📚 A-Level Physics: Circular Motion of the Earth and Orbital Characteristics | A-Level 物理：地球环绕运动与轨道特征
When we observe the Moon, satellites, and planets moving around the Earth or the Sun, we are witnessing one of the most elegant applications of mechanics: orbital motion. At A-Level, you are expected to understand the physics of objects moving in circular paths under the influence of gravity, and to be able to analyse orbital characteristics quantitatively. This article consolidates the essential theory, equations, and exam-relevant insights for CIE A-Level Physics.
当我们观察月球、人造卫星以及行星绕地球或太阳运动时，我们目睹的是力学中最优雅的应用之一：轨道运动。在 A-Level 阶段，你应当理解物体在引力作用下沿圆周路径运动的物理规律，并能定量分析轨道特征。本文系统梳理 CIE A-Level 物理的核心理论、方程和考试要点。
1. Uniform Circular Motion: Key Quantities | 匀速圆周运动的基本量
An object moving in a circle at constant speed is said to be in uniform circular motion. Although the speed is constant, the velocity is not constant because the direction changes continuously. Two fundamental quantities describe this motion: angular displacement θ measured in radians, and angular velocity ω related to the linear speed v by v = rω.
物体以恒定速率沿圆周运动称为匀速圆周运动。虽然速率恒定，但由于方向不断改变，速度矢量并不恒定。描述这种运动的两个基本量为：以弧度（rad）为单位的角位移 θ，以及角速度 ω；线速度 v 与角速度的关系为 v = rω。
The angular velocity can also be expressed in terms of the period T (time for one complete revolution) and the frequency f (number of revolutions per second):
角速度也可以用周期 T（完成一整圈所需时间）和频率 f（每秒转过的圈数）来表示：
ω = 2π / T = 2πf
Here ω is measured in radians per second (rad s⁻¹), T in seconds, and f in hertz (Hz). A full circle corresponds to an angle of 2π radians, so a quarter circle is π/2 rad.
这里 ω 的单位为弧度每秒（rad s⁻¹），T 的单位为秒，f 的单位为赫兹（Hz）。一整圈对应 2π 弧度，因此四分之一圈为 π/2 弧度。
2. Centripetal Acceleration and Force | 向心加速度与向心力
In uniform circular motion, the acceleration is directed towards the centre of the circle. This is called centripetal acceleration. Its magnitude can be derived from the geometry of the velocity vectors and is given by:
在匀速圆周运动中，加速度始终指向圆心，这称为向心加速度。其大小可由速度矢量的几何关系推导得出：
a = v² / r = rω²
Since a resultant force is required to produce an acceleration, the net force on the object must also point towards the centre. This is the centripetal force:
由于产生加速度需要合力作用，物体所受的合外力也必须指向圆心，这就是向心力：
F = ma = m v² / r = m rω²
It is important to recognise that “centripetal force” is not a new kind of force; it is the name given to the resultant force that causes circular motion. In an orbital context, that resultant force is provided by gravitational attraction.
必须认识到，“向心力”并不是一种新的力，而是使物体做圆周运动的合外力的一种称谓。在轨道的场景中，这个合外力由万有引力提供。
3. Newton’s Law of Gravitation | 牛顿万有引力定律
Newton’s law of gravitation states that any two point masses attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of their separation:
牛顿万有引力定律指出：任何两个质点之间都存在相互吸引的力，该力与两者质量的乘积成正比，与它们之间距离的平方成反比：
F = G M m / r²
Here G is the universal gravitational constant, G = 6.67 × 10⁻¹¹ N m² kg⁻²; M and m are the two masses; and r is the distance between their centres. The gravitational force is always attractive, and it acts along the line joining the two masses.
其中 G 为万有引力常量，G = 6.67 × 10⁻¹¹ N m² kg⁻²；M 和 m 为两个物体的质量；r 为两者中心之间的距离。万有引力始终是吸引力，作用方向沿两物体中心的连线。
For a satellite of mass m orbiting a much larger body of mass M (such as the Earth), we can take r to be the distance from the centre of the Earth to the satellite. This approximation is excellent because the Earth’s radius is much greater than the height of most satellites above the surface, and the Earth’s mass can be treated as concentrated at its centre.
对于绕质量远大于自身的天体 M（如地球）运行的质量为 m 的卫星，可将 r 视为地球中心到卫星的距离。这一近似非常精确，因为地球半径远大于大多数卫星离地面的高度，且可认为地球质量集中于其中心。
4. From Gravity to Orbital Motion | 从万有引力到轨道运动
For an object orbiting the Earth in a circular path of radius r, the gravitational force provides the required centripetal force. Equating the two gives:
对于沿半径为 r 的圆轨道绕地球运行的物体，万有引力恰好提供所需的向心力。令两者相等可得：
G M m / r² = m v² / r
Notice that the mass m of the orbiting body cancels out immediately. This shows that the orbital speed does not depend on the mass of the satellite; it depends only on the mass of the central body and the orbital radius.
注意到轨道物体的质量 m 会立即消去。这表明轨道速度与卫星本身的质量无关，而只取决于中心天体的质量和轨道半径。
Rearranging the equation gives an expression for the orbital speed:
整理该方程可得轨道速度的表达式：
v = √(G M / r)
This is a key result. It shows that the closer a satellite is to the Earth, the faster it must travel to remain in a stable circular orbit. Conversely, satellites at greater altitudes move more slowly.
这是一个关键结论。它表明卫星离地球越近，要维持稳定的圆轨道就必须运动得越快。反之，位于更高高度的卫星运动得更慢。
5. Orbital Period and Kepler’s Third Law | 轨道周期与开普勒第三定律
The orbital period T is the time taken for one complete revolution. Since the circumference of the orbit is 2πr and the speed is v, we have:
轨道周期 T 是完成一整圈所需的时间。由于轨道周长为 2πr，速度为 v，因此有：
T = 2πr / v = 2πr / √(G M / r)
Squaring both sides and simplifying gives a very useful relationship:
两边平方并化简，得出一个非常有用的关系式：
T² = (4π² / G M) r³
This is the mathematical form of Kepler’s third law for circular orbits: the square of the orbital period is proportional to the cube of the orbital radius. The constant of proportionality depends only on the mass of the central body.
这就是圆轨道下开普勒第三定律的数学形式：轨道周期的平方与轨道半径的立方成正比。比例常数仅取决于中心天体的质量。
For the Earth, M = 5.97 × 10²⁴ kg. Hence the constant 4π² / G M has a numerical value of approximately 9.9 × 10⁻¹⁴ s² m⁻³. This means that if you know the period of any Earth satellite, you can calculate its orbital radius, and vice versa.
对于地球，M = 5.97 × 10²⁴ kg。因此常数 4π² / G M 的数值约为 9.9 × 10⁻¹⁴ s² m⁻³。这意味着如果你知道任何地球卫星的周期，就能计算其轨道半径，反之亦然。
6. Energy of an Orbiting Body | 轨道物体的能量
A satellite in a circular orbit possesses both kinetic energy and gravitational potential energy. The gravitational potential energy of a mass m at a distance r from the centre of the Earth is:
在圆轨道上运行的卫星同时具有动能和引力势能。质量为 m 的物体在距地球中心 r 处的引力势能为：
Eₚ = − G M m / r
The negative sign indicates that the gravitational potential energy is zero at infinite separation and decreases (becomes more negative) as the object approaches the Earth. The kinetic energy of the satellite is found from its orbital speed:
负号表示引力势能在无穷远处为零，并且当物体靠近地球时势能减小（变得更负）。卫星的动能可由其轨道速度求出：
Eₖ = ½ m v² = G M m / (2r)
Therefore the total mechanical energy of the satellite is:
因此卫星的总机械能为：
E = Eₖ + Eₚ = G M m / (2r) − G M m / r = − G M m / (2r)
Notice that the total energy is negative. This indicates that the satellite is bound to the Earth; it cannot escape unless additional energy is supplied. Furthermore, the kinetic energy is exactly half the magnitude of the potential energy. This is a special property of circular orbits.
注意总能量为负值。这表明卫星被地球束缚，除非额外提供能量，否则无法逃逸。此外，动能恰好等于势能大小的一半。这是圆轨道的一个特殊性质。
7. Geostationary Orbits | 地球同步轨道
A geostationary satellite is one that remains at a fixed point directly above the equator. It has an orbital period exactly equal to the rotational period of the Earth about its own axis, which is approximately 24 hours. In addition, its orbit must be circular, lie in the equatorial plane, and the satellite must travel in the same direction as the Earth’s rotation (west to east).
地球同步卫星是指在赤道正上方某一固定点保持不动的卫星。其轨道周期恰好等于地球自转周期，约为 24 小时。此外，其轨道必须是圆形的、位于赤道平面内，并且卫星必须与地球自转同向（自西向东）运动。
Using T = 24 h = 86 400 s in Kepler’s third law, the orbital radius can be calculated:
将 T = 24 h = 86 400 s 代入开普勒第三定律，可计算出轨道半径：
r³ = G M T² / (4π²)
This gives r ≈ 4.23 × 10⁷ m. Subtracting the Earth’s radius Rₑ ≈ 6.38 × 10⁶ m gives an altitude of about 3.59 × 10⁷ m, or roughly 36 000 km above the Earth’s surface.
由此得到 r ≈ 4.23 × 10⁷ m。减去地球半径 Rₑ ≈ 6.38 × 10⁶ m，得到卫星距地面高度约为 3.59 × 10⁷ m，即大约 36 000 km。
Geostationary satellites are widely used for communications, weather monitoring, and broadcasting because they appear stationery relative to the ground, allowing fixed antennas to maintain a continuous link without tracking the satellite.
地球同步卫星广泛应用于通信、气象监测和广播电视，因为它们相对于地面保持静止，固定天线无需追踪即可持续保持连接。
8. Apparent Weight and Weightlessness | 视重与失重
Astronauts inside an orbiting spacecraft appear to float. This is often described as “weightlessness”, but it is more accurate to say that they experience zero apparent weight. The gravitational force still acts on them; if it did not, they would not remain in orbit. What happens is that the spacecraft and everything inside it are falling towards the Earth with the same acceleration due to gravity, so the normal reaction force between the astronauts and the spacecraft is zero.
轨道飞行器内的宇航员看起来像是漂浮着。这常被称为“失重”，但更准确的说法是他们感受到的视重为零。引力仍然作用在他们身上；如果没有引力，他们就不会保持在轨道上。实际情况是：航天器和其中的一切都在以相同的重力加速度向地球下落，因此宇航员与飞船之间的支持力为零。
The centripetal acceleration required for a satellite at radius r is exactly g(r) = G M / r². Because both the satellite and the astronaut have the same acceleration, no contact force is needed to keep them moving together. The astronaut feels weightless, even though their true weight (the gravitational force) is not zero.
在半径 r 处，卫星所需的向心加速度恰好为 g(r) = G M / r²。由于卫星和宇航员具有相同的加速度，不需要接触力就能保持它们一起运动。因此宇航员感觉不到重量，尽管其真实重量（万有引力）并不为零。
This principle is frequently tested in exams. A common misconception is that gravity is absent in orbit. In fact, at the altitude of the International Space Station (about 400 km), the gravitational acceleration is still approximately 8.7 m s⁻², only slightly less than the surface value of 9.81 m s⁻².
这一原理在考试中经常出现。一个常见的误解是认为轨道上不存在引力。事实上，在国际空间站的高度（约 400 km），重力加速度仍约为 8.7 m s⁻²，仅略小于地面值 9.81 m s⁻²。
9. Escape Velocity | 逃逸速度
Escape velocity is the minimum speed an object must have at a given distance from the centre of the Earth to just be able to escape to infinity with zero residual speed. At escape speed, the total mechanical energy of the object becomes zero:
逃逸速度是物体在距地球中心一定距离处，恰好能够逃逸到无穷远且最终速度为零所需的最小速度。在逃逸速度下，物体的总机械能为零：
½ m vₑₛ꜀² − G M m / r = 0
Solving for the escape speed gives:
解出逃逸速度为：
vₑₛ꜀ = √(2 G M / r) = √2 × vₒᵣᵦ
where vₒᵣᵦ is the circular orbital speed at the same radius. For an object launched from the Earth’s surface, vₑₛ꜀ ≈ 11.2 km s⁻¹, whereas the circular orbital speed near the surface is about 7.9 km s⁻¹.
其中 vₒᵣᵦ 是同一半径下的圆轨道速度。对于从地球表面发射的物体，vₑₛ꜀ ≈ 11.2 km s⁻¹，而近地圆轨道速度约为 7.9 km s⁻¹。
Escape velocity does not depend on the direction of projection, as long as the trajectory does not collide with the planet. It also does not depend on the mass of the object. This is because the kinetic energy and gravitational potential energy both scale with the mass m, so m cancels.
逃逸速度与发射方向无关，只要轨迹不与行星相撞即可。它也与物体质量无关。这是因为动能和引力势能都与质量 m 成正比，所以 m 被消去了。
10. Common Exam Pitfalls and Tips | 常见考试陷阱与技巧
Many students lose marks by confusing the gravitational force with weight at the surface, or by forgetting that the radius r in Newton’s law must be measured from the centre of the Earth, not from the surface. Always add the Earth’s radius to the altitude when calculating orbital parameters.
许多学生因为混淆万有引力与地面重力，或者忘记牛顿定律中的 r 必须从地球中心量起而不是从地面量起而丢分。在计算轨道参数时，务必使用地球半径加上高度。
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UseSI units throughout. Convert kilometres to metres, hours to seconds, and days to seconds before substituting into equations.
全程使用国际单位。代入方程前，将千米换算为米、小时换算为秒、天换算为秒。
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Remember that v = rω only applies when ω is in radians per second. Degrees per second will give incorrect results.
记住 v = rω 仅在 ω 以弧度每秒为单位时成立。使用度每秒会得到错误结果。
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Do not say “centrifugal force” when describing circular motion in an inertial frame. The only real force towards the centre is the centripetal force; the apparent outward push is a pseudo-force experienced in a rotating reference frame.
在惯性系中描述圆周运动时不要使用“离心力”一词。真正的指向圆心的力是向心力；向外推的感觉是转动参考系中感受到的假想力。
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Check the direction of the acceleration. In circular motion, acceleration is always towards the centre, never along the tangent.
注意加速度的方向。在圆周运动中，加速度始终指向圆心，绝不沿切线方向。
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When comparing two orbits, use Kepler’s third law. If one satellite has a period 8 times larger, its orbital radius is 2³ = 8 times smaller? No — because T² ∝ r³, a period 8 times larger means r is 8^(2/3) = 4 times larger.
比较两段轨道时，使用开普勒第三定律。如果一颗卫星的周期是另一颗的 8 倍，那么 T² ∝ r³ 意味着轨道半径是后者的 8^(2/3) = 4 倍，而不是 8 倍。
11. Worked Example: Calculating a Satellite’s Speed and Period | 例题：计算卫星的速度与周期
A weather satellite orbits the Earth at an altitude of 500 km. Given that Mₑ = 5.97 × 10²⁴ kg, Rₑ = 6.38 × 10⁶ m, and G = 6.67 × 10⁻¹¹ N m² kg⁻², determine (a) the orbital speed, and (b) the orbital period.
一颗气象卫星在地球上方 500 km 的高度运行。已知 Mₑ = 5.97 × 10²⁴ kg，Rₑ = 6.38 × 10⁶ m，G = 6.67 × 10⁻¹¹ N m² kg⁻²。求 (a) 轨道速度和 (b) 轨道周期。
(a) The orbital radius is r = 6.38 × 10⁶ + 500 × 10³ = 6.88 × 10⁶ m. Using v = √(G M / r):
(a) 轨道半径为 r = 6.38 × 10⁶ + 500 × 10³ = 6.88 × 10⁶ m。利用 v = √(G M / r)：
v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.88 × 10⁶)] ≈ 7.61 × 10³ m s⁻¹
(b) The period is T = 2πr / v:
(b) 周期为 T = 2πr / v：
T = 2π × 6.88 × 10⁶ / (7.61 × 10³) ≈ 5.68 × 10³ s ≈ 94.7 min
This is a typical low-Earth-orbit period. The satellite completes roughly 15 orbits per day. Notice how quickly the orbit is completed compared with a geostationary satellite’s 24-hour period.
这是典型的近地轨道周期。该卫星每天大约绕地球 15 圈。注意它比地球同步卫星的 24 小时周期快得多。
12. Summary of Key Equations | 关键公式总结
| Quantity | Equation | 备注 |
| Angular velocity | ω = 2π / T = 2πf | 单位：rad s⁻¹ |
| Linear speed | v = rω | 仅适用于 ω 以 rad s⁻¹ 制 |
| Centripetal acceleration | a = v² / r = rω² | 方向始终指向圆心 |
| Centripetal force | F = m v² / r = m rω² | 由实际力（如引力）提供 |
| Newton’s law of gravitation | F = G M m / r² | G = 6.67 × 10⁻¹¹ N m² kg⁻² |
| Orbital speed | v = √(G M / r) | 与卫星质量无关 |
| Kepler’s third law | T² = (4π² / G M) r³ | 适用于圆轨道 |
| Gravitational potential energy | Eₚ = − G M m / r | 取无穷远处为零 |
| Total energy in circular orbit | E = − G M m / (2r) | Eₖ = − ½ Eₚ |
| Escape speed | vₑₛ꜀ = √(2 G M / r) | 约为圆轨道速度的 √2 倍 |
Master these equations and understand their physical meaning, and you will be well prepared for any CIE A-Level question on circular motion and orbital mechanics. Pay special attention to the underlying assumptions, such as circular orbits and point masses, and always check your units.
掌握这些公式并理解其物理意义，你就能从容应对 CIE A-Level 中关于圆周运动与轨道力学的任何问题。请特别注意隐含假设（如圆轨道、质点模型），并始终检查单位。
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