Question

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

Answer 1

Hint:One can check what is the unit of g in the above equation after substituting the values of G, M and R? In the above equation, G is the universal gravitational constant and it is constant for every celestial object.

Complete answer:
We know that the acceleration due to gravity of the planet is expressed as,
\[g = G \times \dfrac{M}{{{R^2}}}\]
Here, G is the universal gravitational constant, M is the mass of the planet and R is the radius of the planet. The value of universal gravitational constant is \[6.67 \times {10^{ - 11}}\,{{\text{m}}^3}{\text{k}}{{\text{g}}^{ - 1}}{{\text{s}}^{ - 2}}\].

From the above equation, we can observe that the acceleration due to gravity of the planet is proportional to its mass and inversely proportional to its radius.Therefore, we can say that planets with the mass greater than the mass of earth but of the same size will have greater acceleration due to gravity than that of the earth.

We know that the acceleration due to gravity of the earth is \[9.8\,m/{s^2}\] which can be obtained by substituting \[6.67 \times {10^{ - 11}}\,{{\text{m}}^3}{\text{k}}{{\text{g}}^{ - 1}}{{\text{s}}^{ - 2}}\] for G, \[6 \times {10^{24}}\,kg\] for M and \[6.4 \times {10^6}\,m\] for R in the above equation.
\[\Rightarrow g = \left( {6.67 \times {{10}^{ - 11}}\,{{\text{m}}^3}{\text{k}}{{\text{g}}^{ - 1}}{{\text{s}}^{ - 2}}} \right) \times \dfrac{{\left( {6 \times {{10}^{24}}\,kg} \right)}}{{{{\left( {6.4 \times {{10}^6}\,m} \right)}^2}}}\]
\[ \therefore g \approx 9.8\,m/{s^2}\]

Note:The equation, \[g = G \times \dfrac{M}{{{R^2}}}\] implies that the acceleration due to gravity is proportional to the mass of the planet but also it implies that the acceleration due to gravity is proportional to the density of the planets. We see outer planet Saturn has comparable acceleration due to gravity as that of the earth even though its size is very large compared to the size of earth. It is because the density of Saturn is less than the density of earth.