Problem 53
Question
(II) Take into account the Earth's rotational speed (1 rev/day) and determine the necessary speed, with respect to Earth, for a rocket to escape if fired from the Earth at the equator in a direction \((a)\) eastward; \((b)\) westward; \((c)\) vertically upward.
Step-by-Step Solution
Verified Answer
For eastward: \(v_{east} = v_{esc} - v_e\); for westward: \(v_{west} = v_{esc} + v_e\); for upward: \(v_{vert} = v_{esc}\).
1Step 1: Calculate Earth's Rotational Speed
The Earth completes one revolution per day, which is equal to 24 hours. The circumference of the Earth can be calculated as \(2\pi R\), where \(R \approx 6400 \text{ km}\). Thus, the rotational speed of the Earth is given by \(v_e = \frac{2\pi \times 6400}{24}\).
2Step 2: Compute Escape Velocity
The escape velocity from Earth's surface is given by the formula \(v_{esc} = \sqrt{\frac{2GM}{R}}\), where \(G\) is the universal gravitational constant and \(M\) is the mass of the Earth. For Earth, this velocity is approximately 11.2 km/s.
3Step 3: Determine Necessary Speed for Eastward Launch
When a rocket is launched eastward, it benefits from the Earth's rotational speed. Thus, the necessary speed is \(v_{east} = v_{esc} - v_e\). Compute this by subtracting Earth's rotational speed from the escape velocity.
4Step 4: Determine Necessary Speed for Westward Launch
For a westward launch, the rocket is launched against Earth's rotation, so the necessary speed is \(v_{west} = v_{esc} + v_e\). Add the Earth's rotational speed to the escape velocity.
5Step 5: Determine Necessary Speed for Vertical Launch
For a launch directly upward, the Earth's rotational speed does not affect the necessary escape speed. Therefore, the required speed remains \(v_{vert} = v_{esc}\).
Key Concepts
Earth's Rotational SpeedEscape Velocity CalculationRocket Launch Directions
Earth's Rotational Speed
The Earth's rotational speed at the equator is crucial to understanding how it impacts rocket launches. We know that the Earth completes one full rotation every 24 hours. This rotation helps define the eastward velocity you'll experience at the equator. To figure out just how quickly the Earth is spinning, we calculate it using the formula for circumference, which is
When launching rockets, this speed is a starting boost if you go eastward, but a hurdle if you head west. For vertical launches, though, it doesn't play a major role as far as the velocity itself is concerned.
- Formula: \[ v_e = \frac{2\pi \times 6400}{24} \]
- Where the radius \( R \approx 6400 \text{ km} \)
When launching rockets, this speed is a starting boost if you go eastward, but a hurdle if you head west. For vertical launches, though, it doesn't play a major role as far as the velocity itself is concerned.
Escape Velocity Calculation
Escape velocity is the minimum speed required for an object to break free from a planet's gravitational pull without further propulsion. For Earth, we use the escape velocity formula:
This speed ensures that a rocket can overcome the gravitational force pulling it back. For different launch directions, this base value of 11.2 km/s adjusts according to Earth's rotational impact.
- Formula: \[ v_{esc} = \sqrt{\frac{2GM}{R}} \]
- \( G \) stands for the gravitational constant, and \( M \) is Earth's mass.
This speed ensures that a rocket can overcome the gravitational force pulling it back. For different launch directions, this base value of 11.2 km/s adjusts according to Earth's rotational impact.
Rocket Launch Directions
When discussing rocket launches, the direction significantly affects the necessary speed to achieve escape velocity.
- Eastward Launch: Since the rocket is launched with Earth's rotation, it benefits from the Earth's rotational speed. Therefore, you reduce the escape velocity by Earth's rotational velocity: \[ v_{east} = v_{esc} - v_e \]
- Westward Launch: Here, the rocket is launched against the rotation, thus requiring more speed: \[ v_{west} = v_{esc} + v_e \]
- Vertical Launch: In this case, the Earth's rotation doesn't add or subtract from the necessary velocity functionally, maintaining the requirement of the escape velocity. The speed remains: \[ v_{vert} = v_{esc} \]
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