📚 Derivation of Escape Velocity Formula (v = √2GM/R) | 逃逸速度公式 (v = √2GM/R) 推导
The concept of escape velocity is fundamental to astrophysics and space exploration. It represents the minimum speed an object must achieve to break free from a celestial body's gravitational pull without any further propulsion. In this article, we will derive the escape velocity formula step by step using the principle of energy conservation, a core topic in Unit 5 of International A Level Physics.
逃逸速度的概念是天体物理学和太空探索的基础。它表示一个物体在不借助额外推进力的情况下，脱离天体引力束缚所需的最小速度。在本文中，我们将利用能量守恒原理，逐步推导逃逸速度公式，这属于国际 A Level 物理第五单元的核心内容。
1. Understanding Gravitational Potential Energy | 理解引力势能
of a planet of mass M, the gravitational potential energy U of the system is given by U = – gravity to separate the masses to infinity, where the potential energy is defined as zero.
当质量为 m 的物体距离质量为 M 的行星中心距离为 r 时，系统的引力势能 U 由 U = –GMm/r 给出，其中 G 是万有引力常数。负号表示为了将物体完全拉离行星引力场、移至无限远处（势能定义为零），必须克服引力做功。
2. Kinetic Energy at the Surface | 表面的动能
, this kinetic energy must be sufficient to overcome the negative gravitational potential energy and reach a final state where both kinetic and potential energy are zero at an infinite distance.
在行星表面（半径为 R），一个以速度 v 发射的物体具有动能 K = ½ mv²。为了逃逸，这个动能必须足以克服负的引力势能，并最终在无限远处达到动能和势能均为零的状态。
3. The Principle of Conservation of Mechanical Energy | 机械能守恒原理
ct at the moment of launch is the sum of its kinetic and potential energies: E = ½ mv² + (– l energy is zero and we want the kinetic energy to be zero as well (minimum escape speed).
物体在发射瞬间的总机械能 E 是其动能与势能之和：E = ½ mv² + (–GMm/R)。为了使物体恰好能够逃逸，我们要求其总能量为零，因为在无限远处势能为零，且我们希望物体刚好到达那里时动能也为零（最小逃逸速度）。
4. Setting Total Energy Equal to Zero | 令总能量等于零
By setting the total mechanical energy to zero, we obtain the equation ½ mv² – ars in both terms, so it can be cancelled, showing that escape velocity does not depend on the mass of the escaping object—a truly elegant result of the equivalence principle.
通过令总机械能等于零，我们得到方程 ½ mv² – GMm/R = 0。物体的质量 m 在两项中都出现，因此可以约去，这表明逃逸速度与逃逸物体的质量无关——这是等效原理的一个非常优雅的结果。
5. Solving for the Escape Velocity v | 求解逃逸速度 v
2GM/R. Taking the square root of both sides gives the celebrated formula for escape speed.
vesc = √2GM/R
重新整理方程：½ mv² = GMm/R → v² = 2GM/R。两边取平方根，就得到著名的逃逸速度公式。
vesc = √2GM/R
6. Interpreting the Formula | 公式解读
arger radii (for a given mass) lower the required speed because gravity weakens with distance. Note that G is a constant, 6.67 × 10⁻¹¹ N m² kg⁻².
逃逸速度仅取决于行星（或恒星）的质量 M 及其半径 R。质量越大的天体逃逸速度越高，而在质量一定时，半径越大所需速度越小，因为引力随距离减弱。注意 G 是常数，6.67 × 10⁻¹¹ N m² kg⁻²。
7. Independence from Launch Direction | 与发射方向无关
Escape velocity is a scalar quantity; as long as the object’s initial kinetic energy meets the threshold, the direction of launch does not matter (ignoring atmospheric drag or obstacles). However, in practice, trajectories that are not purely radial may require slightly different considerations, but the minimum speed remains the same.
逃逸速度是一个标量；只要物体的初始动能达到阈值，发射方向并不重要（忽略大气阻力或障碍物）。然而在实践中，非完全径向的轨道可能需要略微不同的考量，但最小速度保持不变。
8. Worked Example: Earth | 计算示例：地球
For Earth, M = 5.97 × 10²⁴ kg and R = 6.371 × 10⁶ m. Plugging in the numbers: vesc = √2 × 6.67×10⁻¹¹ × 5.97×10²⁴ / 6.371×10⁶ ≈ √1.25 × 10⁸ t 11.2 km s⁻¹. This is the speed a rocket must reach to leave Earth’s gravitational field.
对于地球，M = 5.97 × 10²⁴ kg，R = 6.371 × 10⁶ m。代入数字：vesc = √2 × 6.67×10⁻¹¹ × 5.97×10²⁴ / 6.371×10⁶ ≈ √1.25 × 10⁸ ≈ 1.12 × 10⁴ m s⁻¹，即约 11.2 km s⁻¹。这是火箭脱离地球引力场所需达到的速度。
9. Link to the Schwarzschild Radius | 与史瓦西半径的联系
warzschild radius, a key concept in general relativity that appears in the same PH05 syllabus when discussing stellar evolution and black holes.
如果令逃逸速度等于光速 c，我们可以解出一个使天体成为黑洞的半径：R = 2GM/c²。这就是史瓦西半径，是广义相对论中的一个关键概念，在讨论恒星演化和黑洞的同一 PH05 教学大纲中出现。
10. Common Student Misconceptions | 学生常见误区
Many learners mistakenly think that escape velocity is the speed needed to reach orbit, or that it depends on the mass of the escaping object. It is essential to clarify that orbital speed is vorbit = √GM/r, which is smaller than escape velocity by a factor of √2. Also, the mass m cancels out, so a feather and a hammer would need the same escape speed in a vacuum.
许多学习者误以为逃逸速度是进入轨道所需的速度，或者认为它取决于逃逸物体的质量。需要澄清的是，轨道速度 vorbit = √GM/r，它比逃逸速度小一个 √2 因子。此外，质量 m 被约去，因此在真空中一根羽毛和一把锤子需要相同的逃逸速度。
11. Derivation in the Context of PH05 Syllabus | PH05 大纲中的推导背景
This derivation directly supports learning outcomes in the International A Level Physics Unit 5 (WPH05), where students study gravitational fields, orbital mechanics, and universal gravitation. Being able to reproduce the escape velocity derivation from energy conservation is a typical examination skill tested in questions like those on "PH05-QP-InternationalPhysics-A-21Jun23".
这一推导直接支撑国际 A Level 物理第五单元（WPH05）的学习目标，其中学生要研究引力场、轨道力学和万有引力。能够使用能量守恒再现逃逸速度推导是一项典型的考试技能，经常在类似“PH05-QP-InternationalPhysics-A-21Jun23”的题目中进行测试。
12. Summary and Key Takeaways | 总结与核心要点
quating kinetic energy to the magnitude of gravitational potential energy. The result, v = √2GM/R, highlights the role of a planet’s mass and radius. It is independent of the escaping body’s mass, applies in the absence of non-conservative forces, and paves the way for understanding black hole formation. Mastering this derivation is both a mathematical exercise and a conceptual milestone in physics.
逃逸速度来源于动能与引力势能大小相等的简洁等式。结果 v = √2GM/R 凸显了行星质量和半径的作用。它与逃逸物体质量无关，适用于无非保守力的情况，并为理解黑洞形成铺平了道路。掌握这一推导既是数学训练，也是物理学习中的概念里程碑。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图