📚 The Orbital Period | 轨道周期
In A-Level Physics, the orbital period is the time taken for a satellite, planet, or other body to complete one full revolution around a central mass. For circular orbits, it can be derived directly from Newton's law of gravitation and centripetal force. This topic links circular motion, gravitational fields, and Kepler's laws, and it appears frequently in CIE structured questions.
在 A-Level 物理中，轨道周期是卫星、行星或其他天体绕中心质量完成一整圈公转所需的时间。对于圆轨道，它可以直接由牛顿万有引力定律和向心力推导得出。该主题把圆周运动、引力场和开普勒定律联系起来，在 CIE 结构化试题中经常出现。
1. What is the Orbital Period? | 什么是轨道周期？
The orbital period T is the time for one complete orbit. In SI units it is measured in seconds (s). For planets and satellites, T can range from about 90 minutes for a low Earth orbit to 27.3 days for the Moon, and about 365.25 days for Earth around the Sun.
轨道周期 T 是完成一次完整轨道运动所需的时间。在国际单位制中，它以秒（s）为单位。对于行星和卫星，T 的范围可以是从近地轨道的大约 90 分钟，到月球的 27.3 天，再到地球绕太阳公转的约 365.25 天。
T = 2π / ω. This basic relationship is often a useful starting point in orbital problems.
在圆周运动中，一整圈对应 2π 弧度的角位移。如果角速度为 ω，则 T = 2π / ω。这个基本关系通常是解决轨道问题时的有用起点。
2. Forces in a Circular Orbit | 圆轨道中的力
For a body of mass m moving in a circular orbit of radius r around a central mass M, the gravitational force provides the centripetal force. This condition allows us to link gravitational and circular motion equations.
对于质量为 m 的天体，在半径为 r 的圆轨道上绕中心质量 M 运动时，万有引力提供向心力。这一条件使我们能够把引力方程和圆周运动方程联系起来。
F = GMm / r² = mv² / r
Here G is the gravitational constant, v is the orbital speed, and m cancels from both sides of the equation. This cancellation is important because it shows that the orbital motion of a small body does not depend on its own mass.
这里 G 是引力常量，v 是轨道速度，m 在等式两边可以约去。这一约分非常重要，因为它表明小天体的轨道运动与其自身质量无关。
3. Deriving Orbital Speed | 推导轨道速度
o obtain v² = GM / r. Taking the square root gives the orbital speed for a circular orbit.
从力的方程出发，约去 m 并乘以 r，得到 v² = GM / r。开平方后即可得到圆轨道的轨道速度。
v = √GM / r
This equation shows that the orbital speed decreases as the orbital radius increases. A satellite in a low orbit moves faster than one in a high orbit. The speed does not depend on the satellite mass m.
这个方程表明，轨道速度随轨道半径增大而减小。低轨道卫星运动得比高轨道卫星更快。速度与卫星质量 m 无关。
4. Deriving the Orbital Period | 推导轨道周期
ravel one circumference, 2πr, at constant speed v. Therefore T = 2πr / v. Substituting v = √GM / r gives the period formula.
周期是以恒定速度 v 运动一个圆周 2πr 所需的时间。因此 T = 2πr / v。代入 v = √GM / r 即可得到周期公式。
T = 2π√r³ / GM
t useful form for solving numerical problems because it avoids square roots until the final step.
或者，将两边平方得到 T² = 4π²r³ / (GM)。这是解决数值计算问题最有用的形式，因为它可以在最后一步之前避免开平方。
5. Kepler's Third Law | 开普勒第三定律
For circular orbits, the period equation can be written as T² ∝ r³. This is a special case of Kepler's third law. The constant of proportionality depends only on the central mass M.
对于圆轨道，周期方程可以写成 T² ∝ r³。这是开普勒第三定律的一个特例。比例常数只取决于中心质量 M。
T² / r³ = 4π² / (GM)
All satellites orbiting the same central body have the same value of T² / r³. This allows you to compare two orbits without knowing G or M, provided the central mass is the same.
所有绕同一中心天体运行的卫星都有相同的 T² / r³ 值。这使你在中心质量相同的前提下，无需知道 G 或 M 就能比较两个轨道。
6. Worked Example: Low Earth Orbit | 例题：近地轨道
ove the Earth. Take the Earth mass as 5.97 × 10²⁴ kg and the Earth radius as 6.37 × 10⁶ m.
计算一颗距地面 300 km 高度处卫星的轨道周期。取地球质量为 5.97 × 10²⁴ kg，地球半径为 6.37 × 10⁶ m。
First find the orbital radius: r = 6.37 × 10⁶ + 3.00 × 10⁵ = 6.67 × 10⁶ m. Then use T = 2π√r³ / GM.
先求轨道半径：r = 6.37 × 10⁶ + 3.00 × 10⁵ = 6.67 × 10⁶ m。然后使用 T = 2π√r³ / GM。
T = 2π√[(6.67 × 10⁶)³ / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)] ≈ 5.4 × 10³ s
This is about 90 minutes, which is typical for a low Earth orbit. Notice that the radius must include the Earth radius plus the altitude, not just the altitude.
这大约是 90 分钟，是近地轨道的典型周期。注意轨道半径必须包括地球半径加高度，而不仅仅是高度。
7. Geostationary Satellites | 地球同步卫星
A geostationary satellite has an orbital period of exactly 24 hours and remains above the same point on the equator. For this to happen, the satellite must orbit in the equatorial plane and in the same direction as the Earth rotates.
地球同步卫星的轨道周期正好是 24 小时，并保持在赤道同一点的上方。要实现这一点，卫星必须在赤道平面内运行，并且与地球自转方向相同。
× 10⁷ m. Subtracting the Earth radius gives a height of about 3.58 × 10⁷ m, or 35,800 km.
将 T = 24 × 3600 = 86400 s 代入 T² = 4π²r³ / GM 并求解 r。得到轨道半径约为 4.22 × 10⁷ m。减去地球半径后得到高度约为 3.58 × 10⁷ m，即 35,800 km。
Geostationary satellites are used for communications and weather monitoring because their position in the sky appears fixed to an observer on the ground.
地球同步卫星用于通信和气象监测，因为对地面观察者而言，它们在天空中的位置看上去是固定的。
8. Energy in Circular Orbits | 圆轨道中的能量
For a circular orbit, the kinetic energy is positive and the gravitational potential energy is negative. The total mechanical energy is half the potential energy and is also negative.
对于圆轨道，动能为正，引力势能为负。总机械能是势能的一半，也是负值。
Ek = GMm / (2r) , Ep = -GMm / r , Etotal = -GMm / (2r)
As the orbital radius r increases, the kinetic energy decreases, the potential energy becomes less negative, and the total energy increases. A satellite in a higher orbit has a longer period and a greater total energy, even though its speed is lower.
随着轨道半径 r 增大，动能减小，势能负值变小，总能量增大。高轨道卫星周期更长、总能量更大，尽管它的速度更低。
9. Elliptical Orbits and Kepler's Laws | 椭圆轨道与开普勒定律
Real planetary orbits are elliptical rather than circular. Kepler's first law states that a planet orbits the Sun in an ellipse with the Sun at one focus. The period formula still works if the radius r is replaced by the semi-major axis a.
实际的行星轨道是椭圆形而不是正圆形。开普勒第一定律指出，行星以椭圆轨道绕太阳运行，太阳位于一个焦点上。如果将半径 r 换成半长轴 a，周期公式仍然适用。
T² ∝ a³
Kepler's second law states that a line joining a planet to the Sun sweeps out equal areas in equal times. This means the planet moves faster when nearer to the Sun and slower when farther away, but the orbital period is determined by the semi-major axis.
开普勒第二定律指出，行星与太阳的连线在相等时间内扫过相等的面积。这意味着行星在靠近太阳时运动更快，远离太阳时更慢，但轨道周期由半长轴决定。
10. Common Exam Mistakes and Tips | 常见考试错误与提示
Many students forget to add the planet radius to the altitude of a satellite. Always convert all distances to metres and periods to seconds before substituting into the formula.
许多学生忘记将行星半径加到卫星的高度上。代入公式之前，务必把所有距离换算成米，把周期换算成秒。
Another common error is confusing the central mass M with the satellite mass m. In the period formula, M is the mass of the body being orbited, not the mass of the orbiting body.
另一个常见错误是混淆中心质量 M 与卫星质量 m。在周期公式中，M 是被绕行天体的质量，而不是绕行天体的质量。
When using Kepler's third law to compare two orbits, ensure both bodies orbit the same central mass. Also, do not use the diameter when the radius is required.
当使用开普勒第三定律比较两个轨道时，要确保两个天体绕同一中心质量运行。此外，需要半径时不要使用直径。
11. Summary of Key Equations | 关键方程总结
The following equations summarise the circular orbit topic. You should be able to derive them from Newton's law of gravitation and the centripetal force equation.
以下方程总结了圆轨道主题。你应当能够从牛顿万有引力定律和向心力方程推导出它们。
| Quantity | Equation | 中文释义 |
|---|---|---|
| Orbital speed | v = √GM / r | 轨道速度 |
| Orbital period | T = 2π√r³ / GM | 轨道周期 |
| Kepler's third law | T² ∝ r³ | 开普勒第三定律 |
| Total energy | Etotal = -GMm / (2r) | 总机械能 |
12. Exam Strategy | 考试策略
In CIE Physics exams, questions on the orbital period often ask you to calculate T, r, or v, or to explain why a geostationary satellite has a fixed position. Start by writing the gravitational force equals centripetal force, then manipulate the equations step by step.
在 CIE 物理考试中，关于轨道周期的题目通常要求计算 T、r 或 v，或者解释为什么地球同步卫星位置固定。先写出万有引力等于向心力，然后逐步处理方程。
cube root only after substituting values. Show every step clearly to secure method marks.
如果题目给出周期并要求求半径，可将 T² = 4π²r³ / (GM) 重新整理为 r³ = GMT² / (4π²)。代入数值后再开三次方。每一步都写清楚，以确保获得方法分。
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