📚 Gravitational Fields and Satellites | 引力场与卫星（国际 A-level 专属考点）
This article covers the complete AQA International A-level specification topic "Gravitational fields and satellites", including Newton's law of gravitation, gravitational field strength, gravitational potential, orbital motion of satellites, geostationary orbits and escape velocity. Worked examples and exam-style reasoning are provided throughout.
本文系统梳理 AQA 国际 A-level 考纲中"引力场与卫星"这一专题，涵盖牛顿万有引力定律、引力场强度、引力势、卫星圆周运动、地球同步轨道与逃逸速度。全文贯穿典型例题与考试推理思路，帮助考生建立完整的知识框架。
1. Concept of Gravitational Field | 引力场的概念
A gravitational field is a region of space where a mass experiences a force due to the presence of another mass. Every object with mass generates a gravitational field around itself, and any other mass placed in that field experiences an attractive force.
引力场是指空间中某一质量因另一质量的存在而受到力的作用的区域。任何有质量的物体都会在其周围产生引力场，置于该场中的其他质量会受到吸引力。
The gravitational field is a vector field: at every point, it has both magnitude and direction. By convention, the direction of the field is the direction of the force on a small test mass placed at that point. Since gravitational forces are always attractive, field lines always point towards the mass that creates the field.
引力场是矢量场：在场中每一点既有大小也有方向。按惯例，某点场的方向就是置于该点的试探质量所受力的方向。由于引力始终是吸引力，因此引力场线总是指向产生该场的质量。
- Uniform field: field lines are parallel and equally spaced — valid near the Earth's surface.
- Radial field: field lines converge at the centre of mass — valid outside a spherical object.
- 匀强场：场线平行且等间距——适用于地球表面附近。
- 径向场：场线会聚于质心——适用于球形物体外部。
g = F / m
Here g is the gravitational field strength in N kg⁻¹, F is the gravitational force in newtons, and m is the mass of the test object in kilograms. Note that gravitational field strength and acceleration due to gravity are numerically equal at the same location.
式中 g 为引力场强度，单位 N kg⁻¹；F 为引力，单位 N；m 为试探质量，单位 kg。注意引力场强度与重力加速度在同一位置数值相等。
2. Newton's Law of Gravitation | 牛顿万有引力定律
Newton's law of gravitation states that the gravitational force between two point masses 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：两个质点之间的引力与它们质量的乘积成正比，与它们质心之间距离的平方成反比。
F = G m₁m₂ / r²
kg⁻²), m₁ and m₂ are the two masses (kg), and r is the distance between their centres (m).
其中 F 为万有引力大小（N），G 为引力常量（6.67 × 10⁻¹¹ N m² kg⁻²），m₁ 和 m₂ 为两个质量（kg），r 为两质心间距离（m）。
Although the law is stated for point masses, it also applies to uniform spheres if the distance is measured from the centre of one sphere to the centre of the other. This result, proved by Newton, allows us to treat planets and stars as point masses when calculating orbital motion.
虽然该定律是针对质点表述的，但只要距离从球心算到球心，它也适用于均匀球体。这一由牛顿证明的结论，使我们计算轨道运动时可以把行星和恒星视为质点。
3. Gravitational Field Strength in a Radial Field | 径向场中的引力场强度
For a point mass M, the gravitational field strength at a distance r from its centre can be derived by combining Newton's law of gravitation with the definition of field strength.
对于质量为 M 的质点，距离其中心 r 处的引力场强度可由牛顿万有引力定律与场强定义结合导出。
g = GM / r²
This equation shows an inverse square relationship: if the distance from the centre of mass doubles, the gravitational field strength falls to one quarter of its original value. This applies only outside the mass; inside a uniform sphere, g increases linearly from zero at the centre to a maximum at the surface.
该式呈现平方反比关系：若距质心距离加倍，引力场强度降至原来的四分之一。此公式仅适用于质量外部；在均匀球体内部，g 从球心处的零线性增加至球面处最大。
| Gravitational field strength at Earth's surface | g = 9.81 N kg⁻¹ |
| At altitude h above ground | g = GM / (R + h)² |
| At height equal to Earth's radius (h = R) | g = 9.81 / 4 = 2.45 N kg⁻¹ |
Examiners frequently test the inverse square nature of g with distance. Remember that altitude is always added to the Earth's radius before squaring.
考官经常考查 g 随距离的平方反比特征。切记：高度必须加上地球半径后再平方，绝不能直接用高度平方代入。
4. Gravitational Potential and Potential Energy | 引力势与引力势能
Gravitational potential V at a point is defined as the work done per unit mass in bringing a test mass from infinity to that point. Since gravity is attractive, this work is done by the field, so potential is negative.
引力势 V 定义为单位质量从无穷远处移至某点过程中引力所做的功。由于引力是吸引力，这一过程由场做功，因此势能为负值。
V = − GM / r
Here V is measured in J kg⁻¹. The negative sign is essential: potential increases from a large negative value near the mass towards zero at infinity. A body at infinity has zero gravitational potential energy, and as it falls towards a mass, its potential energy becomes more negative while its kinetic energy increases correspondingly.
式中 V 的单位为 J kg⁻¹。负号至关重要：势能从质量附近的大负值逐渐增大至无穷远处的零。物体在无穷远处引力势能为零；当它向质量下落时，势能变得更负，动能相应增加。
Gravitational potential energy is related to potential by:
引力势能与引力势的关系为：
Eₚ = mV = − GMm / r
For a satellite of mass m in orbit at radius r, this energy is negative. The total mechanical energy of an orbiting satellite is the sum of its kinetic and potential energies, which we shall analyse in Section 8.
对于质量为 m、轨道半径为 r 的卫星，其引力势能为负值。卫星的总机械能是动能与势能之和，我们将在第 8 节详细分析。
5. Kepler's Laws and Orbital Motion | 开普勒定律与轨道运动
Johannes Kepler formulated three laws describing planetary motion based on Tycho Brahe's observations. These laws also apply to all satellites orbiting a central body.
开普勒根据第谷的观测数据总结出三大行星运动定律。这些定律同样适用于绕中心天体运行的所有卫星。
- First law: planets move in elliptical orbits with the Sun at one focus.
- Second law: the line joining a planet to the Sun sweeps out equal areas in equal times.
- Third law: the square of the orbital period is proportional to the cube of the semi-major axis.
- 第一定律：行星沿椭圆轨道运动，太阳位于一个焦点上。
- 第二定律：行星与太阳的连线在相等时间内扫过相等的面积。
- 第三定律：轨道周期的平方与半长轴的立方成正比。
For a circular orbit of radius r, the third law takes the form:
对于半径为 r 的圆轨道，第三定律可写为：
T² = (4π² / GM) r³
This combined relation is derived by equating the gravitational force to the centripetal force required for circular motion. The mass M is that of the central body, not the satellite.
该组合关系式通过令引力等于圆周运动所需的向心力导出。其中 M 为中心天体质量，而非卫星质量。
6. Satellites in Circular Orbit | 圆轨道卫星的运动分析
A satellite in a stable circular orbit is in a state of continuous free fall: gravity provides exactly the centripetal force needed for circular motion at that radius. Equating the two forces gives:
沿稳定圆轨道运行的卫星处于持续自由落体状态：引力恰好提供了该半径下圆周运动所需的向心力。令两力相等可得：
GMm / r² = mv² / r
Cancelling m and rearranging gives the orbital speed:
消去 m 并整理得轨道速度：
v = √GM / r
Key conclusions from this expression:
由此式可得出关键结论：
- Orbital speed is independent of the satellite's mass — any satellite at the same radius moves at the same speed.
- Orbital speed decreases with increasing orbital radius — more distant satellites move more slowly.
- 轨道速度与卫星质量无关——任何卫星在同一半径处速度相同。
- 轨道速度随轨道半径增大而减小——越远的卫星运动越慢。
From v = 2πr / T, we can substitute into v = √GM / r to obtain Kepler' : T² = (4π² / GM) r³. This confirms the consistency between Newtonian mechanics and Kepler's empirical laws.
由 v = 2πr / T，代入 v = √GM / r 即可得开普勒第三定律：T² = (4π² / GM) r³。这印证了牛顿力学与开普勒经验定律的一致性。
7. Geostationary Satellites | 地球同步轨道卫星
A geostationary satellite orbits above the equator with a period of 24 hours, matching the Earth's rotational period. Viewed from a fixed point on Earth, the satellite appears permanently stationary in the sky. This makes it ideal for telecommunications, broadcasting and weather monitoring.
地球同步卫星在赤道正上方运行，周期为 24 小时，与地球自转周期相同。从地球固定点观察，卫星在天空中看起来静止不动。这使其成为通信、广播和气象监测的理想选择。
For a geostationary orbit, the required conditions are:
地球同步轨道需满足以下条件：
- The orbital plane must coincide with the equatorial plane.
- The direction of revolution must be the same as Earth's rotation (west to east).
- The orbital period must equal 24 hours.
- 轨道平面必须与赤道平面重合。
- 公转方向必须与地球自转方向相同（自西向东）。
- 轨道周期必须等于 24 小时。
Using T = 86 400 s in Kepler's third law with M = 5.97 × 10²⁴ kg:
将 T = 86 400 s 和 M = 5.97 × 10²⁴ kg 代入开普勒第三定律：
r³ = GMT² / 4π² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 86 400²) / 4π² = 4.23 × 10²² m³
r ≈ 4.23 × 10⁷ m = 42 300 km
Subtracting Earth's radius gives an altitude of approximately 36 000 km. This calculation is a favourite exam question — memorise the method, not just the answer.
减去地球半径后，轨道高度约为 36 000 km。这是考试热门计算题——务必掌握计算方法，而不只是背答案。
8. Energy of an Orbiting Satellite | 轨道卫星的能量分析
For a satellite in a circular orbit at radius r, both kinetic and potential energies can be expressed in terms of r.
对于半径为 r 的圆轨道卫星，动能和势能都可以用 r 表达。
From v² = GM/r, the kinetic energy is:
由 v² = GM/r，动能：
Eₖ = ½mv² = GMm / 2r
The gravitational potential energy is:
引力势能：
Eₚ = − GMm / r
Therefore the total mechanical energy E = Eₖ + Eₚ:
因此总机械能 E = Eₖ + Eₚ：
E = GMm / 2r − GMm / r = − GMm / 2r
Since E is negative, the satellite is bound to Earth. To move to a higher orbit, energy must be supplied, increasing E towards zero. When a satellite moves from a lower to a higher orbit, its speed decreases but its potential energy increases more, so the total energy rises.
由于 E 为负值，卫星被地球束缚。要使卫星进入更高轨道，需要输入能量，使 E 趋近于零。当卫星从低轨道移至更高轨道时，速度减小，但势能增加更多，因此总能量上升。
9. Escape Velocity | 逃逸速度
Escape velocity is the minimum speed an object must have at the surface of a planet to escape the gravitational field entirely, that is, to reach infinity with zero kinetic energy.
逃逸速度是物体在行星表面为完全脱离该引力场所需的最小速度，即到达无穷远处时动能为零。
Applying conservation of energy: total energy at the surface equals total energy at infinity.
应用能量守恒：表面的总能量等于无穷远处的总能量。
½mv² − GMm / R = 0
Solving for v gives:
解得 v：
v = √2GM / R
For Earth, using M = 5.97 × 10²⁴ kg and R = 6.37 × 10⁶ m:
对于地球，代入 M = 5.97 × 10²⁴ kg，R = 6.37 × 10⁶ m：
v = √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.37 × 10⁶) ≈ 11.2 km s⁻¹
Note that escape velocity is independent of the mass of the escaping object and of the direction of launch (as long as it does not pass through the Earth). A rocket launched from Earth's surface must reach this speed to leave Earth's gravitational influence without further propulsion.
注意逃逸速度与逃逸物体质量无关，也与发射方向无关（只要轨迹不穿过地球即可）。火箭从地球表面发射时，若不再继续推进，必须达到该速度才能脱离地球引力影响。
10. Weightlessness and Orbital Motion | 失重状态与轨道运动
Astronauts inside the International Space Station appear weightless, but gravity at that altitude (about 400 km) is still roughly 90% of its surface value. The sensation of weightlessness arises because both the astronauts and the station are in continuous free fall around Earth.
国际空间站内的宇航员看似失重，但在约 400 km 高度处，重力仍是地表值的约 90%。失重感的产生是因为宇航员与空间站都在绕地球持续自由下落。
Since the station and everything inside it accelerate towards Earth at exactly the same rate, the normal reaction force that would normally press a body against the floor falls to zero. There is no sensation of weight because no contact force acts between the astronaut and the station structure.
由于空间站及其内部所有物体以完全相同的加速度落向地球，正常情况下地板对身体的支持力降为零。因为宇航员与空间站结构之间没有接触力，所以没有"重量"的感觉。
This distinction is frequently examined: weightlessness does not mean zero gravity — it means zero apparent weight due to free fall.
这是常考辨析点：失重并不意味着无引力——而是因自由落体导致的表观重量为零。
11. Worked Example | 典型例题精析
of the satellite. Given: G = 6.67 × 10⁻¹¹ N m² kg⁻², M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m.
一颗质量为 2 500 kg 的卫星在距地面 300 km 的高空绕地球运行。求：(a) 轨道速度；(b) 轨道周期；(c) 卫星总能量。已知：G = 6.67 × 10⁻¹¹ N m² kg⁻²，M = 5.97 × 10²⁴ kg，R = 6.37 × 10⁶ m。
Step 1 — Determine the orbital radius:
第一步 — 确定轨道半径：
r = (6.37 × 10⁶) + (300 × 10³) = 6.67 × 10⁶ m
Step 2 — Calculate orbital speed:
第二步 — 计算轨道速度：
v = √GM / r = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / 6.67 × 10⁶] = 7.73 × 10³ m s⁻¹
Step 3 — Calculate orbital period:
第三步 — 计算轨道周期：
T = 2πr / v = (2π × 6.67 × 10⁶) / 7.73 × 10³ = 5.42 × 10³ s ≈ 90 minutes
Step 4 — Calculate total energy:
第四步 — 计算总能量：
= − GMm / 2r = − (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 2 500) / (2 × 6.67 × 10⁶) = −7.47 × 10¹⁰ J
Note the systematic approach: always convert altitude into orbital radius first, and always check units before substituting into equations.
注意系统化解题思路：永远先将高度转化为轨道半径，并在代入公式前检查单位。
12. Exam Focus and Common Misconceptions | 考点聚焦与常见误区
ning gravitational and centripetal forces, geostationary orbit calculations, and the negative sign of gravitational potential.
本专题最高频考点包括：平方反比定律、g = GM/r² 的推导、引力与向心力的结合、地球同步轨道计算，以及引力势的负号问题。
Common errors to avoid:
常见错误需避免：
- Using diameter instead of radius — always identify the distance between centres.
- Forgetting that r in satellite equations is the orbital radius measured from the planet's centre, not the altitude above the surface.
- Writing positive potential energies — gravitational potential and potential energy are always negative in bound systems.
- Confusing g (field strength) with the gravitational constant G.
- Applying v = √GM/r only for circular orbits — it was derived assuming circular motion.
- 使用直径而非半径——务必确定质心间距离。
- 忘记卫星公式中的 r 是从行星中心算起的轨道半径，而非距地表高度。
- 将势能写为正——束缚系统中引力势和引力势能必为负。
- 混淆 g（场强）与引力常量 G。
- 注意 v = √GM/r 仅适用于圆轨道——该式推导时假设为圆周运动。
Master these core ideas, practise the derivation chain from F = GMm/r² to Kepler' red for any question in this section of the AQA International A-level Physics examination.
掌握这些核心概念，勤练从 F = GMm/r² 到开普勒第三定律的推导链条，你就能从容应对 AQA 国际 A-level 物理考试中本节的所有问题。
Published by TutorHao | Physics Revision Series | aleveler.com
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