Consider two solid spheres of radii $$R_1 = 1$$ m, $$R_2 = 2$$ m and masses $$M_1$$ and $$M_2$$, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $$\frac{M_1}{M_2}$$ is:

Minimum Required Theory

• Gravitational Field of a Solid Sphere: The gravitational field intensity ($$E$$) at a distance $$r$$ from the center of a solid sphere of mass $$M$$ and radius $$R$$ follows two regimes:
  • Inside ($$r < R$$): $$E \propto r$$ (linear increase)
• Outside ($$r \ge R$$): $$E = \frac{GM}{r^2}$$ (inverse-square decrease)
• Peak Value: The maximum gravitational field occurs exactly at the surface of the sphere ($$r = R$$), where:
• Radius $$R_1 = 1\text{ m}$$
• Peak gravitational field value at its surface ($$r = 1\text{ m}$$) is $$E_1 = 2$$
• Radius $$R_2 = 2\text{ m}$$
• Peak gravitational field value at its surface ($$r = 2\text{ m}$$) is $$E_2 = 3$$

$$E_{max} = \frac{GM}{R^2}$$

Step-by-Step Solution

Step 1: Extract data from the graph for Sphere (1)

Using the surface field formula:

$$2 = \frac{GM_1}{(1)^2} \implies 2 = GM_1 \quad \text{--- (Equation 1)}$$

Step 2: Extract data from the graph for Sphere (2)

Using the surface field formula:

$$3 = \frac{GM_2}{(2)^2} \implies 3 = \frac{GM_2}{4} \implies 12 = GM_2 \quad \text{--- (Equation 2)}$$

Step 3: Calculate the ratio $$\frac{M_1}{M_2}$$

Divide Equation 1 by Equation 2:

$$\frac{GM_1}{GM_2} = \frac{2}{12}$$

Cancel the universal gravitational constant $$G$$:

$$\frac{M_1}{M_2} = \frac{1}{6}$$

Final Answer

The value of $$\frac{M_1}{M_2}$$ is $$\frac{1}{6}$$ (or $$0.167$$).