Define Universal gravitational constant G. What is the dimensional formula G?
Answer
637.2k+ views
Hint: 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.
$F \propto \dfrac{{{m_1}{m_{_2}}}}{{{r^2}}}$
To remove the proportionality we always need to introduce a constant of proportionality so here newton inserted a constant called Universal Gravitational constant $\left( G \right)$
\[F = G\dfrac{{{m_1}{m_{_2}}}}{{{r^2}}}\]
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 $\left( G \right)$measured in\[\left( {\dfrac{{N{m^2}}}{{k{g^2}}}} \right)\].
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 $\left( G \right)$
\[{m_1},{m_{_2}} = 1\]Unit
$r = 1$Unit
$\therefore F = G$
Dimensional formula of Universal Gravitational Constant$\left( G \right)$
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: \[\left[ {{M^a}{L^b}{T^c}} \right]\]
Dimensional formula for basic physical quantities,
Mass = $\left[ M \right]$
Distance = \[\left[ L \right]\]
Time = $\left[ T \right]$
Velocity: It is the distance covered per unit time so $v = \left( {\dfrac{d}{t}} \right)$
From here we can say that is the ratio of distance and times so the dimensional formula will be:
$\dfrac{L}{T}$ (Or) $\left[ {L{T^{ - 1}}} \right]$
Acceleration = it is the rate of change of velocity with time so,
We can say that $a = \dfrac{v}{t}$
Here velocity is divided by time again we will substitute the dimensional formulas of velocity band time to get the dimensional formula for acceleration
\[
\therefore a = \left[ {\dfrac{{L{T^{ - 1}}}}{T}} \right] \\
\Rightarrow a = \left[ {L{T^{ - 2}}} \right] \\
\]
Force: It is defined as the mass time of acceleration so,
$F = M \times A$
Now we will be using the dimensional formula of acceleration and mass to find out the dimensional formula of force
$\because F = M \times A$
The dimensional formula of force will also be the product of the dimensional formula of the other two quantities
So, $
F = M \times L{T^{ - 2}} \\
\Rightarrow F = \left[ {ML{T^{ - 2}}} \right] \\
$
So now as we know,
\[
F = G\dfrac{{{m_1}{m_{_2}}}}{{{r^2}}} \\
\therefore G = \dfrac{{F{r^2}}}{{{m_1}{m_{_2}}}} \\
\]
Now for the dimensional formula of $\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. $\left( L \right)$
\[
G = \left[ {\dfrac{{\left( {ML{T^{ - 2}}} \right)\left( {{L^2}} \right)}}{{\left( {M \times M} \right)}}} \right] \\
\Rightarrow G = \left[ {\dfrac{{{L^3}{T^{ - 2}}}}{M}} \right] \\
\Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] \\
\]
On solving we get,
Dimensional formula for $\left[ G \right]$=$\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.
Final answer is, the dimensional formula for universal gravitation constant is $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.
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.
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.
$F \propto \dfrac{{{m_1}{m_{_2}}}}{{{r^2}}}$
To remove the proportionality we always need to introduce a constant of proportionality so here newton inserted a constant called Universal Gravitational constant $\left( G \right)$
\[F = G\dfrac{{{m_1}{m_{_2}}}}{{{r^2}}}\]
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 $\left( G \right)$measured in\[\left( {\dfrac{{N{m^2}}}{{k{g^2}}}} \right)\].
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 $\left( G \right)$
\[{m_1},{m_{_2}} = 1\]Unit
$r = 1$Unit
$\therefore F = G$
Dimensional formula of Universal Gravitational Constant$\left( G \right)$
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: \[\left[ {{M^a}{L^b}{T^c}} \right]\]
Dimensional formula for basic physical quantities,
Mass = $\left[ M \right]$
Distance = \[\left[ L \right]\]
Time = $\left[ T \right]$
Velocity: It is the distance covered per unit time so $v = \left( {\dfrac{d}{t}} \right)$
From here we can say that is the ratio of distance and times so the dimensional formula will be:
$\dfrac{L}{T}$ (Or) $\left[ {L{T^{ - 1}}} \right]$
Acceleration = it is the rate of change of velocity with time so,
We can say that $a = \dfrac{v}{t}$
Here velocity is divided by time again we will substitute the dimensional formulas of velocity band time to get the dimensional formula for acceleration
\[
\therefore a = \left[ {\dfrac{{L{T^{ - 1}}}}{T}} \right] \\
\Rightarrow a = \left[ {L{T^{ - 2}}} \right] \\
\]
Force: It is defined as the mass time of acceleration so,
$F = M \times A$
Now we will be using the dimensional formula of acceleration and mass to find out the dimensional formula of force
$\because F = M \times A$
The dimensional formula of force will also be the product of the dimensional formula of the other two quantities
So, $
F = M \times L{T^{ - 2}} \\
\Rightarrow F = \left[ {ML{T^{ - 2}}} \right] \\
$
So now as we know,
\[
F = G\dfrac{{{m_1}{m_{_2}}}}{{{r^2}}} \\
\therefore G = \dfrac{{F{r^2}}}{{{m_1}{m_{_2}}}} \\
\]
Now for the dimensional formula of $\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. $\left( L \right)$
\[
G = \left[ {\dfrac{{\left( {ML{T^{ - 2}}} \right)\left( {{L^2}} \right)}}{{\left( {M \times M} \right)}}} \right] \\
\Rightarrow G = \left[ {\dfrac{{{L^3}{T^{ - 2}}}}{M}} \right] \\
\Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] \\
\]
On solving we get,
Dimensional formula for $\left[ G \right]$=$\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.
Final answer is, the dimensional formula for universal gravitation constant is $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.
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.
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