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Okay, so I need to figure out how to calculate the escape velocity from Earth. Let me start by recalling what escape velocity means. From what I remember, it's the minimum speed an object needs to break free from the gravitational pull of a planet without any further propulsion. Right, so if something reaches that speed, it can escape the gravity without needing to keep burning fuel.
Now, Earth's escape velocity... I think the formula involves the gravitational constant, the mass of the Earth, and the radius of the Earth. Wait, the formula is something like the square root of (2GM/r), where G is the gravitational constant, M is the mass of the Earth, and r is the radius. Let me check if that makes sense. The units should work out. G is in m³ kg⁻¹ s⁻², M is in kg, r is in meters. So multiplying G and M gives m³ s⁻², then dividing by r gives m²/s², and the square root of that is m/s, which is velocity. Yeah, that seems right.
But wait, why is there a factor of 2? Let me think. The potential energy at the surface of the Earth is -GMm/r, where m is the mass of the escaping object. To escape, the object needs enough kinetic energy to overcome that potential energy. So kinetic energy (1/2 mv²) should equal the gravitational potential energy (GMm/r). Setting them equal: 1/2 mv² = GMm/r. The m cancels out, so 1/2 v² = GM/r, which leads to v = sqrt(2GM/r). Ah, that's where the 2 comes from. Makes sense.
Now, plugging in the numbers. Let me look up the values. The gravitational constant G is approximately 6.674×10^-11 m³ kg⁻¹ s⁻². The mass of the Earth M is about 5.972×10^24 kg. The average radius of the Earth r is 6.371×10^6 meters. Let me compute 2GM first. So 2 * 6.674e-11 * 5.972e24. Let me calculate that step by step.
First, 6.674e-11 * 5.972e24. Let's multiply 6.674 by 5.972. 6 * 5.972 is 35.832, 0.674*5.972 is approximately 4.025, so total around 35.832 + 4.025 = 39.857. Then the exponents: 10^-11 * 10^24 = 10^13. So 39.857e13. Then multiply by 2: 79.714e13, which is 7.9714e14.
Now divide that by the radius r = 6.371e6 meters. So 7.9714e14 / 6.371e6. Let's compute that. 7.9714 / 6.371 ≈ 1.251. Then the exponents: 10^14 / 10^6 = 10^8. So 1.251e8. Then take the square root of that. The square root of 1.251e8 is sqrt(1.251) * sqrt(1e8) = approximately 1.118 * 10^4 = 11180 m/s. Wait, that's about 11.2 km/s. I've heard that number before. So the escape velocity from Earth is approximately 11.2 kilometers per second.
But let me double-check the calculation to make sure I didn't make any mistakes. Let's redo the multiplication step. 2 * G * M. G is 6.674e-11, M is 5.972e24. So 6.674e-11 * 5.972e24 = (6.674 * 5.972) * 10^( -11 +24 ) = approx 39.86 * 10^13 = 3.986e14. Then multiply by 2: 7.972e14. Then divide by 6.371e6: 7.972e14 / 6.371e6 ≈ 1.251e8. Square root of 1.251e8 is sqrt(1.251)1e4 ≈ 1.1181e4 = 11180 m/s. Yep, that's 11.18 km/s, which rounds to about 11.2 km/s. Okay, that seems right.
Wait, but why is the radius used here? Is it the distance from the center of the Earth? Yes, because gravitational potential energy depends on the distance from the center. So if you're starting from the surface, the radius is Earth's radius. If you were starting from a higher altitude, you'd use that altitude plus Earth's radius. But the standard escape velocity is from the surface, so we use Earth's radius.
Also, this assumes no air resistance, right? In reality, escaping Earth's gravity through the atmosphere would require a higher speed because of drag, but the theoretical escape velocity doesn't consider that. It's purely based on gravity.
Another thing to consider: escape velocity is a scalar quantity, so direction doesn't matter. As long as the object reaches that speed in any direction, it will escape, provided there's no other forces acting. But in practice, rockets are launched tangentially to take advantage of Earth's rotation, which gives a bit of a speed boost. For example, from the equator, the Earth's rotational speed is about 1670 km/h, which is roughly 463 m/s. Since escape velocity is 11,200 m/s, that's a small fraction, but it helps.
Let me also check if there's a simpler way to remember this. Sometimes escape velocity is given by sqrt(2) times the orbital velocity for a circular orbit at the same radius. The orbital velocity v_orbital is sqrt(GM/r). So escape velocity is sqrt(2) times that. For Earth, low Earth orbital velocity is about 7.8 km/s, so sqrt(2)*7.8 ≈ 11 km/s, which matches. That's a good cross-check.
So putting it all together, the escape velocity from Earth is calculated using the formula v = sqrt(2GM/r), plugging in the known values for G, M, and r, and the result is approximately 11.2 km/s. This is the minimum speed needed to escape Earth's gravitational pull without additional propulsion.
The escape velocity from Earth is the minimum speed required for an object to break free from Earth's gravitational pull without further propulsion. It is calculated using the formula:
[
v = \sqrt{\frac{2GM}{r}}
]
Where:
( G ) is the gravitational constant ((6.674 \times 10^{-11} , \text{m}^3 \text{kg}^{-1} \text{s}^{-2})),
( M ) is Earth's mass ((5.972 \times 10^{24} , \text{kg})),
( r ) is Earth's radius ((6.371 \times 10^6 , \text{m})).
Step-by-Step Calculation:
Compute ( 2GM ):
[
2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24} \approx 7.972 \times 10^{14} , \text{m}^3/\text{s}^2
]
Divide by Earth's radius:
[
\frac{7.972 \times 10^{14}}{6.371 \times 10^6} \approx 1.251 \times 10^8 , \text{m}^2/\text{s}^2
]
Take the square root:
[
\sqrt{1.251 \times 10^8} \approx 11,180 , \text{m/s} , \text{or} , 11.2 , \text{km/s}
]
Key Points:
The factor of 2 arises because the kinetic energy must equal the gravitational potential energy at Earth's surface.
This velocity assumes no atmospheric drag and starts from Earth's surface.
It is approximately (\sqrt{2}) times the orbital velocity for a low Earth orbit ((7.8 , \text{km/s})).
Final Answer:
The escape velocity from Earth is approximately (\boxed{11.2 , \text{km/s}}).
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发表于 2025-11-3 22:01:15