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 of equal time periods on the surface of a planet and the Earth, given that the diameter and density of the planet are twice that of the Earth.

### Step 1: Understand the relationship between the length 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 equation for equal time periods
Since the time periods of the pendulums on the planet and the Earth are equal, we can write:
\[ T_{planet} = T_{earth} \]
This implies:
\[ 2\pi \sqrt{\frac{l_{planet}}{g_{planet}}} = 2\pi \sqrt{\frac{l_{earth}}{g_{earth}}} \]
By squaring both sides and simplifying, we get:
\[ \frac{l_{planet}}{g_{planet}} = \frac{l_{earth}}{g_{earth}} \]
Thus, we can express the ratio of the lengths as:
\[ \frac{l_{planet}}{l_{earth}} = \frac{g_{planet}}{g_{earth}} \]

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

### Step 4: Find the mass of the planet and Earth
The mass \( M \) can be expressed in terms of density \( \rho \) and volume \( V \):
\[ M = \rho V \]
For a sphere, the volume \( V \) is:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass becomes:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 5: Substitute the mass into the gravity formula
Substituting for mass in the gravity formula gives:
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} G \pi \rho R \]
This shows that \( g \) is proportional to \( \rho R \).

### Step 6: Apply the given conditions
Given that the diameter and density of the planet are twice that of the Earth, we have:
- Density of the planet, \( \rho_{planet} = 2 \rho_{earth} \)
- Radius of the planet, \( R_{planet} = 2 R_{earth} \)

### Step 7: Calculate the ratio of gravitational accelerations
Now, we can calculate the ratio of the gravitational accelerations:
\[ \frac{g_{planet}}{g_{earth}} = \frac{(2 \rho_{earth})(2 R_{earth})}{\rho_{earth} R_{earth}} = \frac{4 \rho_{earth} R_{earth}}{\rho_{earth} R_{earth}} = 4 \]

### Step 8: Find the ratio of the lengths of the pendulums
Now substituting back into the ratio of lengths:
\[ \frac{l_{planet}}{l_{earth}} = \frac{g_{planet}}{g_{earth}} = 4 \]

### Final Answer
The ratio of the lengths of the two simple pendulums is:
\[ \frac{l_{planet}}{l_{earth}} = 4 \]