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Question: If the value of G, the gravitational constant in the C.G.S system is \[6.70\times {{10}^{-8}}\], the...

If the value of G, the gravitational constant in the C.G.S system is 6.70×10−86.70\times {{10}^{-8}}, then its value in the S.I unit?

Explanation

Solution

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 Nm2kg-2\text{N}{{\text{m}}^{\text{2}}}\text{k}{{\text{g}}^{\text{-2}}}.

Complete step by step answer:
In the problem, it is specified that the value of the gravitational constant in C.G.S unit is 6.70×10−86.70\times {{10}^{-8}} (dyne)(cm)2(g)−2\left( dyne \right){{\left( cm \right)}^{2}}{{\left( g \right)}^{-2}}. In order to convert the value of gravitational constant given in the C.G.S unit into SI unit, 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) to newton, metre and kilogram respectively (corresponding units in SI system). The conversion factors are given by,
1 Newton=105 Dyne 1 Metre=102 Centimetre 1 Kilogram=103 Gram \begin{aligned} & 1\text{ Newton}={{10}^{5}}\text{ Dyne} \\\ & \text{1 Metre}={{10}^{2}}\text{ Centimetre} \\\ & \text{1 Kilogram}={{10}^{3}}\text{ Gram} \\\ \end{aligned}
Substituting this conversion in the value of G in the C.G.S system,
G=6.7×10−8(10−5Newton)(10−2m)2(10−3kg)−2G=6.7\times {{10}^{-8}}\dfrac{\left( {{10}^{-5}}Newton \right){{\left( {{10}^{-2}}m \right)}^{2}}}{{{\left( {{10}^{-3}}kg \right)}^{-2}}}
⇒G=6.7×10−8(10−5Newton)(10−4m)(10−6kg)\Rightarrow G=6.7\times {{10}^{-8}}\dfrac{\left( {{10}^{-5}}Newton \right)\left( {{10}^{-4}}m \right)}{\left( {{10}^{-6}}kg \right)}
∴G=6.7×10−8×10−3(Newton)(m)(kg)\therefore G=6.7\times {{10}^{-8}}\times {{10}^{-3}}\dfrac{\left( Newton \right)\left( m \right)}{\left( kg \right)}
So, the value of the gravitational constant in SI unit is G=6.7×10−11(Newton)(m)(kg)G=6.7\times {{10}^{-11}}\dfrac{\left( Newton \right)\left( m \right)}{\left( kg \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.