earth satelite moves in circular orbit around earth 915km above surface with period of 103.1 min. what are the speed of and magnitude of the centripetal accelerationof the satellite?

To find the **speed** and **magnitude of the centripetal acceleration** of the satellite, we will apply the following formulas:

1. **Speed of the satellite**:
   \[
   v = \frac{2\pi r}{T}
   \]
   where:
   - \(v\) is the orbital speed,
   - \(r\) is the orbital radius (distance from the center of Earth to the satellite),
   - \(T\) is the orbital period (in seconds).

2. **Centripetal acceleration**:
   \[
   a_c = \frac{v^2}{r}
   \]
   where:
   - \(a_c\) is the centripetal acceleration,
   - \(v\) is the orbital speed (calculated from the previous formula),
   - \(r\) is the orbital radius.

### Step 1: Determine the orbital radius \(r\)
The orbital radius is the distance from the **center of the Earth** to the satellite. This is the sum of the **Earth's radius** and the **altitude of the satellite**.

- Radius of Earth (\(R_E\)) = 6371 km
- Altitude of the satellite = 915 km

Thus, the orbital radius is:
\[
r = R_E + \text{altitude} = 6371\, \text{km} + 915\, \text{km} = 7286\, \text{km} = 7.286 \times 10^6\, \text{m}
\]

### Step 2: Convert the period \(T\) to seconds
Given the period is 103.1 minutes:
\[
T = 103.1 \times 60 = 6186\, \text{seconds}
\]

### Step 3: Calculate the speed of the satellite
Using the formula for speed:
\[
v = \frac{2\pi r}{T} = \frac{2\pi (7.286 \times 10^6)}{6186}
\]

### Step 4: Calculate the centripetal acceleration
Once the speed \(v\) is known, we can calculate the centripetal acceleration using:
\[
a_c = \frac{v^2}{r}
\]

Let me calculate these values for you.The **speed** of the satellite is approximately \(7400.47 \, \text{m/s}\), and the **magnitude of the centripetal acceleration** is approximately \(7.52 \, \text{m/s}^2\).

Would you like further details or clarification on any step?

Here are 5 related questions:
1. How would the speed change if the satellite were at a higher altitude?
2. What would happen to the centripetal acceleration if the orbital period increased?
3. How does the speed of this satellite compare to that of the International Space Station?
4. What role does gravitational force play in maintaining this satellite's orbit?
5. How is orbital velocity related to escape velocity?

**Tip:** Centripetal acceleration always points toward the center of the circular path, keeping an object in orbit.

Earth satellite moves in circular orbit around Earth 915km above surface with period of 103.1 min. What are the speed and magnitude of the centripetal acceleration of the satellite?

This problem explains how to calculate the speed and centripetal acceleration of a satellite orbiting 915 km above Earth's surface with a period of 103.1 minutes. Key concepts include circular motion, orbital mechanics, and centripetal acceleration. Suitable for high school students in Grades 10-12.

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