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Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from Earth’s surface is R/α. Find the value of α.
Correct answer is '3'. Can you explain this answer?
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Maximum height reached by a rocket fired with a speed equal to 50% of ...
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The correct answer is: 3
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To calculate the maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from Earth, we need to know the formula for escape velocity and the height equation for projectile motion.
The escape velocity from Earth can be calculated using the formula:
V_escape = sqrt((2 * G * M) / R)
Where:
- V_escape is the escape velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- R is the radius of the Earth (approximately 6,371 km)
Substituting the known values:
V_escape = sqrt((2 * 6.67430 × 10^-11 * 5.972 × 10^24) / (6,371 × 10^3))
V_escape ≈ 11,186 m/s
To calculate the maximum height reached by the rocket fired with a speed equal to 50% of the escape velocity, we need to use the height equation for projectile motion:
h = (v^2 * sin^2θ) / (2g)
Where:
- h is the maximum height
- v is the initial velocity (50% of the escape velocity, so v = 0.5 * 11,186 m/s = 5,593 m/s)
- θ is the launch angle (we assume it to be 90° for maximum height)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the known values:
h = (5,593^2 * sin^2(90°)) / (2 * 9.8)
h ≈ (5,593^2 * 1) / 19.6
h ≈ 1,560,506 m
Therefore, the maximum height reached by the rocket is approximately 1,560,506 meters or 1,560.5 kilometers.
The escape velocity from Earth can be calculated using the formula:
V_escape = sqrt((2 * G * M) / R)
Where:
- V_escape is the escape velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- R is the radius of the Earth (approximately 6,371 km)
Substituting the known values:
V_escape = sqrt((2 * 6.67430 × 10^-11 * 5.972 × 10^24) / (6,371 × 10^3))
V_escape ≈ 11,186 m/s
To calculate the maximum height reached by the rocket fired with a speed equal to 50% of the escape velocity, we need to use the height equation for projectile motion:
h = (v^2 * sin^2θ) / (2g)
Where:
- h is the maximum height
- v is the initial velocity (50% of the escape velocity, so v = 0.5 * 11,186 m/s = 5,593 m/s)
- θ is the launch angle (we assume it to be 90° for maximum height)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the known values:
h = (5,593^2 * sin^2(90°)) / (2 * 9.8)
h ≈ (5,593^2 * 1) / 19.6
h ≈ 1,560,506 m
Therefore, the maximum height reached by the rocket is approximately 1,560,506 meters or 1,560.5 kilometers.
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Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from Earth’s surface isR/α.Find the value ofα.Correct answer is '3'. Can you explain this answer?
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