The diameter and density of a planet are twice that of the earth. Assuming the earth and the planet to be of uniform density, find the ratio of the lengths of two simple pendulums, of equal time periods, on the surface of the planet and the earth.

Answer

To solve the problem, we need to find the ratio of the lengths of two simple pendulums (L1 for Earth and L2 for the planet) that have equal time periods. Given that the diameter and density of the planet are twice that of Earth, we can use the following steps:

### Step 1: Understand the relationship between the time period of a pendulum and the acceleration due to gravity.
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Set up the equations for the two pendulums.
Let \( T_1 \) be the time period of the pendulum on Earth and \( T_2 \) be the time period of the pendulum on the planet. Since the time periods are equal, we have:
\[ T_1 = T_2 \]
This implies:
\[ 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L_2}{g_2}} \]
Cancelling \( 2\pi \) from both sides gives:
\[ \sqrt{\frac{L_1}{g_1}} = \sqrt{\frac{L_2}{g_2}} \]
Squaring both sides results in:
\[ \frac{L_1}{g_1} = \frac{L_2}{g_2} \]
Thus, we can express the ratio of the lengths as:
\[ \frac{L_1}{L_2} = \frac{g_1}{g_2} \]

### Step 3: Calculate the acceleration due to gravity on Earth and the planet.
The acceleration due to gravity \( g \) on the surface of a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

#### For Earth:
- Let the radius of Earth be \( R_e \) and the density be \( \rho_e \).
- The mass of Earth \( M_e \) can be expressed as:
\[ M_e = \rho_e \cdot V_e = \rho_e \cdot \frac{4}{3} \pi R_e^3 \]
Thus:
\[ g_1 = \frac{G M_e}{R_e^2} = \frac{G \rho_e \cdot \frac{4}{3} \pi R_e^3}{R_e^2} = \frac{4}{3} \pi G \rho_e R_e \]

#### For the planet:
- The radius of the planet is \( R_p = 2R_e \) and the density is \( \rho_p = 2\rho_e \).
- The mass of the planet \( M_p \) can be expressed as:
\[ M_p = \rho_p \cdot V_p = 2\rho_e \cdot \frac{4}{3} \pi (2R_e)^3 = 2\rho_e \cdot \frac{4}{3} \pi \cdot 8R_e^3 = \frac{64}{3} \pi \rho_e R_e^3 \]
Thus:
\[ g_2 = \frac{G M_p}{R_p^2} = \frac{G \cdot \frac{64}{3} \pi \rho_e R_e^3}{(2R_e)^2} = \frac{G \cdot \frac{64}{3} \pi \rho_e R_e^3}{4R_e^2} = \frac{16}{3} \pi G \rho_e R_e \]

### Step 4: Find the ratio of \( g_1 \) and \( g_2 \).
Now we can find the ratio:
\[ \frac{g_1}{g_2} = \frac{\frac{4}{3} \pi G \rho_e R_e}{\frac{16}{3} \pi G \rho_e R_e} = \frac{4}{16} = \frac{1}{4} \]

### Step 5: Substitute back to find the ratio of lengths.
Using the ratio of the accelerations due to gravity in the length ratio:
\[ \frac{L_1}{L_2} = \frac{g_1}{g_2} = \frac{1}{4} \]

### Final Answer:
The ratio of the lengths of the two simple pendulums is:
\[ \frac{L_1}{L_2} = \frac{1}{4} \]