- Acceleration due to gravity on moon is 1 / 6 of the...

- Acceleration due to gravity on moon is $1 / 6$ of the acceleration due to gravity on earth. If the ratio of densities of earth $\rho_{\mathrm{e}}$ and moon $\rho_{\mathrm{m}}$ is $\left(\rho_{\mathrm{e}} / \rho_{\mathrm{m}}\right)=5 / 3$ then radius of moon $\mathrm{R}_{\mathrm{e}}$ in terms of $\mathrm{R}_{\mathrm{e}}$ will be (A) $(5 / 18) \mathrm{R}_{\mathrm{e}}$ (B) $(1 / 6) \mathrm{R}_{\mathrm{e}}$ (C) $(3 / 16) \mathrm{R}_{\mathrm{e}}$ (D) $[1 /(2 \sqrt{3})] R_{e}$

- Acceleration due to gravity on moon is $1 / 6$ of the acceleration due to gravity on earth. If the ratio of densities of earth $\rho_{\mathrm{e}}$ and moon $\rho_{\mathrm{m}}$ is $\left(\rho_{\mathrm{e}} / \rho_{\mathrm{m}}\right)=5 / 3$ then radius of moon $\mathrm{R}_{\mathrm{e}}$ in terms of $\mathrm{R}_{\mathrm{e}}$ will be
(A) $(5 / 18) \mathrm{R}_{\mathrm{e}}$
(B) $(1 / 6) \mathrm{R}_{\mathrm{e}}$
(C) $(3 / 16) \mathrm{R}_{\mathrm{e}}$
(D) $[1 /(2 \sqrt{3})] R_{e}$

Key Concepts

Scaling Laws in Gravitation

Scaling laws help to understand how changes in one physical parameter (such as radius or density) can affect another (such as gravitational acceleration). By recognizing that g is proportional to the product of density and radius, one can scale the known properties of one object to predict or compare the properties of another, forming the basis for many astronomical and physical calculations.

Proportional Relationships

Understanding proportional relationships is key when comparing physical quantities such as gravitational acceleration, density, and radius. In problems like this, setting up ratios allows one to relate the parameters of two different bodies, leading to a direct method for solving for an unknown variable when the ratios of other variables are known.

Density

Density is defined as mass per unit volume and is a fundamental property that influences the gravitational field of a body. When comparing gravitational accelerations, the ratio of densities becomes important because it directly affects the mass and thus the gravitational pull experienced at the surface.

Gravitational Acceleration on a Spherical Body

This concept explains that the acceleration due to gravity at the surface of a spherical object is given by g = (4/3)?G?R, where G is the gravitational constant, ? is the density of the object, and R is its radius. This formula reveals that the gravitational acceleration depends linearly on both the density and the radius, which is crucial when comparing different celestial bodies.

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