Consider two solid spheres of radii R_{1} = 1 | vedclass

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(D) The gravitational field $I_{g}$ on the surface of a solid sphere of mass $M$ and radius $R$ is given by $I_{g} = \frac{GM}{R^{2}}$.From the given

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$\frac{1}{6}$

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(D) The gravitational field $I_{g}$ on the surface of a solid sphere of mass $M$ and radius $R$ is given by $I_{g} = \frac{GM}{R^{2}}$.From the given graph,the maximum value of the gravitational field occurs at the surface of the sphere.For sphere $(1)$,the radius is $R_{1} = 1 \; m$ and the maximum gravitational field is $I_{g1} = 2$.Thus,$\frac{GM_{1}}{(1)^{2}} = 2 \implies GM_{1} = 2$.For sphere $(2)$,the radius is $R_{2} = 2 \; m$ and the maximum gravitational field is $I_{g2} = 3$.Thus,$\frac{GM_{2}}{(2)^{2}} = 3 \implies \frac{GM_{2}}{4} = 3 \implies GM_{2} = 12$.Dividing the two equations:$\frac{GM_{1}}{GM_{2}} = \frac{2}{12} = \frac{1}{6}$.Therefore,$\frac{M_{1}}{M_{2}} = \frac{1}{6}$.

Consider two solid spheres of radii $R_{1} = 1 \; m$ and $R_{2} = 2 \; m$ and masses $M_{1}$ and $M_{2}$,respectively. The gravitational field due to sphere $(1)$ and $(2)$ are shown in the graph. The value of $\frac{M_{1}}{M_{2}}$ is

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