Question

Derive the unit of G

Answer 1

Your answer is m³/s²kg or Newton-metre²/kg² in SI and dyne-cm²/gm² in CGS.

Now let me explain,
Newton stated that if there is two body of mass m1 and m2 at the distance 'd' then the gravitational force between them is proportional to their masses and inversely proportional to the square of their distances.
Therefore gravitational force F proportional to m1m2 when distance is constant.
And F proportional to 1/d² when masses are constant.
So when both are variable then F is proportional to m1m2/d²
So F = Gm1m2/d² where G is the constant of proportionality.
Now let the masses of the two body be 1kg and the distance between them be 1m.
So the equation should look like
F = G1kg×1kg/1m²
Therefore,when mass and distances are 1 then F=G
plotting the unit of F we get
Kg-m/s²=Gkg²/m²
G=(kg-m/s² )m²/kg²
G=m³/s²kg
or Newton-metre²/kg²
It is the SI unit of Gravitational constant.
The CGS unit is Dyne-cm²/gm².
And the value of G is 6.67×10`¹¹ N-m²/kg²
It is the amount of force which acts between two bodys of mass 1kg at a distance of 1m.

Hope it helped.
If yes show your gratitude.
Have a nice day.

Answer 2

Answer:

As, we know that

$f = g \times \frac{m1 \times m2}{ {r}^{2} } \\ \\ g = f \times \frac{ {r}^{2} }{m1 \times m2} \\ \\ putting \: unit \: of \: length \: and \: mass \\ \\ g = \frac{f \: \times {m}^{2} }{ {kg}^{2} }$

Now , Force, = mass × acceleration

$= \: \frac{kg \: m}{ {s}^{2} }$

so,

$g = \: \frac{kgm \times {m}^{2} }{ {s}^{2} \times {kg}^{2} } \\ \\ = \frac{ {m}^{3} }{ {s}^{2} kg }$

So, answer is m³/s² kg