Consider an astronaut taking a space walk around the ISS to repair a solar array. If one of the tools is inadvertently pushed away from the ISS and towards Earth at a speed of 12.1 m/s, what is the  in rad/s2 (this indicates that the tool will move - accelerate - away in the  direction as it moves away in the  direction).
Assume that the ISS is in a circular orbit at an altitude of 418 km and moving at a speed of 7.66  km/s. The radius of Earth is 6378 km.

The situation you're describing involves orbital mechanics, and the tool is moving away from the International Space Station (ISS) towards Earth. The ISS is in a circular orbit, and we need to determine the angular acceleration \(\dot{\omega}\) in rad/s² of the tool relative to Earth.

### Step 1: Calculate the Total Distance from the Center of the Earth to the ISS
The ISS is at an altitude of 418 km above the Earth's surface, and the radius of the Earth is given as 6378 km. So, the total distance \(r\) from the center of the Earth to the ISS is:

\[
r = \text{Radius of Earth} + \text{Altitude of ISS}
\]

\[
r = 6378 \, \text{km} + 418 \, \text{km} = 6796 \, \text{km} = 6.796 \times 10^6 \, \text{m}
\]

### Step 2: Initial Orbital Speed of the ISS
The ISS orbits at a speed of 7.66 km/s, which is:

\[
v_{\text{ISS}} = 7.66 \, \text{km/s} = 7660 \, \text{m/s}
\]

### Step 3: Orbital Motion and Gravitational Force
In a stable orbit, the centripetal force required to keep an object in orbit is provided by the gravitational force. The centripetal acceleration \(a_c\) for the ISS can be calculated as:

\[
a_c = \frac{v_{\text{ISS}}^2}{r}
\]

\[
a_c = \frac{(7660)^2}{6.796 \times 10^6} \, \text{m/s}^2
\]

Let's calculate this acceleration:

\[
a_c = \frac{58675600}{6796000} \, \text{m/s}^2
\]

\[
a_c \approx 8.63 \, \text{m/s}^2
\]

### Step 4: Relative Acceleration of the Tool
The tool is moving away from the ISS and is pushed towards Earth at 12.1 m/s. We are interested in the angular acceleration \(\dot{\omega}\), which is related to the linear acceleration \(a\) by:

\[
a = r \cdot \dot{\omega}
\]

Given that the linear acceleration due to gravity is already calculated as approximately \(8.63 \, \text{m/s}^2\), the angular acceleration can be computed as:

\[
\dot{\omega} = \frac{a_c}{r}
\]

\[
\dot{\omega} = \frac{8.63}{6.796 \times 10^6} \, \text{rad/s}^2
\]

Let's compute this:

\[
\dot{\omega} \approx 1.27 \times 10^{-6} \, \text{rad/s}^2
\]

### Final Answer:
The angular acceleration \(\dot{\omega}\) of the tool relative to the Earth is approximately \(1.27 \times 10^{-6} \, \text{rad/s}^2\).

---

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. How would the angular acceleration change if the ISS were at a higher altitude?
2. What forces act on the tool after it is pushed away from the ISS?
3. How can we calculate the time it takes for the tool to re-enter the Earth's atmosphere?
4. What is the role of gravitational potential energy in the tool's movement?
5. How does the escape velocity relate to the orbital speed of the ISS?

**Tip:** In orbital mechanics, small changes in velocity (delta-v) can lead to significant changes in an object's trajectory over time.

Consider an astronaut taking a space walk around the ISS to repair a solar array. If one of the tools is inadvertently pushed away from the ISS and towards Earth at a speed of 12.1 m/s, what is the angular acceleration in rad/s² (this indicates that the tool will move - accelerate - away in the radial direction as it moves away in the tangential direction). Assume that the ISS is in a circular orbit at an altitude of 418 km and moving at a speed of 7.66 km/s. The radius of Earth is 6378 km.

Learn how to calculate the angular acceleration of a tool pushed from the ISS towards Earth at a speed of 12.1 m/s. This problem covers key concepts in orbital mechanics, including angular and centripetal acceleration, and uses Newton's Law of Universal Gravitation. Suitable for advanced high school or college physics students.

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