity has both the magnitude as well as direction. Hence, it is a vector quantity. The SI unit of g is m/s.
Then consider the gravitational force of attraction. That is,
$F=\dfrac{-GMm}{{{r}^{2}}}$
F is the gravitational force
G is the gravitational constant
M is the mass of earth
m is the mass of plane
r is the radius of circular orbit.
Thus by rearranging them we get,
$\Rightarrow G=\dfrac{-F{{r}^{2}}}{Mm}$
Then the dimensional formula for G is given by,
$G=\dfrac{\left[ ML{{T}^{-2}} \right]{{\left[ L \right]}^{2}}}{\left[ M \right]\left[ M \right]}$
$\Rightarrow G=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]$
Thus the SI unit of G is $k{{g}^{-1}}{{m}^{3}}{{s}^{-2}}$.
And $g=\dfrac{GM}{{{r}^{2}}}$
Thus the dimension of g is,
$g=\dfrac{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]\left[ M \right]}{{{\left[ L \right]}^{2}}}$
$\therefore g=\left[ L{{T}^{-2}} \right]$
Hence SI unit of g is $m{{s}^{-2}}$.

Note:
When a body is falling freely downwards, there will be an acceleration due to gravity towards the earth. The value of acceleration due to gravity is generally equal to $9.8m/s$. This can be denoted by the letter ‘g’. The value of acceleration due to gravity may vary with altitude. Its value generally depends on its gravitational environments. The value of acceleration due to gravity has maximum value at poles and minimum value at equator. And the gravitational force of attraction is always negative.