Orbital Period: Definition & Calculation | 轨道周期的定义与计算

📚 Orbital Period: Definition & Calculation | 轨道周期的定义与计算

The orbital period is a fundamental concept in astrophysics and classical mechanics, frequently tested in CIE A-Level Physics. It is the time taken for an object to complete one full orbit around another object. Let’s break down its definition, the equations that govern it, and how to apply them in exams.

轨道周期是天体物理学和经典力学中的一个基本概念，也是 CIE A-Level 物理的常考考点。它指的是一个物体围绕另一个物体完成一整圈公转所需的时间。让我们来详细拆解它的定义、支配它的方程，以及如何在考试中应用这些方程。

1. Definition and Basic Formulas | 定义与基本公式

The orbital period (T) is defined as the time a satellite or planet takes to complete one complete revolution around a central body. In uniform circular motion, this is related to the orbital speed (v) and the radius of the orbit (r).

轨道周期 (T) 定义为卫星或行星围绕中心天体完成一整圈公转所需的时间。在匀速圆周运动中，它与轨道速度 (v) 和轨道半径 (r) 有关。

The formula connecting these quantities is v = 2πr / T. Therefore, we can write T = 2πr / v. It can also be defined using angular speed (ω), which is a measure of how quickly the object sweeps out angle as it orbits.

连接这些物理量的公式是 v = 2πr / T。因此，我们可以写出 T = 2πr / v。也可以用角速度 (ω) 来定义它，角速度衡量的是物体在轨道上扫过角度的快慢。

T = 2πr / v = 2π / ω

A common mistake students make in exams is confusing orbital radius with orbital height. Always ensure ‘r’ represents the distance from the centre of the central mass, not simply the altitude above its surface.

学生在考试中常犯的错误是混淆轨道半径与轨道高度。务必确保 ‘r’ 代表到中心天体中心的距离，而不仅仅是其表面上方的高度。

2. Kepler’s Third Law | 开普勒第三定律

For planets orbiting the Sun, or satellites orbiting a large central mass, the square of the orbital period is directly proportional to the cube of the semi-major axis (average radius) of the orbit. This is one of the most powerful relationships in celestial mechanics.

对于绕太阳运行的行星或绕大质量中心天体运行的卫星，其轨道周期的平方与轨道半长轴（平均半径）的立方成正比。这是天体力学中最强大的关系之一。

This is expressed by Kepler’s Third Law: T² ∝ r³. To use this as an equation, we introduce the gravitational constant (G) and the mass of the central body (M), giving us a precise formula that can be used for calculations.

这就是开普勒第三定律：T² ∝ r³。为了将其用作方程，我们引入万有引力常量 (G) 和中心天体的质量 (M)，得到一个可用于计算的精确公式。

T² = (4π² / GM) × r³

Here, G is the gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²), M is the mass of the central body in kilograms, and r is the orbital radius in meters.

其中，G 是万有引力常量（6.67 × 10⁻¹¹ N m² kg⁻²），M 是中心天体的质量（单位：千克），r 是轨道半径（单位：米）。

3. Deriving T² = (4π² / GM) × r³ | 推导 T² = (4π² / GM) × r³

This formula is derived from Newton’s Law of Gravitation and the centripetal force requirement. For a stable orbit, the gravitational force (F = GMm / r²) provides the necessary centripetal force (F = mv² / r) to keep the satellite moving in a circle.

这个公式由牛顿万有引力定律和向心力条件推导而来。对于稳定轨道，万有引力 (F = GMm / r²) 提供了所需的向心力 (F = mv² / r)，使卫星保持圆周运动。

Setting them equal gives: GMm / r² = mv² / r. It is crucial to notice that the mass of the satellite (m) cancels out on both sides of the equation. This shows that the orbital period does not depend on the mass of the satellite itself, only on the central mass.

将两者相等得到：GMm / r² = mv² / r。关键在于注意卫星的质量 (m) 在方程两边被约掉了。这表明轨道周期与卫星本身的质量无关，只取决于中心天体的质量。

Substituting v = 2πr / T into the equation gives GM / r² = (2πr / T)² / r, which simplifies to GM / r² = 4π²r / T². Rearranging this to solve for T² yields the final formula T² = (4π² / GM) × r³.

将 v = 2πr / T 代入方程，得到 GM / r² = (2πr / T)² / r，简化后为 GM / r² = 4π²r / T²。重新整理以求解 T²，即可得到最终公式 T² = (4π² / GM) × r³。

4. Calculating Orbital Radius | 计算轨道半径

We can rearrange Kepler’s Third Law to solve for the orbital radius. Given the mass of the central body (M) and the orbital period (T), the radius can be found by isolating r in the equation.

我们可以重新整理开普勒第三定律来求解轨道半径。已知中心天体的质量 (M) 和轨道周期 (T)，通过在方程中分离出 r，即可求出半径。

r = ∛(GMT² / 4π²)

A very common exam question involves finding the altitude of a satellite above a planet’s surface. This requires the simple subtraction: h = r – R_planet, where R_planet is the radius of the planet and h is the height above the surface.

一个非常常见的考试题型是求卫星距行星表面的高度。这需要做一个简单的减法：h = r – R_行星，其中 R_行星是行星的半径，h 是距地面的高度。

5. Worked Example 1: Finding T | 例题 1：求周期

A satellite orbits the Earth at an average height of 300 km. Given that g = 9.81 m/s² and the Earth’s radius R_E = 6.37 × 10⁶ m, calculate the orbital period of the satellite.

一颗卫星在距地面平均高度 300 km 处绕地球运行。已知 g = 9.81 m/s²，地球半径 R_地 = 6.37 × 10⁶ m，计算该卫星的轨道周期。

First, recall that for Earth, the product GM can be calculated using g and R_E: GM = gR². So, GM = 9.81 × (6.37 × 10⁶)² ≈ 3.98 × 10¹⁴ m³/s². The orbital radius is r = R_E + h = 6.37 × 10⁶ + 300 × 10³ = 6.67 × 10⁶ m.

首先，记住对于地球，GM 的乘积可以用 g 和 R_地计算：GM = gR²。所以，GM = 9.81 × (6.37 × 10⁶)² ≈ 3.98 × 10¹⁴ m³/s²。轨道半径为 r = R_地 + h = 6.37 × 10⁶ + 300 × 10³ = 6.67 × 10⁶ m。

Using the derived formula: T² = (4π² / GM) × r³ = (4π² / 3.98 × 10¹⁴) × (6.67 × 10⁶)³ ≈ 2.94 × 10⁷ s². Taking the square root gives T ≈ 5424 s, which is approximately 90 minutes.

使用推导出的公式：T² = (4π² / GM) × r³ = (4π² / 3.98 × 10¹⁴) × (6.67 × 10⁶)³ ≈ 2.94 × 10⁷ s²。开平方得 T ≈ 5424 s，大约 90 分钟。

6. Worked Example 2: Finding Central Mass (M) | 例题 2：求中心天体质量

The Moon orbits the Earth with a period of approximately 27.3 days (2.36 × 10⁶ s) at an average distance of 3.84 × 10⁸ m. Use this information to estimate the mass of the Earth.

月球绕地球运行的周期约为 27.3 天（2.36 × 10⁶ s），平均距离为 3.84 × 10⁸ m。利用这些信息来估算地球的质量。

We can rearrange the formula T² = (4π² / GM) × r³ to make M the subject of the equation: M = 4π²r³ / GT². This is a standard rearrangement that appears frequently in CIE A-Level exam papers.

我们可以重新整理公式 T² = (4π² / GM) × r³，使 M 成为方程的主项：M = 4π²r³ / GT²。这是 CIE A-Level 试卷中经常出现的标准变形。

Substituting the values: M = 4π² × (3.84 × 10⁸)³ / (6.67 × 10⁻¹¹ × (2.36 × 10⁶)²) ≈ 6.02 × 10²⁴ kg. This matches the known mass of the Earth very closely.

代入数值：M = 4π² × (3.84 × 10⁸)³ / (6.67 × 10⁻¹¹ × (2.36 × 10⁶)²) ≈ 6.02 × 10²⁴ kg。这与已知的地球质量非常接近。

7. Geostationary Orbits | 地球同步轨道

A geostationary satellite orbits the Earth directly above the equator, moving in the same direction as the Earth’s rotation, with a period of exactly 24 hours (T = 86,400 s). This is a classic example where the orbital period is specified, and you must find the radius or altitude.

地球同步卫星位于赤道正上方，运行方向与地球自转方向相同，周期精确为 24 小时（T = 86,400 s）。这是一个经典例子，题目给定轨道周期，你需要求出轨道半径或高度。

Using the formula with M = 5.97 × 10²⁴ kg, we find that the orbital radius is approximately 4.22 × 10⁷ m. Subtracting the Earth’s radius (6.37 × 10⁶ m) gives the altitude of about 3.58 × 10⁷ m, which is roughly 36,000 km.

将 M = 5.97 × 10²⁴ kg 代入公式，我们得到轨道半径约为 4.22 × 10⁷ m。减去地球半径（6.37 × 10⁶ m）后，得到轨道高度约为 3.58 × 10⁷ m，约合 36,000 km。

This specific orbit is crucial for weather monitoring, global communications, and television broadcasting, as the satellite remains fixed relative to a point on the equator.

这种特殊的轨道对于气象

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Orbital Period: Definition & Calculation | 轨道周期的定义与计算