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Question: Explain what is meant by equation: \(g=G\times \dfrac{M}{{{R}^{2}}}\) Where the symbols have thei...

Explain what is meant by equation: g=G×MR2g=G\times \dfrac{M}{{{R}^{2}}}
Where the symbols have their usual meanings.

Explanation

Solution

The given formula is used to expression the acceleration due to gravity (g)\left( g \right) by substituting values of G,MG,M and R.R. Where GG is the gravitational constant of value 6.67×10−11m3kg−1s−26.67\times {{10}^{-11}}{ }{{{m}}^{3}}k{{g}^{-1}}{{s}^{-2}}. The acceleration due to gravity is not the same on different planets.

Complete answer:
The below equation is used to express the acceleration due to gravity of the planet.
g=G×MR2g=G\times \dfrac{M}{{{R}^{2}}}
Where,
g=g= acceleration due to gravity of planet
G=G= Universal gravitational constant
M=M= mass of the planet
R=R= radius of the planet
The value of universal gravitational constant is 6.67×10−11m3kg−1s−26.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}.
If we go through the equation, we can say that the acceleration due to gravity of the planet is directly proportional to its mass and inversely proportional to its radius. Hence, if any planet is having the mass greater than the mass of earth but having the same size, then the acceleration due to gravity of that particular planet is greater too.
We know that, the acceleration due to gravity of the earth is 9.8m/s29.8{ m/}{{{s}}^{2}} which can be obtained by substituting 6.67×10−11m3kg−1s−26.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}} for as,
∴G=6.67×10−11m3kg−1s−2\therefore G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}
Put, M=6×1024kgM=6\times {{10}^{24}}kg (mass of earth)
R=6.4×106mR=6.4\times {{10}^{6}}m (Radius of earth)
In above equation
Hence, we get, g=(6.67×10−11)×(6×1024)(6.4×106)unitg=\left( 6.67\times {{10}^{-11}} \right)\times \dfrac{\left( 6\times {{10}^{24}} \right)}{\left( 6.4\times {{10}^{6}} \right)}unit
∴g≈9.8m/s2\therefore g\approx 9.8{ m/}{{{s}}^{2}}

Note: The equation, g=GMR2g =\dfrac{GM}{{{R}^{2}}} implies that not only the mass of the planet but also the density of the planet is directly proportional.
The outer planet, Saturn is of less density than the earth’s density, its acceleration due to gravity is comparable to that of earth. Regardless of the size of Saturn which is very large. The acceleration due to gravity is not the same for all heights even when we are on the planet. However the height with respect to the surface of the planet is negligible when compared to the radius of the planet and so we neglect it and conventionally assume the acceleration due to gravity to be the same.