Titan completes one (circular) orbit of Saturn in 16.0 days. The distance from Saturn to Titan is 1.22×109m. What is the centripetal acceleration of Titan?

To calculate the centripetal acceleration \( a_c \) of Titan as it orbits Saturn, we can use the formula:

\[
a_c = \frac{v^2}{r}
\]

Where:
- \( v \) is the orbital velocity of Titan.
- \( r \) is the radius of the orbit (the distance from Saturn to Titan).

We can find the orbital velocity \( v \) using the formula:

\[
v = \frac{2\pi r}{T}
\]

Where:
- \( T \) is the orbital period (time taken to complete one orbit).

### Step 1: Convert the orbital period to seconds
Given that the period is \( T = 16.0 \, \text{days} \), we first convert this to seconds:

\[
T = 16.0 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} = 1,382,400 \, \text{seconds}
\]

### Step 2: Calculate the orbital velocity
Now, using the orbital radius \( r = 1.22 \times 10^9 \, \text{m} \) and the orbital period \( T \), we calculate the velocity \( v \):

\[
v = \frac{2\pi (1.22 \times 10^9)}{1,382,400} \approx \frac{7.66 \times 10^9}{1,382,400} \approx 5544 \, \text{m/s}
\]

### Step 3: Calculate the centripetal acceleration
Now, using the velocity \( v \) and the radius \( r \), we can find the centripetal acceleration \( a_c \):

\[
a_c = \frac{(5544)^2}{1.22 \times 10^9} \approx \frac{3.07 \times 10^7}{1.22 \times 10^9} \approx 0.0251 \, \text{m/s}^2
\]

Thus, the centripetal acceleration of Titan is approximately:

\[
a_c \approx 0.0251 \, \text{m/s}^2
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the centripetal force relate to centripetal acceleration in orbital motion?
2. What is the relationship between the period of orbit and the velocity in circular motion?
3. How does the mass of Titan affect its orbit around Saturn?
4. How would the centripetal acceleration change if Titan were further from Saturn?
5. What role does gravitational force play in maintaining Titan’s orbit?

**Tip:** In orbital motion, the centripetal acceleration is always directed toward the center of the orbiting body's path, in this case, Saturn.

Titan completes one (circular) orbit of Saturn in 16.0 days. The distance from Saturn to Titan is 1.22×10^9m. What is the centripetal acceleration of Titan?

Discover how to calculate the centripetal acceleration of Titan as it orbits Saturn. This example problem involves key concepts in circular motion and orbital mechanics, using the formulas for orbital velocity and centripetal acceleration. Suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation