📚 Gravitation for IB & Edexcel Physics | IB 与 Edexcel 物理：万有引力考点精讲
Gravitation is a central topic in both IB Physics (for Higher Level and Standard Level) and Edexcel A-Level Physics, linking celestial mechanics, fields, and energy. This revision guide covers the essential concepts, formulas, and skills you need to master, from Newton’s law to satellite orbits and escape velocity, with particular attention to exam-style reasoning and common pitfalls.
万有引力是 IB 物理（无论高水平还是标准水平）和 Edexcel A-Level 物理的核心主题，它将天体力学、场与能量紧密联系起来。本考点精讲涵盖从牛顿定律到卫星轨道与逃逸速度的关键概念、公式和能力，特别关注考试推导思路和常见失分点。
1. Newton’s Law of Universal Gravitation | 牛顿万有引力定律
Newton’s law states that every point mass attracts every other point mass 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. In equation form, the magnitude is F = G m₁ m₂ / r², where G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant. The force acts along the line joining the two centres and is always attractive.
牛顿万有引力定律指出，任何两个质点间均存在引力，其大小与两质点质量的乘积成正比，与它们之间距离的平方成反比。表达式为 F = G m₁ m₂ / r²，其中引力常量 G = 6.67 × 10⁻¹¹ N m² kg⁻²。力的方向沿两质心连线，且始终为吸引力。
For extended spherical bodies with uniform density (or spherical symmetry), the entire mass can be treated as if concentrated at the centre. This assumption is vital when dealing with planets and stars. The inverse‑square nature means that doubling the separation reduces the force to one‑quarter of its original value.
对于密度均匀（或具有球对称）的球体，整个物体的质量可视为集中于球心。处理行星与恒星问题时这一假设至关重要。平方反比特性意味着间距增大为原来的两倍，力将减小为原来的四分之一。
F = G m₁ m₂ / r²
2. Gravitational Field Strength | 引力场强度 g
Gravitational field strength g at a point is defined as the gravitational force per unit mass experienced by a small test mass placed at that point: g = F / m. It is a vector quantity whose direction is the same as the force on a positive test mass – towards the source mass. The SI unit is N kg⁻¹, which is equivalent to m s⁻², so g also represents the acceleration due to gravity.
引力场强度 g 定义为置于该点的小检验质量所受万有引力与其质量之比：g = F / m。它是一个矢量，方向与该点正检验质量所受引力方向一致——指向场源质量。国际单位是 N kg⁻¹，这等价于 m s⁻²，因此 g 也代表重力加速度。
For a point mass or a spherical body of mass M, the field strength at a distance r from its centre is g = GM / r². On Earth’s surface this gives g ≈ 9.81 N kg⁻¹. In orbit, g is not zero; it is simply less than at the surface and provides the required centripetal acceleration for satellites.
对于一个质点或质量为 M 的球体，在距其中心 r 处的场强为 g = GM / r²。在地球表面，由此计算出的 g 约为 9.81 N kg⁻¹。在轨道上，g 并不为零，只是小于地表值，并为卫星提供所需的向心加速度。
g = GM / r²
3. Gravitational Potential | 引力势
Gravitational potential V at a point in a field is the work done per unit mass in bringing a small test mass from infinity to that point, with the sign taken as negative because gravity does work on the mass as it falls inwards. The defining equation is V = −GM / r, where the zero of potential is taken at infinity. Potential is a scalar quantity and depends only on the position relative to the source mass.
引力势 V 定义为把单位质量从无穷远处移到该点外力所做的功，并取负号，因为向场内移动时引力做正功。定义式为 V = −GM / r，零势点选在无穷远处。引力势是标量，只与相对于场源的位置有关。
Equipotential surfaces are surfaces on which the potential is constant. Around a point mass they form concentric spheres. Moving along an equipotential requires no work against the gravitational field. The gradient of the potential (−dV/dr) gives the field strength g, a relationship that appears in both IB and Edexcel specifications.
等势面是势值处处相等的曲面。点质量周围的等势面是一系列同心球面。沿等势面移动时不需要克服引力场做功。势的梯度（−dV/dr）给出场强 g，IB 与 Edexcel 大纲均要求掌握这一关系。
V = −GM / r
4. Gravitational Potential Energy | 引力势能
The gravitational potential energy U of a system of two point masses (or two spherical bodies) separated by a distance r is U = −GMm / r. Like potential, potential energy is taken to be zero when the masses are infinitely far apart. The negative sign indicates that work must be done against gravity to separate the masses.
相距为 r 的两个质点（或两个球体）所组成的系统的引力势能为 U = −GMm / r。与势一样，我们规定当两物体相距无穷远时势能为零。负号表示要将它们分开必须克服引力做功。
For a satellite of mass m moving from one orbit to another, the change in potential energy ΔU = m (V₂ − V₁). Examiners often ask you to calculate the work done to lift a satellite from the Earth’s surface into orbit; this involves both the change in kinetic and potential energy.
对于从一条轨道转移到另一条轨道的质量为 m 的卫星，势能变化为 ΔU = m (V₂ − V₁)。考官常要求计算将卫星从地表送入轨道所需做的功，此时需同时考虑动能和势能的变化。
U = −GMm / r
5. Kepler’s Laws and Orbital Mechanics | 开普勒定律与轨道力学
Kepler’s three empirical laws describe planetary motion: (1) planets move in ellipses with the Sun at one focus; (2) a line segment joining a planet and the Sun sweeps out equal areas in equal times; (3) the square of the orbital period T is proportional to the cube of the semi‑major axis a: T² ∝ a³. For circular orbits, a is simply the orbital radius r.
开普勒三大定律描述了行星运动：(1) 行星沿椭圆轨道运行，太阳位于一个焦点上；(2) 行星与太阳的连线在相等时间内扫过相等的面积；(3) 轨道周期 T 的平方与半长轴 a 的立方成正比：T² ∝ a³。对于圆轨道，a 就是轨道半径 r。
For a circular orbit, we can derive Kepler’s third law from Newtonian mechanics: setting the gravitational force equal to centripetal force, GMm / r² = m (2π / T)² r, yields T² = (4π² / GM) r³. This relationship allows the determination of planetary masses and the analysis of binary star systems, both of which are common exam contexts.
对于圆轨道，可以从牛顿力学推导出开普勒第三定律：令引力等于向心力，即 GMm / r² = m (2π / T)² r，可得 T² = (4π² / GM) r³。该关系式可用于计算行星质量和分析双星系统，这些都是常见的考题背景。
T² = (4π² / GM) r³
6. Circular Orbits and Satellite Motion | 圆轨道与卫星运动
For a satellite in a stable circular orbit, the gravitational force provides exactly the centripetal force. This lets us derive three crucial expressions: orbital speed v = √(GM / r), angular speed ω = √(GM / r³), and period T = 2π √(r³ / GM). Note that v decreases with increasing r – a counter‑intuitive result for many students.
对于在稳定圆轨道上运行的卫星，引力恰好提供所需要的向心力。由此可以推导出三个关键公式：轨道速度 v = √(GM / r)，角速度 ω = √(GM / r³)，周期 T = 2π √(r³ / GM)。值得注意的是，v 随 r 的增大而减小——这与不少同学的直觉相反。
Geostationary satellites have a special orbit: they must sit in the equatorial plane, move from west to east, and have an orbital period equal to one sidereal day (about 23 h 56 min). Calculation using Kepler’s third law gives an orbital radius of roughly 4.23 × 10⁷ m (height ≈ 3.6 × 10⁷ m). These satellites appear stationary relative to the Earth’s surface, making them ideal for communications and weather monitoring.
地球同步卫星有特殊的轨道要求：必须位于赤道平面，自西向东运行，且轨道周期等于一个恒星日（约 23 小时 56 分）。利用开普勒第三定律可算出轨道半径大约为 4.23 × 10⁷ m（高度约 3.6 × 10⁷ m）。这类卫星相对地面静止，非常适合作通信和气象监测用途。
v = √(GM / r), T = 2π √(r³ / GM)
7. Escape Velocity | 逃逸速度
Escape velocity is the minimum speed an unpowered projectile must have at the surface of a celestial body in order to escape its gravitational field entirely, reaching infinity with zero final speed. Using conservation of energy, initial kinetic energy + initial potential energy = 0: ½ mv² + (−GMm / R) = 0, which gives v = √(2GM / R).
逃逸速度是指物体从天体表面出发、仅靠初始动能且无需额外动力便能完全脱离该天体引力场到达无穷远处的最小速度。利用能量守恒，初动能 + 初势能 = 0，即 ½ mv² + (−GMm / R) = 0，可得 v = √(2GM / R)。
The escape velocity depends only on the mass and radius of the body, not on the mass of the escaping object. For Earth it is approximately 11.2 km s⁻¹, while for the Moon it is only 2.4 km s⁻¹. This explains why rockets require such enormous energy to leave Earth but lunar modules needed much less fuel to return from the Moon’s surface.
逃逸速度只取决于天体本身的质量和半径，与逃逸物体的质量无关。地球的逃逸速度约为 11.2 km s⁻¹，而月球的仅有 2.4 km s⁻¹。这就解释了为什么火箭需要携带巨大能量才能离开地球，而登月舱从月球表面返回时所需燃料少得多。
v = √(2GM / R)
8. Energy of an Orbiting Satellite | 轨道卫星的能量
For a satellite in a circular orbit, the kinetic energy K = ½ mv² = GMm / (2r), while the potential energy U = −GMm / r. Hence the total mechanical energy E = K + U = −GMm / (2r). The negative total energy indicates that the satellite is bound to the central body; energy must be supplied to remove it to infinity.
在圆轨道上，卫星的动能为 K = ½ mv² = GMm / (2r)，势能为 U = −GMm / r。因此总机械能 E = K + U = −GMm / (2r)。总能量为负表明卫星被束缚于中心天体；要把它移至无穷远必须向其提供能量。
A transfer from a lower orbit to a higher orbit requires an increase in total energy, even though the kinetic energy actually decreases. This counter‑intuitive point is frequently tested. In elliptical orbits the total energy is given by E = −GMm / (2a), where a is the semi‑major axis; the circular formula is therefore just a special case with a = r.
从低轨道转移到高轨道需要增加总能量，尽管动能实际上会减小——这个反直觉的结论经常受到考查。在椭圆轨道中，总能量由 E = −GMm / (2a) 给出，其中 a 是半长轴；圆轨道公式只是当 a = r 时的特殊情况。
E = −GMm / (2r) (circular orbit)
9. Weightlessness in Orbit | 轨道中的失重现象
Astronauts in a space station experience apparent weightlessness not because there is no gravity – in low Earth orbit g is still about 90 % of its surface value – but because they, and the station, are freely falling together around the Earth. The floor of the station provides no reaction force, so they feel weightless.
空间站中的宇航员感受到失重现象，并非因为那里没有引力——在近地轨道，g 仍约为地表值的 90%——而是因为他们和空间站一起在引力作用下绕地球自由下落。空间站地板不提供支持力，因此他们感觉不到重量。
Apparent weight can be understood from the normal contact force measured by a scale. When the centripetal acceleration is exactly provided by gravity, the contact force drops to zero. This same principle explains why objects in a falling lift appear weightless. Both exam boards expect you to connect these scenarios with Newton’s laws and circular motion.
视重可以通过弹簧秤测得的支持力来理解。当引力恰好提供全部向心加速度时，支持力降为零。同样的原理也解释了失重电梯中的现象。两个考试局都要求你能将这些情景与牛顿定律和圆周运动联系起来。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Avoid using the height of a satellite above Earth’s surface directly as r in gravitational formulas; r must be measured from the centre of the Earth: r = R_earth + h. Double‑check the sign in potential and potential energy expressions – forgetting the negative sign can completely alter your answer and reasoning.
避免直接将卫星距地面的高度代入引力公式中的 r；r 必须从地心量起：r = R_earth + h。注意引力势和势能公式中的负号——忘记负号会彻底改变你的答案和论证。
In geostationary orbit calculations, use the sidereal day (86 164 s), not the solar day (86 400 s), because the Earth rotates once relative to the stars in that shorter period. For energy‑related questions, establish whether the system is isolated and identify the zero reference point of potential energy before writing any equation.
在同步卫星计算中，务必使用恒星日（86 164 秒）而非太阳日（86 400 秒），因为地球相对于恒星自转一圈的周期更短。对于能量类问题，在列方程前先判断系统是否孤立，并确认势能零参考点。
Practise deriving Kepler’s third law and satellite speed equations from first principles, because many exam questions start from GMm / r² = m v² / r. Finally, keep track of units – for example, converting km to m and days to seconds – and always ask yourself whether your numerical answer makes physical sense, such as whether an orbital period is plausible for a given radius.
要练习从基本原理推导开普勒第三定律和卫星速度方程，因为很多考题都是从 GMm / r² = m v² / r 入手的。最后，要注意单位转换——比如把 km 转换为 m，把天转换为秒——并始终问自己：计算出的数值在物理上是否合理，比如给定半径后轨道周期是否与预期相符。
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