📚 Deriving Satellite Orbital Period: PH04 June 2023 Exam Formula Derivation | 推导卫星轨道周期：PH04 2023年6月考试公式推导
In the June 2023 International Physics A (PH04) examination, one of the high-mark questions asked students to derive the orbital period of a satellite moving in a circular path around a planet. This classic derivation combines Newton’s law of gravitation with circular motion principles, and mastering it not only secures exam marks but also deepens your understanding of celestial mechanics. We will break it down step by step using clear reasoning and show exactly how the formula T² = (4π²r³)/(GM) emerges from first principles.
在2023年6月的国际物理A（PH04）考试中，一道高分题要求学生推导卫星绕行星圆轨道运动的周期公式。这一经典推导将牛顿万有引力定律与圆周运动原理相结合，掌握它不仅能确保考试得分，还能加深你对天体力学本质的理解。我们将从基本原理出发，逐步拆解推理过程，清晰展示公式 T² = (4π²r³)/(GM) 是如何得出的。
1. Understanding the Question Context | 理解问题背景
The PH04 paper often presents a scenario: a satellite of mass m orbits a planet of mass M at a distance r from the planet’s centre. The question states that the orbit is circular and asks candidates to derive an expression for the orbital period T in terms of r, M, and the gravitational constant G. Ignoring atmospheric drag and other perturbations simplifies the problem to a two-body central force motion.
PH04试卷通常给出这样的场景：一颗质量为 m 的卫星，在距离行星中心 r 处围绕质量为 M 的行星运行。题目明确轨道为圆形，要求考生用 r、M 和引力常量 G 推导轨道周期 T 的表达式。忽略大气阻力和其他摄动，问题简化为二体有心力运动。
2. Key Physical Principles | 关键物理原理
Two fundamental laws govern this derivation. First, Newton’s law of universal gravitation gives the attractive force between the planet and satellite: Fgrav = GMm / r². Second, for an object in uniform circular motion, the centripetal force required to keep it moving in a circle of radius r at speed v is Fc = mv² / r. In a stable orbit, the gravitational force provides exactly this centripetal force.
两个基本定律支配着整个推导。第一，牛顿万有引力定律给出行星与卫星之间的吸引力：Fgrav = GMm / r²。第二，对于做匀速圆周运动的物体，维持其在半径 r 处以速率 v 运动所需的向心力为 Fc = mv² / r。在稳定轨道中，万有引力恰好提供这一向心力。
3. Equating Gravitational and Centripetal Forces | 令引力等于向心力
We set the gravitational force equal to the centripetal force because no other radial forces act on the satellite. This yields the equation:
GMm / r² = mv² / r
Notice that the satellite’s mass m appears on both sides. This is a crucial simplification that reveals the orbital motion is independent of the satellite’s mass, a fact that will later lead to Kepler’s third law.
我们令引力等于向心力，因为卫星在径向不受其他外力。由此得到方程：
GMm / r² = mv² / r
注意到卫星质量 m 出现在等式两边。这是一个关键简化，表明轨道运动与卫星质量无关，这一事实随后将导出开普勒第三定律。
4. Cancellation and Solving for v² | 约去质量并解出 v²
Cancel m from both sides and multiply through by r to isolate the square of the orbital speed:
GM / r = v²
Thus, the orbital speed squared is directly proportional to the planet’s mass and inversely proportional to the orbital radius. This intermediate result is valuable on its own: it shows that satellites closer to the planet must travel faster.
两边约去 m，再乘以 r 以分离轨道速率的平方：
GM / r = v²
因此，轨道速率的平方与行星质量成正比，与轨道半径成反比。这个中间结果本身就很有价值：它表明离行星越近的卫星必须运行得越快。
5. Relating Speed to Period | 建立速率与周期的关系
In circular motion, the satellite travels a distance equal to the circumference 2πr in one period T. Therefore, its constant speed can be expressed as v = 2πr / T. Substituting this relation will introduce the period into our equation.
在圆周运动中，卫星在一个周期 T 内经过的距离等于圆周长 2πr。因此，其恒定速率可表示为 v = 2πr / T。将此关系代入，就能将周期引入方程。
6. Substituting v in Terms of T | 用 T 表示 v 并代入
Replace v in the equation GM / r = v² with 2πr / T:
GM / r = (2πr / T)²
Expand the square on the right-hand side carefully:
GM / r = 4π²r² / T²
Now we have an equation linking the gravitational parameters to the geometry of the orbit and the period.
将方程 GM / r = v² 中的 v 替换为 2πr / T：
GM / r = (2πr / T)²
仔细展开右边的平方：
GM / r = 4π²r² / T²
现在我们得到了一个联系引力参数与轨道几何及周期的方程。
7. Rearranging to Solve for T² | 整理方程解出 T²
Multiply both sides by T² and then by r to bring T² to one side and collect the radius terms:
GM × T² = 4π²r³
Finally, divide by GM to obtain the target expression:
T² = (4π²r³) / (GM)
This is the desired relationship. The period squared is proportional to the cube of the orbital radius, a statement of Kepler’s third law for the special case of circular orbits.
两边同乘 T²，再乘以 r，将 T² 单独移到一边并合并半径项：
GM × T² = 4π²r³
最后，除以 GM 便得到目标表达式：
T² = (4π²r³) / (GM)
这就是所求的关系式。周期平方与轨道半径立方成正比，此即为圆形轨道情形下的开普勒第三定律。
8. Verifying the Result: Kepler’s Third Law | 验证结果：开普勒第三定律
Kepler’s third law states that T² ∝ r³ for all planets orbiting the same central body. Our derived formula shows the constant of proportionality is 4π²/(GM). Since G and M are constants for a given planetary system, the derivation confirms the law. Students often gain extra marks by explicitly linking the final expression to Kepler’s work.
开普勒第三定律指出，对于绕同一中心天体运行的所有行星，T² ∝ r³。我们推导出的公式显示比例常数为 4π²/(GM)。由于 G 和 M 在给定行星系统中是常数，该推导证实了该定律。考生往往可以通过明确将最终表达式与开普勒的工作联系起来获得额外分数。
9. Worked Numerical Example from PH04 Style | PH04风格数值示例
A typical follow-up question might be: ‘Calculate the orbital period of a satellite 7000 km from Earth’s centre. Take G = 6.67×10⁻¹¹ N m² kg⁻² and Earth’s mass as 5.97×10²⁴ kg.’ Using T = √(4π²r³/GM), first compute r³ = (7.0×10⁶ m)³ = 3.43×10²⁰ m³. Then 4π²r³ ≈ 1.35×10²², and GM = 3.98×10¹⁴. Dividing gives T² ≈ 3.39×10⁷ s², so T ≈ 5820 s or about 1.62 hours. Always check units and convert distances to metres.
典型的后续问题可能是：“计算一颗距地心7000 km的卫星的轨道周期。取 G = 6.67×10⁻¹¹ N m² kg⁻²，地球质量 5.97×10²⁴ kg。”使用 T = √(4π²r³/GM)，首先计算 r³ = (7.0×10⁶ m)³ = 3.43×10²⁰ m³，接着 4π²r³ ≈ 1.35×10²²，GM = 3.98×10¹⁴。相除得 T² ≈ 3.39×10⁷ s²，因此 T ≈ 5820 s 或约1.62小时。务必检查单位并将距离换算为米。
10. Common Mistakes in Derivation | 推导中的常见错误
One frequent error is forgetting to square the entire term (2πr/T) when substituting, leading to a missing 4π² factor. Another is mixing up r as the radius of the orbit versus altitude above the surface; always use the distance from the centre of the planet. Students also sometimes cancel the radius incorrectly, ending up with r² instead of r³. A logical checklist can prevent these slip-ups.
一个常见错误是在代入时忘记对整个项 (2πr/T) 进行平方，导致遗漏 4π² 因子。另一个错误是混淆轨道半径 r 与距地表高度；始终使用到行星中心的距离。学生有时还会错误地约分半径，最终得到 r² 而非 r³。一份逻辑清晰的自检清单可以防止这些失误。
11. Exam Technique for PH04 Derivations | PH04推导题应试技巧
When answering a derivation question, present your steps clearly and label each equation. Start by stating the two relevant force equations. Show cancellations explicitly. Write intermediate expressions like v² = GM/r before introducing the period. If asked to ‘hence’ derive Kepler’s law, comment on the proportionality. Even if the derivation is partially incomplete, displaying correct physics principles earns method marks.
回答推导题时，要清晰地展示步骤并为每个方程做标注。从陈述两个相关的力公式开始。明确写出约分过程。在引入周期之前先写下像 v² = GM/r 这样的中间表达式。若要求“由此”导出开普勒定律，要对比例关系加以评论。即使推导部分不完整，展示正确的物理原理也能获得方法分。
12. Conclusion: Building Confidence in Derivation | 结语：建立推导信心
The satellite period derivation is a cornerstone of gravitational field topics in International Physics A. By internalising the logical flow—equate forces, cancel m, introduce v = 2πr/T, and rearrange—you equip yourself to handle variations, such as deriving the geostationary orbit radius or comparing two satellites. Practice this sequence until it becomes second nature, and you will approach the PH04 paper with assurance.
卫星周期推导是国际物理A中引力场话题的基石。通过内化这一逻辑流程——令力相等、约去 m、引入 v = 2πr/T、整理方程——你就能处理各种变体，如推导地球同步轨道半径或比较两颗卫星。反复练习这一序列，直到它成为第二本能，你就能自信地应对 PH04 试卷。
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