Applications of Newton's Law of Gravitation | 万有引力定律的应用

📚 Applications of Newton’s Law of Gravitation | 万有引力定律的应用

Newton’s law of gravitation is one of the cornerstones of classical physics and a recurring theme in A-Level examinations. It explains not only the motion of planets and satellites, but also provides the mathematical framework for determining planetary mass, orbital velocity, escape speed, and the behaviour of objects in orbit. This article covers every key application required for your exam, with equations presented in a clear, non-LaTeX format.

万有引力定律是经典物理学的基石之一，也是 A-Level 考试中反复出现的核心主题。它不仅解释了行星和卫星的运动，还为我们提供了测定天体质量、计算环绕速度、逃逸速度以及理解轨道中物体行为的数学框架。本文将逐一讲解考试所需的所有关键应用，所有公式均以清晰的非 LaTeX 格式呈现。

1. Kepler’s Laws of Planetary Motion | 开普勒行星运动定律

Before Newton formulated his law of universal gravitation, Johannes Kepler derived three empirical laws describing the motion of planets around the Sun. These laws are essential for understanding orbital mechanics and are often tested in A-Level physics papers.

在牛顿提出万有引力定律之前，开普勒已经总结出了描述行星绕太阳运动的三大经验定律。这三条定律是理解轨道力学的关键，也是 A-Level 物理试卷中的常考内容。

First Law (Law of Ellipses): Every planet moves in an elliptical orbit with the Sun at one focus.

第一定律（椭圆定律）： 每一颗行星都沿椭圆轨道运动，太阳位于椭圆的一个焦点上。

Second Law (Law of Equal Areas): A line joining a planet and the Sun sweeps out equal areas in equal intervals of time. This implies that planets move faster when they are closer to the Sun (perihelion) and slower when farther away (aphelion).

第二定律（面积定律）： 行星与太阳的连线在相等时间内扫过相等的面积。这意味着行星在靠近太阳（近日点）时运动较快，在远离太阳（远日点）时运动较慢。

Third Law (Law of Periods): The square of the orbital period T is proportional to the cube of the semi-major axis r of the orbit:

T² ∝ r³

For a circular orbit, this becomes T² = (4π² / GM) × r³, where G is the gravitational constant (6.674 × 10⁻¹¹ N·m²·kg⁻²) and M is the central mass. The third law is particularly useful because it provides a direct method for measuring the mass of the central body.

第三定律（周期定律）： 行星公转周期的平方与轨道半长轴的立方成正比：

T² ∝ r³

对于圆形轨道，该式可写为 T² = (4π² / GM) × r³，其中 G 为万有引力常量（6.674 × 10⁻¹¹ N·m²·kg⁻²），M 为中心天体的质量。第三定律特别有用，因为它提供了一种直接测定中心天体质量的方法。

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

Newton’s law of universal gravitation states that any two masses attract each other 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 of mass.

牛顿万有引力定律指出：任意两个物体之间都彼此吸引，引力的大小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 = Gm₁m₂ / r²

Key points for the exam:

考试关键要点：

The gravitational force is always attractive, acting along the line joining the two masses.
The force is mutual: the force on m₁ due to m₂ is equal in magnitude and opposite in direction to the force on m₂ due to m₁ (Newton’s third law).
G is a universal constant (6.674 × 10⁻¹¹ N·m²·kg⁻²), the same everywhere in the universe.
The distance r is measured from the centre of mass of each object — for uniform spheres, this is the distance between their centres.

万有引力始终是吸引力，方向沿两物体质心的连线。
力是相互的：m₁ 对 m₂ 的引力与 m₂ 对 m₁ 的引力大小相等、方向相反（牛顿第三定律）。
G 是普适常量（6.674 × 10⁻¹¹ N·m²·kg⁻²），在宇宙任何地方都相同。
距离 r 从每个物体的质心量起——对于均匀球体，r 就是两球心之间的距离。

3. Gravitational Field Strength | 引力场强度

Gravitational field strength g at a point is defined as the gravitational force per unit mass acting on a small test mass placed at that point. It is a vector quantity pointing toward the centre of the mass creating the field.

引力场强度 g 的定义为：放在某一点上的小测试质量所受到的引力与测试质量之比。g 是矢量，方向指向产生该引力场的质量中心。

g = F / m

For a point mass or a spherical body of mass M, the gravitational field strength at a distance r from its centre is:

对于质量为 M 的质点或球体，在距其中心距离为 r 处的引力场强度为：

g = GM / r²

On the surface of the Earth, this gives g₀ = GMₑ / Rₑ² ≈ 9.81 m/s², where Mₑ is the mass of the Earth and Rₑ is its radius. A common exam question involves determining g at a height h above the Earth’s surface:

在地球表面，该式给出 g₀ = GMₑ / Rₑ² ≈ 9.81 m/s²，其中 Mₑ 是地球质量，Rₑ 是地球半径。一个常见考题是求地球表面上方高度 h 处的引力场强度：

g = GMₑ / (Rₑ + h)²

Notice how g decreases with the square of the distance from Earth’s centre. When h is small compared with Rₑ, the change in g is negligible — this is why g is often treated as constant near the Earth’s surface.

注意 g 随距地心距离的平方而减小。当 h 远小于 Rₑ 时，g 的变化可以忽略不计——这就是为什么在地表附近 g 常被视为常数的原因。

4. Motion of Satellites in Circular Orbits | 卫星的圆周轨道运动

For a satellite of mass m in a circular orbit of radius r around a central mass M, the gravitational force provides the required centripetal force. This is one of the most frequently tested applications in A-Level physics.

对于质量为 m、环绕中心天体 M 作半径为 r 的圆周运动的卫星，万有引力提供了所需的向心力。这是 A-Level 物理中最常考的应用之一。

GMm / r² = mv² / r

Cancelling m from both sides and solving for v, we obtain the orbital velocity:

消去等式两边的 m，解出 v 即得到环绕速度：

v = √(GM / r)

Three important conclusions follow from this equation:

The orbital speed is independent of the satellite’s mass.
The orbital speed depends only on the central mass M and the orbital radius r.
As r increases, the orbital speed decreases.

从该方程可以得出三个重要结论：

环绕速度与卫星本身的质量无关。
环绕速度只取决于中心天体质量 M 和轨道半径 r。
r 越大，环绕速度越小。

The orbital period can be derived by combining v = 2πr / T with v = √(GM / r):

将 v = 2πr / T 与 v = √(GM / r) 联立，即可推导出轨道周期：

T = 2π √(r³ / GM)

This is the equation form of Kepler’s third law for circular orbits, and it shows that a satellite at a higher altitude has a longer orbital period.

这实质上就是圆形轨道下开普勒第三定律的方程形式，表明轨道高度越高的卫星，其运行周期越长。

5. Weightlessness and Free Fall in Orbit | 失重与轨道中的自由落体

A satellite in orbit is in continuous free fall toward the Earth. The satellite and everything inside it share the same acceleration; consequently, objects inside the satellite appear weightless. This does not mean gravity is absent — the gravitational force is still acting, but it is entirely used to maintain circular motion.

轨道上的卫星实际上一直在向地球作自由落体运动。卫星及其内部的一切物体拥有相同的加速度，因此卫星内部的物体看起来处于“失重”状态。这并不意味着引力不存在——引力仍然在起作用，只是它被完全用于维持圆周运动了。

An important exam distinction:

一个重要的考试辨析：

True weightlessness: occurs when gravitational force is truly zero, e.g. in deep space far from any mass.
Apparent weightlessness: occurs when the normal reaction is zero, e.g. inside an orbiting spacecraft. The astronaut still experiences gravity, but no normal force is exerted by the floor.

真正的失重： 指引力确实为零的情况，例如远离任何天体的深空。
表观失重： 指支持力为零的情况，例如在轨道飞行的航天器内部。宇航员仍然受到引力作用，但地板不再对他施加支持力。

From the perspective of the astronaut, objects float because the spacecraft and the objects within it are accelerating at exactly the same rate. A common exam question asks you to explain this phenomenon using the concept of centripetal acceleration.

从宇航员的视角来看，物体飘浮是因为航天器及其中所有物体的加速度完全相同。常见考题会要求考生用向心加速度的概念解释这一现象。

6. Geostationary Satellites | 地球同步静止卫星

Geostationary satellites are satellites that remain fixed above a specific point on the Earth’s equator. They are widely used for telecommunications, broadcasting, and weather monitoring.

地球同步静止卫星是指固定在地球赤道某一地点正上方的卫星，广泛应用于通信、广播和气象监测。

For a satellite to be geostationary, it must satisfy three conditions:

一颗卫星要做到地球同步静止，必须满足三个条件：

It must orbit in the same direction as the Earth’s rotation (west to east).
Its orbital period must equal the Earth’s rotational period (T = 24 h = 86,400 s).
It must orbit directly above the equator in the equatorial plane.

公转方向必须与地球自转方向相同（自西向东）。
轨道周期必须等于地球自转周期（T = 24 小时 = 86,400 秒）。
轨道必须位于赤道平面内、赤道的正上方。

Using T = 2π √(r³ / GM), we can solve for the geostationary orbital radius:

利用 T = 2π √(r³ / GM)，可以解出地球同步静止卫星的轨道半径：

r³ = GMT² / 4π²

Substituting Mₑ = 5.97 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N·m²·kg⁻², and T = 86,400 s gives:

代入 Mₑ = 5.97 × 10²⁴ kg、G = 6.674 × 10⁻¹¹ N·m²·kg⁻² 和 T = 86,400 s，可得：

r ≈ 4.23 × 10⁷ m ≈ 42,300 km

This corresponds to an altitude of approximately 35,800 km above the Earth’s surface (subtracting the Earth’s radius of 6,370 km). The corresponding orbital speed is about 3.07 km/s. These numerical values are frequently required in exam calculations.

这对应卫星距地表约 35,800 km 的高度（减去地球半径 6,370 km 后）。对应的环绕速度约为 3.07 km/s。这些数值在考试计算中经常用到。

7. Determining the Mass of a Planet or Star | 测定行星或恒星的质量

Astronomers cannot directly weigh celestial bodies, but they can determine their masses by observing the motion of objects around them. If a moon or probe orbits a planet with period T at orbital radius r, then from Newton’s law and Kepler’s third law:

天文学家无法直接给天体“称重”，但可以通过观察绕其运动物体的轨道来确定天体的质量。若一颗卫星或月球以周期 T 和轨道半径 r 绕某行星运动，则根据牛顿定律和开普勒第三定律：

M = 4π²r³ / (GT²)

This method works for any system where we can observe a smaller body orbiting a larger one — planets orbiting the Sun, moons orbiting planets, or stars orbiting a common centre in binary systems. In exam questions, you will typically be given three of the four quantities (M, r, T, and g) and asked to find the fourth.

这一方法适用于任何可观察到小天体绕大天体运动的系统——如行星绕太阳、卫星绕行星、或双星系统中的恒星绕共同中心运动。在考题中，通常会给定 M、r、T、g 四个量中的三个，要求求解第四个。

If the gravitational field strength at the surface is known, we can also use g₀ = GM / R² to find the mass:

如果已知天体表面的引力场强度，也可用 g₀ = GM / R² 来求天体质量：

M = g₀R² / G

This is often used for the Earth, where g₀ ≈ 9.81 m/s² and Rₑ = 6.37 × 10⁶ m, giving Mₑ ≈ 5.97 × 10²⁴ kg. The same principle can be applied to find the Sun’s mass using Earth’s orbital radius and period.

该方法常用于计算地球质量，g₀ ≈ 9.81 m/s²、Rₑ = 6.37 × 10⁶ m 时，得出 Mₑ ≈ 5.97 × 10²⁴ kg。同样的原理也可以利用地球的轨道半径和公转周期来计算太阳的质量。

8. Gravitational Potential Energy and Total Energy | 引力势能与总能量

For a mass m at a distance r from a central mass M, the gravitational potential energy is:

对于距中心天体 M 距离为 r 的质量 m，其引力势能为：

U = −GMm / r

The negative sign indicates that the gravitational force is attractive — energy must be added to separate the two masses. The zero of gravitational potential energy is defined at infinity (r → ∞), where the gravitational force is negligible.

负号表示万有引力是吸引力——要将两个物体分开，必须对其做功、输入能量。引力势能的零值定义在无穷远处（r → ∞），因为那里万有引力趋于零。

The total mechanical energy of a satellite in a circular orbit is the sum of its kinetic and potential energy:

圆形轨道上卫星的总机械能是动能与势能之和：

E = K + U = −GMm / 2r

This result comes from the virial theorem: for stable circular orbits, K = −U/2. In other words, the kinetic energy is exactly half the magnitude of the potential energy. The total energy is negative, indicating that the satellite is bound to the central mass and cannot escape unless energy is added to it.

这一结果来自位力定理：在稳定圆周轨道中，K = −U/2。换句话说，动能恰好是势能大小的一半。总能量为负值，说明卫星被束缚于中心天体，若要逃离则必须获得额外能量。

9. Escape Velocity | 逃逸速度

The escape velocity is the minimum speed an object must have at the surface of a planet (or at a given distance from its centre) to escape its gravitational field entirely, reaching infinity with zero residual speed.

逃逸速度是指物体在行星表面（或距中心某一给定距离处）脱离行星引力场所需的最小速度，即到达无穷远处时速度恰好为零。

Setting the total energy of the escaped object at infinity to zero:

令该物体抵达无穷远处时的总能量为零：

½mv² − GMm / R = 0

Solving for v gives the escape velocity:

解出 v 即得逃逸速度：

vₑₛ꜀ = √(2GM / R)

Notice that escape velocity is √2 times larger than the orbital velocity at the same radius. For the Earth’s surface, vₑₛ꜀ ≈ 11.2 km/s. For the Moon, which has much smaller mass, the escape velocity is only about 2.4 km/s — which is why the Moon cannot retain an atmosphere.

注意：同一位置的逃逸速度是环绕速度的 √2 倍。在地球表面，vₑₛ꜀ ≈ 11.2 km/s。月球质量小得多，其逃逸速度仅约 2.4 km/s——这正是月球无法保持大气层的原因。

10. Energy Changes During Orbit Transfer | 轨道转移过程中的能量变化

When a satellite moves from a lower orbit to a higher orbit, work must be done against gravity. This is a common exam context involving multiple spacecraft launches and orbital manoeuvres.

当卫星从低轨道向高轨道转移时，必须克服引力做功。这是涉及航天器发射和轨道机动的一类常见考题情境。

If the orbital radius changes from r₁ to r₂ (where r₂ > r₁):

若轨道半径从 r₁ 变为 r₂（其中 r₂ > r₁）：

ΔE = E₂ − E₁ = (−GMm / 2r₂) − (−GMm / 2r₁)

ΔE = GMm / 2 × (1/r₁ − 1/r₂) > 0

This positive energy input comes from the rocket engines. Note that while the total energy increases, the kinetic energy actually decreases because the orbital speed v = √(GM / r) is smaller at higher altitude. The increase in potential energy more than compensates for the loss in kinetic energy.

这个正值能量由火箭发动机提供。注意：虽然总能量增加了，动能实际上反而减少了，因为较高轨道的环绕速度 v = √(GM / r) 更小。势能的增加量足以补偿动能的减少量。

This counter-intuitive result — that firing a rocket forwards can cause a satellite to decelerate — is often tested conceptually. Students should be able to explain it using energy considerations rather than simple intuition.

这个反直觉的结论——向前点火反而会使卫星减速——是经常考到的概念性问题。考生应当能够从能量的角度来解释它，而非仅凭直觉判断。

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Applications of Newton’s Law of Gravitation | 万有引力定律的应用