AIPMT PRELIMS 2004 Physics Planetary Motion MCQ Question

To solve this problem, we need to use the formula for gravity at the surface of a planet, which is given by:

$$g = \frac{GM}{R^2}$$

where:

• $g$ is the acceleration due to gravity at the surface,
• $G$ is the universal gravitational constant,
• $M$ is the mass of the planet, and
• $R$ is the radius of the planet.

Step 1: Relate the Mass and Density of the Planet

Given that the density of the newly discovered planet ($\rho_p$) is twice that of Earth ($\rho_e$), we can write:

$$\rho_p = 2 \rho_e$$

The mass of the planet can be expressed in terms of its density and volume. The volume ($V$) of a sphere is given by:

$$V = \frac{4}{3} \pi R^3$$

Thus, the mass of the planet $M$ can be written as:

$$M = \rho_p V = \rho_p \left(\frac{4}{3} \pi R_p^3\right)$$

Substituting $\rho_p$:

$$M = 2\rho_e \left(\frac{4}{3} \pi R_p^3\right)$$

Step 2: Apply the Condition for Gravity

Since the acceleration due to gravity at the surface of the planet is equal to that of the Earth, we have:

$$g_e = g_p$$

For Earth, the acceleration due to gravity is given by:

$$g_e = \frac{GM_e}{R^2}$$

For the newly discovered planet, we have:

$$g_p = \frac{GM}{R_p^2}$$

Setting $g_e = g_p$:

$$\frac{GM_e}{R^2} = \frac{GM}{R_p^2}$$

Step 3: Substitute Mass in Terms of Density

Substituting the expressions for $M_e$ and $M$:

1. For Earth, $M_e = \rho_e \left(\frac{4}{3} \pi R^3\right)$.
2. For the new planet, $M = 2\rho_e \left(\frac{4}{3} \pi R_p^3\right)$.

Substituting these into the gravity equation gives:

$$\frac{G \left(\rho_e \frac{4}{3} \pi R^3\right)}{R^2} = \frac{G \left(2\rho_e \frac{4}{3} \pi R_p^3\right)}{R_p^2}$$

Step 4: Simplify the Equation

We can cancel $G$, $\frac{4}{3} \pi$, and $\rho_e$ from both sides, leading to:

$$\frac{R^3}{R^2} = \frac{2 R_p^3}{R_p^2}$$

This simplifies to:

$$R = 2R_p$$

Step 5: Solve for $R_p$

Rearranging gives:

R_p = \frac{R}{2} $$ ### Conclusion Thus, the radius of the newly discovered planet is:

R_p = \frac{1}{2} R

This means the correct answer is option **C) $ \frac{1}{2} R $**. ### Clarification of Other Options - **Option A (4R)**: This would imply a much larger planet, which contradicts the gravity condition. - **Option B (¼ R)**: This would imply a significantly smaller planet with a much higher density, which does not satisfy the gravity equality. - **Option D (2R)**: Similar to option A, this suggests a larger radius, leading to a lower gravity than Earth, which is incorrect. Thus, option C is the only correct choice.