Value of universal gravitational constant $G$ in CGS unit is-

6.67 \times 10^{8} c m^{3} g^{1} s

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6.67 \times 10^{8} c m^{3} g^{- 1} s^{- 2}

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6.67 \times 10^{9} c m^{3} g^{1} s^{2}

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6.67 \times 10^{7} c m^{3} g^{- 1} s^{2}

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Solution

The correct option is C $6.67 \times 10^{8} c m^{3} g^{- 1} s^{- 2}$
Newton's law of gravitation:
$F = G \frac{M m}{r^{2}}$
$G = \frac{F r^{2}}{M m}$
In CGS unit system:
r - distance - cm,
M,m - mass - grams,
F - force - dyne - $g c m / s^{2}$
Therefore units of G - $\frac{(g c m / s^{2}) (c m^{2})}{g^{2}} = c m^{3} g^{- 1} s^{- 2}$

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Similar questions

Q. The value of universal gravitational constant G =

6.67 \times 10^{- 11} N m^{2} k g^{- 2}

. The value of G in units of

g^{- 1} c m^{3} s^{- 2}

Q. The value of universal gravitational cosntant G in CGS system is

6.67 \times 10^{- 8}

dyne

{c m}^{2} {g m}^{- 2}

. Its value in SI system is :

Q. The value of gravitation constant is

G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2}

in SI units. Convert it into CGS system of units.

Q. Universal gravitational constant is given by

6.67 \times 10^{- 11} N m^{2} {k g}^{- 2}

. Then the no. of significant figures in it is:

Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart. Find the gravitational force of attraction between them. Take $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$

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Value of universal gravitational constant G in CGS unit is-

The correct option is C 6.67×108cm3g−1s−2Newton's law of gravitation:F=GMmr2 G=Fr2MmIn CGS unit system: r - distance - cm, M,m - mass - grams, F - force - dyne - gcm/s2Therefore units of G - (g cm/s2)(cm2)g2=cm3g−1s−2

Value of universal gravitational constant $G$ in CGS unit is-