Answer
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Hint: The gravitational constant (G) is a proportionality constant used in Newton’s law of gravitation. It relates the gravitational force acting between two masses which are separated by a certain distance. In the S.I. system the unit of G is $\text{N}{{\text{m}}^{\text{2}}}\text{k}{{\text{g}}^{\text{-2}}}$.
Complete step by step answer:
In the problem, we are asked to specify the value of the gravitational constant in the C.G.S unit system. In order to convert the value of gravitational constant given in the C.G.S unit system, we must convert it from the SI unit system, and we need to know the conversion factors of dyne, centimetre and gram (since these are the units that make up the C.G.S system) from newton, metre and kilogram respectively (corresponding units in SI system). The conversion factors are given by,
$\begin{align}
& 1\text{ Newton}={{10}^{5}}\text{ Dyne} \\
& \text{1 Metre}={{10}^{2}}\text{ Centimetre} \\
& \text{1 Kilogram}={{10}^{3}}\text{ Gram} \\
\end{align}$
The value of the gravitational constant in SI unit is $G=6.67\times {{10}^{-11}}\dfrac{\left( Newton \right)\left( m \right)}{\left( kg \right)}$.
Substituting this conversion in the value of G in the SI system,
$G=6.67\times {{10}^{-11}}\dfrac{\left( {{10}^{5}}Dyne \right){{\left( {{10}^{2}}cm \right)}^{2}}}{{{\left( {{10}^{3}}g \right)}^{2}}}$
$\Rightarrow G=6.67\times {{10}^{-11}}\dfrac{\left( {{10}^{5}}Dyne \right)\left( {{10}^{4}}cm \right)}{\left( {{10}^{6}}g \right)}$
$\therefore G=6.67\times {{10}^{-8}}\dfrac{\left( Dyne \right)\left( cm \right)}{\left( g \right)}$
So, the value of the gravitational constant in SI unit is $G=6.7\times {{10}^{-8}}\dfrac{\left( Dyne \right)\left( cm \right)}{\left( g \right)}$.
Additional Information:
The C.G.S unit for force is dyne (Dyn).
The C.G.S unit for pressure is barye (Ba).
The C.G.S unit for velocity is centimetre per second (cm/s).
The C.G.S unit for acceleration is gal (Gal).
The C.G.S unit for mass is gram (g).
The C.G.S unit for wavenumber is kayser (K).
Note: The C.G.S system becomes less intuitive and more confusing to students when we leave the realm of mechanics. When a new concept such as charge is encountered, it seems necessary to introduce a new unit to measure its quantity. Both the coulomb and the Faraday were introduced in this way to define a new concept.
The SI unit system is more flexible and easily usable compared to the C.G.S system.
Complete step by step answer:
In the problem, we are asked to specify the value of the gravitational constant in the C.G.S unit system. In order to convert the value of gravitational constant given in the C.G.S unit system, we must convert it from the SI unit system, and we need to know the conversion factors of dyne, centimetre and gram (since these are the units that make up the C.G.S system) from newton, metre and kilogram respectively (corresponding units in SI system). The conversion factors are given by,
$\begin{align}
& 1\text{ Newton}={{10}^{5}}\text{ Dyne} \\
& \text{1 Metre}={{10}^{2}}\text{ Centimetre} \\
& \text{1 Kilogram}={{10}^{3}}\text{ Gram} \\
\end{align}$
The value of the gravitational constant in SI unit is $G=6.67\times {{10}^{-11}}\dfrac{\left( Newton \right)\left( m \right)}{\left( kg \right)}$.
Substituting this conversion in the value of G in the SI system,
$G=6.67\times {{10}^{-11}}\dfrac{\left( {{10}^{5}}Dyne \right){{\left( {{10}^{2}}cm \right)}^{2}}}{{{\left( {{10}^{3}}g \right)}^{2}}}$
$\Rightarrow G=6.67\times {{10}^{-11}}\dfrac{\left( {{10}^{5}}Dyne \right)\left( {{10}^{4}}cm \right)}{\left( {{10}^{6}}g \right)}$
$\therefore G=6.67\times {{10}^{-8}}\dfrac{\left( Dyne \right)\left( cm \right)}{\left( g \right)}$
So, the value of the gravitational constant in SI unit is $G=6.7\times {{10}^{-8}}\dfrac{\left( Dyne \right)\left( cm \right)}{\left( g \right)}$.
Additional Information:
The C.G.S unit for force is dyne (Dyn).
The C.G.S unit for pressure is barye (Ba).
The C.G.S unit for velocity is centimetre per second (cm/s).
The C.G.S unit for acceleration is gal (Gal).
The C.G.S unit for mass is gram (g).
The C.G.S unit for wavenumber is kayser (K).
Note: The C.G.S system becomes less intuitive and more confusing to students when we leave the realm of mechanics. When a new concept such as charge is encountered, it seems necessary to introduce a new unit to measure its quantity. Both the coulomb and the Faraday were introduced in this way to define a new concept.
The SI unit system is more flexible and easily usable compared to the C.G.S system.
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