Question

The gravitational pull of the moon is $(\frac{1}{6})^{th}$ of the earth and mass of moon is $(\frac{1}{8})^{th}$ of the earth. This implies that the

• A. radius of moon is $(1/4)^{th}$ of the earth's radius.
• B. radius of the earth is $(\sqrt{4/3})^{th}$ of the moon's radius.
• C. moon's radius is half that of the earth.
• D. radius of the earth is $(4/3)^{th}$ of the moon's radius.

Hint:

Gravity is directly proportional to mass and inversely proportional to the square of the radius.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The "gravitational pull" refers to the acceleration due to gravity ($g$) on the surface. We need to find the relationship between the radii of the Earth and the Moon given the ratios of their $g$ values and masses.
Step 2: Key Formula or Approach:
Acceleration due to gravity is:
\[ g = \frac{GM}{R^2} \implies R^2 = \frac{GM}{g} \]
Step 3: Detailed Explanation:
Let $g_e, M_e, R_e$ be Earth's parameters and $g_m, M_m, R_m$ be Moon's.
Given: $g_m = \frac{1}{6} g_e$ and $M_m = \frac{1}{8} M_e$.
Using the formula for $g$:
\[ \frac{g_m}{g_e} = \frac{M_m}{M_e} \times \left(\frac{R_e}{R_m}\right)^2 \]
\[ \frac{1/6 g_e}{g_e} = \frac{1/8 M_e}{M_e} \times \left(\frac{R_e}{R_m}\right)^2 \]
\[ \frac{1}{6} = \frac{1}{8} \times \left(\frac{R_e}{R_m}\right)^2 \]
\[ \left(\frac{R_e}{R_m}\right)^2 = \frac{8}{6} = \frac{4}{3} \]
Taking the square root on both sides:
\[ \frac{R_e}{R_m} = \sqrt{\frac{4}{3}} \]
\[ R_e = \sqrt{\frac{4}{3}} R_m \]
Step 4: Final Answer:
The radius of the Earth is $(\sqrt{4/3})^{th}$ of the Moon's radius.