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Question: Define Universal gravitational constant G. What is the dimensional formula G?...

Define Universal gravitational constant G. What is the dimensional formula G?

Explanation

Solution

Newton’s Law of gravitation:
Newton’s Law of Universal Gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
Fm1m2r2F \propto \dfrac{{{m_1}{m_{_2}}}}{{{r^2}}}
To remove the proportionality we always need to introduce a constant of proportionality so here newton inserted a constant called Universal Gravitational constant (G)\left( G \right)
F=Gm1m2r2F = G\dfrac{{{m_1}{m_{_2}}}}{{{r^2}}}

Complete solution:
Universal gravitational constant
The gravitational constant is the proportionality constant that is used in Newton’s Law of gravitation. The force of attraction between any two unit masses separated by a unit distance is called the universal gravitational constant denoted by (G)\left( G \right)measured in(Nm2kg2)\left( {\dfrac{{N{m^2}}}{{k{g^2}}}} \right).
Mathematically,
It two objects of unit mass and are separated by unit distance then, the force with which they’ll attract each other is called universal gravitational constant (G)\left( G \right)
m1,m2=1{m_1},{m_{_2}} = 1Unit
r=1r = 1Unit
F=G\therefore F = G

Dimensional formula of Universal Gravitational Constant(G)\left( G \right)
The expressions or formulae which tell us how and which of the fundamental quantities are present in a physical quantity are known as the Dimensional Formula of the Physical Quantity.
Suppose there is a physical quantity X which depends on base dimensions M (Mass), L (Length), and T (Time) with respective powers a, b and c, then its dimensional formula is represented as: [MaLbTc]\left[ {{M^a}{L^b}{T^c}} \right]
Dimensional formula for basic physical quantities,
Mass = [M]\left[ M \right]
Distance = [L]\left[ L \right]
Time = [T]\left[ T \right]
Velocity: It is the distance covered per unit time so v=(dt)v = \left( {\dfrac{d}{t}} \right)
From here we can say that is the ratio of distance and times so the dimensional formula will be:
LT\dfrac{L}{T} (Or) [LT1]\left[ {L{T^{ - 1}}} \right]
Acceleration = it is the rate of change of velocity with time so,
We can say that a=vta = \dfrac{v}{t}
Here velocity is divided by time again we will substitute the dimensional formulas of velocity band time to get the dimensional formula for acceleration

a=[LT1T] a=[LT2]  \therefore a = \left[ {\dfrac{{L{T^{ - 1}}}}{T}} \right] \\\ \Rightarrow a = \left[ {L{T^{ - 2}}} \right] \\\

Force: It is defined as the mass time of acceleration so,
F=M×AF = M \times A
Now we will be using the dimensional formula of acceleration and mass to find out the dimensional formula of force
F=M×A\because F = M \times A
The dimensional formula of force will also be the product of the dimensional formula of the other two quantities
So, F=M×LT2 F=[MLT2]  F = M \times L{T^{ - 2}} \\\ \Rightarrow F = \left[ {ML{T^{ - 2}}} \right] \\\
So now as we know,

F=Gm1m2r2 G=Fr2m1m2  F = G\dfrac{{{m_1}{m_{_2}}}}{{{r^2}}} \\\ \therefore G = \dfrac{{F{r^2}}}{{{m_1}{m_{_2}}}} \\\

Now for the dimensional formula of (G),\left( G \right), substitute the dimensional formula of other quantities
As r is the radius so its dimensional formula will be the same as of distance i.e. (L)\left( L \right)

G=[(MLT2)(L2)(M×M)] G=[L3T2M] G=[M1L3T2]  G = \left[ {\dfrac{{\left( {ML{T^{ - 2}}} \right)\left( {{L^2}} \right)}}{{\left( {M \times M} \right)}}} \right] \\\ \Rightarrow G = \left[ {\dfrac{{{L^3}{T^{ - 2}}}}{M}} \right] \\\ \Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] \\\

On solving we get,
Dimensional formula for [G]\left[ G \right]=[M1L3T2]\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right].

Final answer is, the dimensional formula for universal gravitation constant is [M1L3T2]\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right].

Note: 1. Dimensional formula of any quantity can be derived for the fundamental quantities if the relation between them is known.
2. Dimensional Formulas are used to check whether a given formula is dimensionally correct or not.
3. Dimensional Formulae become not defined in the case of the trigonometric, logarithmic, and exponential functions as they are not physical quantities.