Gravitation for IGCSE OCR Physics Exam Essentials | IGCSE OCR 物理：万有引力考点精讲

📚 Gravitation for IGCSE OCR Physics Exam Essentials | IGCSE OCR 物理：万有引力考点精讲

Gravitation is one of the fundamental forces you will encounter in the IGCSE OCR Physics syllabus. This article breaks down the key concepts, equations and typical exam questions around Newton’s law of universal gravitation, gravitational field strength, orbital motion and the distinction between mass and weight. Each section is structured to give you a clear explanation in English followed by its Chinese equivalent, making it perfect for bilingual revision.

引力是你在 IGCSE OCR 物理课程中会遇到的基本力之一。本文拆解了万有引力定律、重力场强度、轨道运动以及质量与重量的区别等关键概念、方程和典型考题。每个部分均先提供英文解释，再紧跟中文对照，非常适合双语复习。

1. Newton’s Law of Universal Gravitation | 牛顿万有引力定律

Newton’s law of universal gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

牛顿万有引力定律指出，宇宙中每一个质点都会吸引其他每一个质点，引力的大小与两质点质量的乘积成正比，与它们中心之间距离的平方成反比。

In equation form, this is written as:

用方程形式表示为：

F = G M m / r²

F is the gravitational force between two masses (N)
G is the universal gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
M and m are the two masses (kg)
r is the distance between the centres of the two masses (m)

F 为两质量间的引力（牛顿）
G 为万有引力常数，6.67 × 10⁻¹¹ 牛·米²/千克²
M 和 m 分别为两个物体的质量（千克）
r 为两物体中心之间的距离（米）

The inverse-square law means that if you double the distance, the gravitational force becomes one quarter of its original value. This principle is examinable both qualitatively and quantitatively.

平方反比定律意味着，如果将距离加倍，引力将变为原来的四分之一。这一原理既会以定性方式考察，也会要求定量计算。

2. Gravitational Field Strength g | 重力场强度 g

Gravitational field strength (g) is defined as the force per unit mass acting on a small test mass placed at a point in the field. Near the Earth’s surface, it is approximately 9.8 N/kg.

重力场强度（g）定义为单位质量的小检验质量在重力场中某一点所受的力。在地球表面附近，它大约为 9.8 牛/千克。

The relationship is:

关系式为：

g = F / m

From this, we can also express the gravitational force on an object of mass m in a gravitational field of strength g as:

由此，我们也可以将质量为 m 的物体在强度为 g 的重力场中所受的引力表示为：

W = m g

This weight equation is distinct from the universal law: W = mg applies locally, whereas F = G M m / r² applies universally. The exam often tests your ability to switch between these viewpoints.

这个重量方程与万有引力定律不同：W = mg 适用于局部区域，而 F = G M m / r² 则适用于整个宇宙。考试中经常会考察你对这两种视角的切换能力。

3. Mass vs Weight | 质量与重量的区别

Mass is a scalar quantity measuring the amount of matter in an object, measured in kilograms (kg). It does not change with location.

质量是标量，用于衡量物体所含物质的多少，单位为千克（kg）。它不会随位置改变。

Weight is a vector quantity representing the gravitational force acting on a mass. It is measured in newtons (N) and depends on the local gravitational field strength g.

重量是矢量，表示作用在某一质量上的重力，单位为牛顿（N），其大小取决于当地的重力场强度 g。

A common exam mistake is to use ‘weight’ and ‘mass’ interchangeably. Remember: on the Moon your mass is unchanged, but your weight is about 1/6 of that on Earth because g on the Moon is approximately 1.6 N/kg.

一个常见的考试错误是混用「重量」和「质量」。请记住：在月球上你的质量不变，但你的重量大约只有地球上的 1/6，因为月球上的 g 约为 1.6 牛/千克。

4. Deriving the Value of g Using Newton’s Law | 利用牛顿定律推导 g 的值

For a planet of mass M and radius R, the gravitational field strength at its surface can be found by equating mg and G M m / R²:

对于质量为 M、半径为 R 的行星，其表面的重力场强度可以通过令 mg 等于 G M m / R² 求得：

m g = G M m / R² → g = G M / R²

This formula shows that g is independent of the test mass m, and depends only on the mass and radius of the planet. You may be asked to calculate the mass of the Earth given G, g and the Earth’s radius.

该公式表明 g 与检验质量 m 无关，只取决于行星的质量和半径。考试中可能会要求你根据 G、g 和地球半径计算地球的质量。

Using g ≈ 9.8 N/kg, R ≈ 6.37 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m² kg⁻², the Earth’s mass works out to be about 6.0 × 10²⁴ kg. This type of calculation reinforces the link between local observations and universal gravitation.

利用 g ≈ 9.8 牛/千克，R ≈ 6.37×10⁶ 米和 G = 6.67×10⁻¹¹ 牛·米²/千克²，可算出地球质量约为 6.0×10²⁴ 千克。这类计算强化了局部观测与万有引力之间的联系。

5. Orbital Motion and Gravitational Force | 轨道运动与引力

A satellite stays in orbit because the gravitational force provides the necessary centripetal force. For a satellite of mass m orbiting a planet of mass M at a radius r with speed v:

卫星之所以能保持在轨道上，是因为引力提供了所需的向心力。对于质量为 m 的卫星以速度 v 在半径 r 的轨道上绕质量为 M 的行星运行：

G M m / r² = m v² / r

Cancelling m and simplifying gives:

消去 m 并化简得：

v² = G M / r or v = √(G M / r)

This tells us that the orbital speed depends only on the mass of the central body and the orbital radius; it does not depend on the satellite’s mass. A common examination task is to calculate the speed or period of a satellite.

这告诉我们，轨道速度只取决于中心天体的质量和轨道半径，与卫星的质量无关。考试中常见的任务是计算卫星的速度或周期。

6. Kepler’s Third Law in the Context of IGCSE | IGCSE 背景下的开普勒第三定律

Kepler’s third law states that the square of the orbital period T is proportional to the cube of the average orbital radius r:

开普勒第三定律指出，轨道周期 T 的平方与平均轨道半径 r 的立方成正比：

T² ∝ r³

For circular orbits under gravity, we can derive this relationship by combining v = 2π r / T with v² = G M / r:

对于引力作用下的圆形轨道，我们可以结合 v = 2π r / T 与 v² = G M / r 推导出这一关系：

(2π r / T)² = G M / r → T² = (4π² / G M) r³

In the OCR exam, you usually do not need to reproduce the full derivation, but you must be able to use the proportionality to compare periods and radii of different planets or satellites.

在 OCR 考试中，通常不需要完整再现推导过程，但你必须能够利用这一比例关系来比较不同行星或卫星的周期和半径。

7. Geostationary Satellites | 地球静止轨道卫星

A geostationary satellite orbits the Earth exactly once every 24 hours and remains above the same point on the equator. Its orbital period equals the Earth’s rotational period.

地球静止轨道卫星每 24 小时绕地球一周，并始终位于赤道上空的同一点上方。它的轨道周期等于地球的自转周期。

For OCR, you need to know the key features:

对于 OCR 课程，你需要知道以下关键特征：

Orbit must be equatorial (above the equator).
Orbit radius approximately 42,300 km from Earth’s centre (about 35,800 km above the surface).
Period is 24 hours.
Used for telecommunications and weather monitoring because the satellite appears stationary from the ground.

轨道必须在赤道面内（赤道上空）。
轨道半径距地心约 42,300 km（距地面约 35,800 km）。
周期为 24 小时。
用于通信和气象监测，因为从地面看卫星是静止的。

You may be asked to explain why a geostationary orbit requires a specific radius. The answer links back to r³ ∝ T² and the fixed period of 24 hours.

考试可能会问你为什么地球静止轨道需要特定的半径。答案与 r³ ∝ T² 以及固定的 24 小时周期有关。

8. Graphs of Gravitational Field Strength vs Distance | 重力场强度与距离的关系图

Outside a planet, g decreases with the square of the distance from the centre:

在行星外部，g 随距地心距离的平方衰减：

g ∝ 1 / r²

Inside a uniform spherical shell the field is zero, and inside a uniform solid planet the field strength increases linearly with distance from the centre (assuming uniform density). However, for IGCSE level, the main expectation is to sketch or interpret the inverse‑square decrease outside the planet.

在均匀球壳内部引力场为零，而在均匀实心行星内部引力场强度随距地心的距离线性增大（假设密度均匀）。但在 IGCSE 阶段，主要要求是能够画出或解释行星外部的平方反比衰减曲线。

A typical graph question provides a plot of g against r and asks you to read values, calculate the mass of the planet, or compare gravitational force at different altitudes.

典型的图表题会给出 g 与 r 的关系图，要求你读取数值、计算行星质量或比较不同高度处的引力大小。

9. Weightlessness and Free Fall | 失重与自由落体

Weightlessness does not mean absence of gravity. Astronauts in an orbiting spacecraft experience apparent weightlessness because they are in free fall around the Earth. Both the astronaut and the spacecraft accelerate towards Earth at the same rate, so the normal contact force feels zero.

失重并不意味着没有重力。轨道飞船中的宇航员体验到的是表观失重，因为他们实际上是在围绕地球自由落体。宇航员和飞船以相同的加速度向地球下落，因此感受到的法向接触力为零。

In the exam, you might be asked to explain why objects float inside the International Space Station even though Earth’s gravity is still about 90% of that at the surface.

在考试中，你可能会被问及为什么国际空间站内的物体会漂浮，尽管那里地球的引力仍然大约是地面的 90%。

The key phrase is ‘in free fall’ – both the station and its contents share the same acceleration due to gravity, resulting in no relative motion and the sensation of weightlessness.

关键短语是「自由落体」——空间站与其内部的物体由于重力作用具有相同的加速度，因此没有相对运动，产生失重感。

10. Common Misconceptions and Exam Pitfalls | 常见误区与考试陷阱

Many students confuse G and g. G is the universal constant 6.67 × 10⁻¹¹ (with units), while g is the local field strength, typically 9.8 N/kg on Earth. They are not the same.

很多学生混淆了 G 和 g。G 是万有引力常数 6.67×10⁻¹¹（带单位），而 g 是局部重力场强度，在地球上通常为 9.8 牛/千克。二者并不相同。

Another pitfall is forgetting to square the distance, or using the diameter instead of the radius. Always check that you are using the centre‑to‑centre distance.

另一个陷阱是忘记把距离平方，或者误用直径而非半径。务必确认使用的是中心到中心的距离。

Finally, confirm unit consistency: mass in kg, distance in m, force in N, and in calculations involving G, the 10⁻¹¹ exponent often leads to arithmetic slips. Use standard form carefully.

最后，确保单位一致：质量用千克，距离用米，力用牛顿；涉及 G 的计算中 10⁻¹¹ 的指数常导致运算错误，请小心使用科学记数法。

11. Application: Calculating the Mass of a Planet from Satellite Data | 应用：根据卫星数据计算行星质量

A classic problem uses the period and radius of a moon to find the mass of the central planet. Equating centripetal force to gravity:

一道典型题目是利用卫星的周期和轨道半径求中心行星的质量。将向心力与引力等起来：

m (2π r / T)² / r = G M m / r²

Solving for M:

解出 M：

M = 4π² r³ / (G T²)

This expression is directly linked to Kepler’s third law. In an OCR structured question, you might be given r, T and G and asked to compute M. Show your substitution and keep powers of ten separate to avoid errors.

这个表达式直接与开普勒第三定律相关。在 OCR 结构化问题中，可能会给你 r、T 和 G，要求计算 M。展示代入过程，并将 10 的幂次分开处理以避免错误。

12. Summary of Key Equations for Revision | 复习关键方程总结

Keep these relationships at your fingertips:

将以下关系熟记于心：

Equation	Use
F = G M m / r²	Universal gravitation
W = m g	Weight near a planet’s surface
g = G M / R²	Field strength at a planet’s surface
v = √(G M / r)	Orbital speed for a circular orbit
T² ∝ r³	Kepler’s third law (relate period and radius)

Be comfortable rearranging these and converting units. Practise past paper questions that combine these ideas, such as comparing the weight of a rover on Mars and Earth.

要熟练地重组这些公式并转换单位。多练历年真题，如比较火星车在火星和地球上的重量等综合题目。

Published by TutorHao | IGCSE Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Gravitation for IGCSE OCR Physics Exam Essentials | IGCSE OCR 物理：万有引力 考点精讲