📚 Applications of Newton’s Law of Universal Gravitation | 牛顿万有引力定律的应用
Newton’s Law of Universal Gravitation is one of the cornerstones of classical physics and a perennial favourite in A-Level examinations. This article systematically explores how this law is applied to solve problems involving planetary motion, satellite orbits, gravitational field strength, and the determination of celestial masses — all essential skills for exam success.
牛顿万有引力定律是经典物理学的基石之一，也是A-Level考试中的常青考点。本文系统探讨如何运用这一定律解决行星运动、卫星轨道、引力场强度及天体质量测定等问题——这些都是考试取得高分的关键技能。
1. The Law Itself | 定律本身
Newton’s Law of Universal Gravitation states that every point mass in the universe 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. The mathematical form is:
牛顿万有引力定律指出：宇宙中每个质点均以与两质量乘积成正比、与两质点中心距离的平方成反比的力吸引其他质点。其数学表达式为：
F = G·m₁·m₂ / r²
where G is the universal gravitational constant, 6.67 × 10⁻¹¹ N·m²·kg⁻². The force acts along the line joining the two masses, and each mass experiences a force of equal magnitude but opposite direction — satisfying Newton’s Third Law.
其中G为万有引力常量，数值为6.67 × 10⁻¹¹ N·m²·kg⁻²。引力沿两质量连线方向作用，两质量各自受到的力大小相等、方向相反——满足牛顿第三定律。
2. Gravitational Field Strength | 引力场强度
Gravitational field strength, denoted g, is defined as the gravitational force per unit mass acting on a small test mass placed at that point. For a point mass M, the field strength at distance r is:
引力场强度（符号g）定义为单位质量在引力场中某点所受的引力。对于质量为M的质点，距离r处的场强为：
g = F/m = G·M / r²
This equation shows that g decreases with the square of distance and is independent of the test mass. On Earth’s surface, this yields the familiar value of approximately 9.81 N·kg⁻¹. Note that field strength is a vector quantity directed towards the mass creating the field.
该方程表明g随距离平方递减，且与测试质量无关。在地球表面，代入数据可得熟悉的9.81 N·kg⁻¹。注意场强是矢量，方向指向产生引力场的质量。
3. Deriving Kepler’s Third Law | 推导开普勒第三定律
For a planet or satellite in a circular orbit around a central mass M, the gravitational force provides exactly the centripetal force required for circular motion. Equating these two forces yields Kepler’s Third Law. Begin with:
对于绕中心质量M做匀速圆周运动的行星或卫星，万有引力恰好提供圆周运动所需的向心力。令两者相等即可推导出开普勒第三定律。首先写出：
G·M·m / r² = m·v² / r
Since v = 2πr/T for circular motion, substituting and simplifying gives:
因为匀速圆周运动中v = 2πr/T，代入并化简得：
T² = (4π² / G·M) × r³
This is Kepler’s Third Law in Newtonian form: the square of the orbital period is proportional to the cube of the orbital radius. The proportionality constant depends only on the mass of the central body. This derivation is a classic exam question — memorise every step.
这即牛顿形式的开普勒第三定律：轨道周期的平方与轨道半径的立方成正比。比例常数仅取决于中心天体质量。此推导是经典考题——务必牢记每一步。
4. Determining the Mass of a Central Body | 测定中心天体质量
Rearranging Kepler’s Third Law allows us to determine the mass of the central body, provided we know the orbital period and radius of a satellite orbiting it:
重新整理开普勒第三定律，可在已知卫星轨道周期和轨道半径的前提下，测定中心天体的质量：
M = 4π²·r³ / (G·T²)
This method is used to calculate the mass of the Sun from Earth’s orbital data (r = 1.50 × 10¹¹ m, T = 3.16 × 10⁷ s), yielding M ≈ 1.99 × 10³⁰ kg. Similarly, by observing the orbital motion of the Moon, one can determine the mass of the Earth. This technique is also applied to measure the masses of planets with natural satellites, and even black holes by studying orbiting companion stars.
此方法可通过地球轨道数据（r = 1.50 × 10¹¹ m，T = 3.16 × 10⁷ s）计算太阳质量，得M ≈ 1.99 × 10³⁰ kg。同样，通过观测月球运动可测定地球质量。该技术也用于测定有天然卫星的行星质量，甚至可通过研究伴星轨道来测定黑洞质量。
5. Orbits and Escape Velocity | 轨道与逃逸速度
For an object to escape the gravitational field of a planet permanently, its kinetic energy must at least equal the gravitational potential energy at its initial position. Setting KE = |GPE| gives:
物体要永久脱离行星的引力场，其动能必须至少等于初始位置的引力势能绝对值。令动能等于引力势能绝对值：
½·m·vₑ² = G·M·m / R
Solving for escape velocity yields:
解出逃逸速度得：
vₑ = √(2·G·M / R)
For Earth, using M = 5.97 × 10²⁴ kg and R = 6.37 × 10⁶ m, the escape velocity is approximately 11.2 km/s. Compare this with the orbital velocity of a satellite just above the atmosphere, which is v = √(G·M/R) ≈ 7.9 km/s — note that escape velocity is √2 times the orbital velocity at the same radius.
对地球，代入M = 5.97 × 10²⁴ kg、R = 6.37 × 10⁶ m，逃逸速度约为11.2 km/s。与大气层外卫星的轨道速度v = √(G·M/R) ≈ 7.9 km/s相比——注意同一半径处逃逸速度是轨道速度的√2倍。
6. Satellites: Geostationary vs Polar | 卫星：地球同步与极地轨道
A geostationary satellite orbits above the Equator with a period of 24 hours, matching Earth’s rotation. Consequently, it remains at a fixed position relative to a point on the Equator. From the equation T² = (4π²/GM)r³, the geostationary orbital radius is approximately 4.23 × 10⁷ m, corresponding to an altitude of about 3.58 × 10⁷ m above the surface.
地球同步卫星在赤道正上方运行，周期为24小时，与地球自转角速度相同。因此它相对于赤道上某点保持固定位置。由T² = (4π²/GM)r³计算，同步轨道半径约为4.23 × 10⁷ m，对应距地表高度约3.58 × 10⁷ m。
Geostationary satellites are ideal for telecommunications, broadcasting and weather monitoring because they are always visible from the same ground station. Polar-orbiting satellites, in contrast, pass over the poles at lower altitudes, completing a full orbit in roughly 90-100 minutes, providing complete coverage of the Earth as the planet rotates beneath them — essential for mapping and climate research.
同步卫星因始终位于同一地面站正上方，非常适合通信、广播和气象监测。相比之下，极地卫星在较低高度穿越两极，约90-100分钟完成一次轨道运行，随着地球在下方自转，可实现对全球的完整覆盖——对测绘和气候研究至关重要。
7. Weightlessness and Apparent Weight | 失重与视重
Astronauts in orbit experience apparent weightlessness — but is gravity actually absent? Not at all. At the orbital altitude of the International Space Station (approximately 400 km), the gravitational field strength is about 8.7 N·kg⁻¹, roughly 89% of the surface value. Astronauts appear weightless because both they and the spacecraft are in continuous free fall around the Earth; the normal reaction force from their surroundings is zero.
轨道中的宇航员经历视重为零——但引力真的消失了吗？完全没有。在国际空间站轨道高度（约400 km）处，引力场强度约为8.7 N·kg⁻¹，约为地表值的89%。宇航员看似失重，是因为他们与航天器都在围绕地球持续自由下落；周围环境对它们的支持力为零。
Gravitational potential energy at a distance r from a mass M is given by V = −G·M·m/r. The negative sign indicates that the potential energy is zero at infinity and decreases (becomes more negative) as objects approach the mass. This concept is essential for solving energy-based problems involving satellite launches and orbital transfers.
距离质量M为r处的引力势能为V = −G·M·m/r。负号表示无穷远处势能为零，且物体靠近质量时势能降低（变得更负）。这一概念对求解卫星发射与轨道变轨的能量问题至关重要。
8. Variation of g | 重力加速度的变化
As altitude increases from Earth’s surface, the local gravitational field strength decreases according to:
当高度从地表增加时，局部引力场强度按以下规律递减：
g(r) = G·M / r²
At height h above the surface, r = R + h. Doubling the distance from Earth’s centre quarter the field strength. At the Earth’s equator, g is slightly less than at the poles because the equatorial radius is larger (the Earth is an oblate spheroid), and also because of the small centrifugal effect due to rotation. These variations, though small, can appear in exam multiple-choice questions.
在距地表高度h处，r = R + h。距地球中心距离加倍，场强变为四分之一。赤道上的g略小于两极，因为赤道半径更大（地球是扁球体），且自转产生微小的离心效应。这些差异虽小，却可能出现在考试选择题中。
9. Worked Example: Finding g on a Distant Planet | 例题：求遥远行星上的g
A planet has mass 3.0 × 10²³ kg and radius 2.0 × 10⁶ m. Calculate the gravitational field strength on its surface, and hence the weight of a 5.0 kg mass on that planet.
某行星质量3.0 × 10²³ kg，半径2.0 × 10⁶ m。求其表面引力场强度，以及该行星上5.0 kg物体所受的重力。
Step 1 — Write down the equation: g = G·M / r²
Step 2 — Substitute values: g = (6.67 × 10⁻¹¹ × 3.0 × 10²³) / (2.0 × 10⁶)²
Step 3 — Calculate: g = (2.001 × 10¹³) / (4.0 × 10¹²) = 5.0 N·kg⁻¹
Step 4 — For the weight: W = m·g = 5.0 × 5.0 = 25 N
第一步——写出方程：g = G·M / r²
第二步——代入数值：g = (6.67 × 10⁻¹¹ × 3.0 × 10²³) / (2.0 × 10⁶)²
第三步——计算：g = (2.001 × 10¹³) / (4.0 × 10¹²) = 5.0 N·kg⁻¹
第四步——求重力：W = m·g = 5.0 × 5.0 = 25 N
10. Worked Example: Satellite Orbital Period | 例题：卫星轨道周期
Calculate the orbital period of a satellite orbiting the Earth at an altitude of 300 km. Use Mₑ = 5.97 × 10²⁴ kg and Rₑ = 6.37 × 10⁶ m.
计算距地面300 km高度绕地运行卫星的轨道周期。已知Mₑ = 5.97 × 10²⁴ kg，Rₑ = 6.37 × 10⁶ m。
Orbital radius: r = 6.37 × 10⁶ + 0.30 × 10⁶ = 6.67 × 10⁶ m.
Using T² = 4π²·r³/(G·M): T² = (4 × π² × (6.67 × 10⁶)³) / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴).
T² = (4 × 9.87 × 2.97 × 10²⁰) / (3.98 × 10¹⁴) = 2.95 × 10⁷ s².
Hence T = √(2.95 × 10⁷) ≈ 5.43 × 10³ s = 90.5 minutes.
轨道半径：r = 6.37 × 10⁶ + 0.30 × 10⁶ = 6.67 × 10⁶ m。
由 T² = 4π²·r³/(G·M)：T² = (4 × π² × (6.67 × 10⁶)³) / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)。
T² = (4 × 9.87 × 2.97 × 10²⁰) / (3.98 × 10¹⁴) = 2.95 × 10⁷ s²。
因此 T = √(2.95 × 10⁷) ≈ 5.43 × 10³ s = 90.5分钟。
11. Common Pitfalls | 常见易错点
Students frequently confuse gravitational field strength g and gravitational potential V. Remember: g is a vector, measured in N·kg⁻¹, representing force per unit mass; V is a scalar, measured in J·kg⁻¹, representing energy per unit mass. They are related by g = −dV/dr.
学生经常混淆引力场强度g与引力势V。切记：g是矢量，单位N·kg⁻¹，表示单位质量的力；V是标量，单位J·kg⁻¹，表示单位质量具有的能量。二者的关系为g = −dV/dr。
Another frequent error involves distances: always measure r from the centre of the celestial body, not from its surface. When a question provides altitude, you must add the radius of the body to obtain the true r. Additionally, remember that gravitational force acts between all masses in the universe — not merely between celestial bodies — and that inside a uniform spherical shell, the net gravitational field is zero (verified experimentally through undergraduate-level analyses).
另一个常见错误涉及距离：r始终从天体中心量起，而非从表面。当题目给出高度时，必须加上天体半径才能得到真正的r。此外，引力作用于宇宙中所有质量之间——不仅限于天体——并且在均匀球壳内部，净引力场为零（由大学阶段的分析可验证）。
12. Summary of Key Equations | 关键公式总结
The table below consolidates the most important equations from this article for quick revision:
下表汇总本文最重要的公式，便于快速复习：
| Physical Quantity | Equation | Notes |
| Gravitational force | F = G·m₁·m₂ / r² | Universal law |
| Field strength | g = G·M / r² | Also g = F/m |
| Kepler’s Third Law | T² = 4π²·r³ / (G·M) | For circular orbits |
| Central mass | M = 4π²·r³ / (G·T²) | Requires T and r |
| Escape velocity | vₑ = √(2·G·M / R) | √2 × orbital velocity |
| Orbital velocity | v = √(G·M / r) | For circular orbits |
| Gravitational potential | V = −G·M / r | Scalar, negative at finite r |
Consistent practice in deriving these equations from first principles will build both confidence and speed in examinations. When tackling gravitational problems, first identify whether the situation involves forces (F = Gm₁m₂/r²), fields (g = GM/r²), energy (V = −GM/r), or orbital mechanics (T² ∝ r³) — clear classification will guide you directly to the correct approach.
持续练习从第一性原理推导这些公式，将同时增强考试中的信心与速度。解题时，先判断情境涉及的是力（F = Gm₁m₂/r²）、场（g = GM/r²）、能量（V = −GM/r），还是轨道力学（T² ∝ r³）——清晰的分类将引导你直达正确解法。
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