📚 Gravitational Potential Energy | 引力势能
Gravitational potential energy is one of the central ideas in the CIE A Level Physics syllabus. In a uniform gravitational field, we use ΔGPE = mgΔh, but in a radial field around a planet or star, the energy depends on distance as −GMm/r. This article builds from the definition of potential energy through to escape velocity and orbital energy, giving you the exam-ready toolkit.
引力势能是 CIE A Level 物理大纲的核心概念之一。在均匀引力场中我们使用 ΔGPE = mgΔh，但在行星或恒星周围的径向场中，能量按 −GMm/r 随距离变化。本文从势能定义出发，一直延伸到逃逸速度和轨道能量，为你提供备考所需的完整工具。
1. Defining Gravitational Potential Energy | 引力势能的定义
Gravitational potential energy is the energy stored in a system of two masses due to their gravitational attraction. For two point masses M and m separated by a distance r, the gravitational potential energy U is defined as the work done by an external agent to bring the mass m from infinity to that point without acceleration.
引力势能是由于两个质量之间的引力吸引而储存在系统中的能量。对于相距 r 的两个点质量 M 和 m，引力势能 U 定义为外界将质量 m 从无穷远不加速地移到该点所做的功。
This definition leads to the key equation for a radial field:
这个定义引出了径向场中的关键方程：
U = −GMm / r
where G is the universal gravitational constant, M is the mass of the central body, m is the smaller mass, and r is the separation from the centre of M.
其中 G 是万有引力常量，M 是中心天体的质量，m 是较小质量，r 是到 M 中心的距离。
2. Reference Point at Infinity | 无穷远参考点
In A Level Physics, the zero of gravitational potential energy is chosen at infinity. This is a convenient reference because the gravitational force approaches zero as r becomes very large. As a mass moves closer to a planet, the gravitational field does positive work on it, reducing the system’s potential energy.
在 A Level 物理中，引力势能的零点选在无穷远。这是一个方便的参考点，因为当 r 非常大时引力趋近于零。当质量靠近行星时，引力场对其做正功，使系统的势能减少。
Since the potential energy at infinity is zero and energy is lost when the mass approaches, U must be negative for any finite separation. The negative sign is not a mathematical trick; it shows that the mass is bound to the central body.
由于无穷远处的势能为零，而质量靠近时能量减少，因此在任何有限距离处 U 一定为负值。负号不是数学技巧；它表明该质量被束缚在中心天体上。
3. Gravitational Potential: Energy per Unit Mass | 引力势：每单位质量的能量
Gravitational potential V is the gravitational potential energy per unit mass at a point in a field. It is a scalar quantity and is measured in joules per kilogram, J kg⁻¹.
引力势 V 是场中某一点处每单位质量的引力势能。它是标量，单位为焦耳每千克，J kg⁻¹。
The relation between potential and potential energy is:
引力势与引力势能之间的关系为：
V = U / m = −GM / r
Because V depends only on the source mass M and the distance r, it is very useful for comparing different positions in a radial field without knowing the mass of the test object.
由于 V 仅取决于源质量 M 和距离 r，因此在不知道试验物体质量的情况下，用它来比较径向场中不同位置非常方便。
4. Work Done and Changes in GPE | 做功与引力势能的变化
When a mass is moved from one point to another in a gravitational field, the change in gravitational potential energy equals the work done by an external force, assuming no change in kinetic energy.
当一个质量在引力场中从一点移动到另一点时，引力势能的变化等于外力所做的功，前提是动能不变。
For a radial field, moving a mass m from r₁ to r₂ gives:
对于径向场，将质量 m 从 r₁ 移到 r₂ 可得：
W = ΔU = GMm(1/r₁ − 1/r₂)
If r₂ is greater than r₁, the term in brackets is positive, so an external force must supply energy to separate the masses. If r₂ is smaller than r₁, the work done is negative, meaning the field does work and the mass loses potential energy.
如果 r₂ 大于 r₁，括号内为正，因此外力必须提供能量来分离两者。如果 r₂ 小于 r₁，做功为负，说明引力场做功，质量损失势能。
5. Relation to Gravitational Field Strength | 与引力场强度的关系
Gravitational field strength g is the force per unit mass at a point. In a radial field, its magnitude is given by g = GM/r². It can also be expressed as the negative gradient of the gravitational potential:
引力场强度 g 是某一点处每单位质量所受的力。在径向场中，其大小为 g = GM/r²。它也可以表示为引力势的负梯度：
g = −dV/dr
This relationship is important in exam questions because it links the field strength with the slope of the potential-distance graph.
这个关系在考试中很重要，因为它将场强与势-距离图像的斜率联系起来。
Near the Earth’s surface, g is approximately constant, so the change in GPE over a small vertical height Δh is simply ΔU = mgΔh. However, this formula breaks down for large height changes where g varies significantly.
在地球表面附近，g 近似恒定，因此在小垂直高度 Δh 上引力势能的变化简化为 ΔU = mgΔh。但当高度变化很大、g 显著变化时，该公式不再适用。
6. Potential Wells and Equipotentials | 势阱与等势面
A graph of gravitational potential V against distance r shows a negative potential well. The curve starts at zero at infinity and becomes more negative as r approaches the centre of the mass. This visual representation helps explain why objects naturally fall towards masses: they move towards lower potential.
引力势 V 随距离 r 变化的图像显示一个负势阱。曲线从无穷远处的零开始，随着 r 接近质量中心而变得越来越负。这种可视化表示有助于解释为什么物体自然地落向质量：它们向较低势的方向运动。
Equipotential surfaces in a radial field are spheres centred on the source mass. No work is done when a mass moves along an equipotential surface because there is no change in gravitational potential.
径向场中的等势面是以源质量为中心的球面。当质量沿等势面移动时不做功，因为引力势没有变化。
Field lines always point radially inward and are perpendicular to equipotential surfaces. The closer the equipotential surfaces are to each other, the stronger the field.
场线始终沿径向向内，并与等势面垂直。等势面越密集，场越强。
7. Conservation of Energy in Orbits | 轨道中的能量守恒
For a satellite of mass m in a circular orbit of radius r around a planet of mass M, the centripetal force is provided by gravity. From this, the orbital speed squared is obtained as v² = GM/r.
对于在半径为 r 的圆形轨道上绕质量 M 的行星运行的质量为 m 的卫星，向心力由引力提供。由此得到轨道速度的平方为 v² = GM/r。
The satellite has both gravitational potential energy and kinetic energy:
卫星同时具有引力势能和动能：
E = U + K = −GMm/r + ½mv²
Substituting v² = GM/r gives the total orbital energy for a circular orbit:
代入 v² = GM/r，得到圆形轨道的总轨道能量：
E = −GMm / (2r)
This shows that higher orbits have less negative total energy, meaning more energy must be supplied to raise a satellite from a lower orbit to a higher one.
这表明较高轨道的总能量负值更小，意味着将卫星从较低轨道提升到较高轨道必须提供更多能量。
8. Escape Velocity from GPE | 由引力势能求逃逸速度
Escape velocity is the minimum launch speed required for an object to escape a planet’s gravitational field and reach infinity with zero final kinetic energy. At the planet’s surface, the object has potential energy U = −GMm/r and initial kinetic energy K = ½mv².
逃逸速度是物体脱离行星引力场并以零最终动能到达无穷远所需的最小发射速度。在行星表面，物体具有势能 U = −GMm/r 和初始动能 K = ½mv²。
By conservation of energy, the total energy at launch must be at least zero, since at infinity both potential and kinetic energy become zero:
根据能量守恒，发射时的总能量必须至少为零，因为在无穷远处势能和动能都为零：
½mv² − GMm/r = 0
Solving for v gives the escape velocity:
解出 v 可得逃逸速度：
v = √(2GM / r)
Notice that escape velocity does not depend on the mass of the escaping object, only on the mass and radius of the planet.
注意逃逸速度不依赖于逃逸物体的质量，只取决于行星的质量和半径。
9. Comparing Uniform and Radial Fields | 均匀场与径向场的比较
Many exam questions ask you to compare gravitational potential energy in a uniform field with that in a radial field. In a uniform field, GPE increases linearly with height because ΔU = mgΔh. In a radial field, GPE varies as −1/r, so it rises more slowly as r increases and approaches zero asymptotically.
许多考题要求你比较均匀场和径向场中的引力势能。在均匀场中，由于 ΔU = mgΔh，引力势能随高度线性增加。在径向场中，引力势能按 −1/r 变化，因此随着 r 增大它上升得更慢，并渐近地趋近于零。
| Field type 场类型 | GPE expression 表达式 | Zero reference 零参考点 |
| Uniform 均匀场 | ΔU = mgΔh | Arbitrary chosen level 任选水平面 |
| Radial 径向场 | U = −GMm/r | Infinity 无穷远 |
Always check which field model applies before choosing a formula. Use mgΔh only for small height changes near a planet’s surface.
在选择题型公式之前，务必先判断适用哪种场模型。仅在行星表面附近的小高度变化时使用 mgΔh。
10. Worked Examples | 例题解析
Example 1: Calculate the gravitational potential energy of a 500 kg satellite at a distance of 2.0 × 10⁷ m from the centre of the Earth. Take G = 6.67 × 10⁻¹¹ N m² kg⁻² and M = 5.97 × 10²⁴ kg.
例题 1：计算质量为 500 kg 的卫星在距地心 2.0 × 10⁷ m 处的引力势能。取 G = 6.67 × 10⁻¹¹ N m² kg⁻²，M = 5.97 × 10²⁴ kg。
Using U = −GMm/r:
使用 U = −GMm/r：
U = −(6.67 × 10⁻¹¹)(5.97 × 10²⁴)(500) / (2.0 × 10⁷) = −9.95 × 10⁹ J
Example 2: Find the escape velocity from the surface of the Earth, given the Earth’s radius r = 6.37 × 10⁶ m.
例题 2：求地球表面的逃逸速度，已知地球半径 r = 6.37 × 10⁶ m。
Using v = √(2GM / r):
使用 v = √(2GM / r)：
v = √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.37 × 10⁶) ≈ 1.12 × 10⁴ m s⁻¹
This is about 11.2 km s⁻¹, which matches the accepted value for Earth’s escape velocity.
这约为 11.2 km s⁻¹，与公认的地球逃逸速度一致。
11. Common Misconceptions and Exam Tips | 常见误区与备考建议
A common error is to write gravitational potential energy as a positive quantity. In the CIE A Level convention, U is zero at infinity and negative everywhere else, so forgetting the minus sign will lead to incorrect answers, especially in calculations involving work or escape velocity.
一个常见错误是把引力势能写成正值。在 CIE A Level 的约定中，U 在无穷远处为零，在其他地方均为负值，因此忘记负号会导致错误答案，尤其是在涉及功或逃逸速度的计算中。
Another misconception is confusing gravitational potential V with gravitational potential energy U. Remember that V is energy per unit mass and does not include the mass m of the test object. Multiplying V by m gives U.
另一个误区是混淆引力势 V 和引力势能 U。记住 V 是每单位质量的能量，不包含试验物体的质量 m。用 V 乘以 m 才得到 U。
When a question asks for the work done to move a satellite between two orbits, use the difference in total energy E = −GMm/(2r) for circular orbits, not just the change in GPE. This accounts for the simultaneous change in kinetic energy.
当题目要求将卫星在两个轨道之间移动所做的功时，应使用圆形轨道总能量 E = −GMm/(2r) 的差值，而不仅仅是引力势能的变化。这样才能同时考虑动能的变化。
Finally, always state the zero reference point when using gravitational potential energy, and ensure that all distances are measured from the centre of the planet, not from its surface.
最后，在使用引力势能时始终说明零参考点，并确保所有距离都从行星中心量起，而不是从表面量起。
12. Summary of Key Formulas | 关键公式总结
The following table summarises the essential equations for CIE A Level Physics questions on gravitational potential energy.
下表总结了 CIE A Level 物理中关于引力势能的核心方程。
| Quantity 物理量 | Formula 公式 | Units 单位 |
| GPE in radial field 径向场中的引力势能 | U = −GMm/r | J |
| Gravitational potential 引力势 | V = −GM/r | J kg⁻¹ |
| Change in GPE 引力势能变化 | ΔU = GMm(1/r₁ − 1/r₂) | J |
| Uniform field GPE 均匀场引力势能 | ΔU = mgΔh | J |
| Orbital total energy 轨道总能量 | E = −GMm/(2r) | J |
| Escape velocity 逃逸速度 | v = √(2GM/r) | m s⁻¹ |
Keep these formulas and their sign conventions clear, and you will be well prepared for gravitational potential energy questions in the CIE A Level Physics examination.
清楚地掌握这些公式及其符号约定，你就能很好地应对 CIE A Level 物理考试中关于引力势能的问题。
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