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Is it true that the value and the unit of gravitational constant $G$ in the CGS system are $6.67 \times 10^{- 8} {dyne cm}^{2} g^{- 2}$ ? (Given in SI system $= 6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}$ )

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Detailed Solution

The above given statement is true.
SI unit of gravitational constant

G

= 6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}

We are asked to find out the value and unit of gravitational constant

G

in the CGS system.
Now we have to look into the unit of

G

in the SI system and find out the equivalent of those terms in the CGS system.
Here for

G

in the SI system,

N

stands for Newton,

m

stands for metres, and

kg

stands for kilograms.
In the CGS system the equivalents of these terms are

1

Newton

=

10^{5}

dyne

1

Metre

=

10^{2}

Centimeter

1

Kilogram

=

10^{3}

gram
Now we can substitute all these conversions in

G

G = 6.67 \times 10^{- 11} \frac{N m^{2}}{{kg}^{2}}

G

= 6.67 \times 10^{- 11} \frac{(10^{5} dyne) {(10^{2} Centimeter)}^{2}}{{(10^{3} gram)}^{2}}

G

= 6.67 \times 10^{- 11} \times \frac{10^{5} \times 10^{4}}{10^{6}} \frac{(dyne) {(Centimeter)}^{2}}{{(gram)}^{2}}

G

= 6.67 \times 10^{- 8} \frac{(dyne) {(Centimeter)}^{2}}{{(gram)}^{2}}

G

= 6.67 \times 10^{- 8} \frac{dyne {Cm}^{2}}{g^{2}}

G

= 6.67 \times 10^{- 8} dyne {Cm}^{2} g^{- 2}

Thus by converting

G

from SI system to CGS system we get

G

= 6.67 \times 10^{- 8} dyne {Cm}^{2} g^{- 2}

Thus the value and unit of gravitational constant

G

in the CGS system is

6.67 \times 10^{- 8} dyne {Cm}^{2} g^{- 2}

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