Orbiting under Gravity | 引力作用下的轨道运动

📚 Orbiting under Gravity | 引力作用下的轨道运动

When a satellite, moon or planet moves in a curved path around a large mass, gravity supplies the force that keeps it in orbit. A-Level CIE Physics expects you to combine Newton’s law of gravitation with circular motion, energy conservation and Kepler’s laws. This article builds the key ideas step by step.

当卫星、月球或行星绕着大质量天体沿曲线路径运动时，引力提供了维持轨道所需的力。A-Level CIE 物理要求你把牛顿万有引力定律与圆周运动、能量守恒和开普勒定律结合起来。本文逐步构建这些核心概念。

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

Newton’s law states that any two point masses attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of their separation.

牛顿万有引力定律指出，任何两个质点都以一种力相互吸引，该力与它们质量的乘积成正比，与它们距离的平方成反比。

F = GMm / r²

Here G is the universal gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻². The force acts along the line joining the two centres of mass.

其中 G 是万有引力常量，6.67 × 10⁻¹¹ N m² kg⁻²。力的方向沿两个质心连线。

At the surface of a planet, the gravitational field strength g is given by g = GM / R², where R is the planet’s radius. This links universal gravitation to the familiar weight W = mg.

在行星表面，引力场强度 g 由 g = GM / R² 给出，其中 R 是行星半径。这把万有引力与常见的重力 W = mg 联系了起来。

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2. Circular Orbits and Centripetal Force | 圆周轨道与向心力

For a satellite in a circular orbit, the only force acting on it is gravity. This gravitational pull must provide the centripetal force required for circular motion.

对于沿圆周轨道运行的卫星，作用在它身上的唯一力是引力。这个引力必须提供圆周运动所需的向心力。

GMm / r² = mv² / r

The satellite’s mass m cancels, showing that orbital motion depends only on the central mass M and the orbital radius r.

卫星质量 m 会被消去，这说明轨道运动只取决于中心天体质量 M 和轨道半径 r。

Since the centripetal acceleration is a = v² / r, we also get a = GM / r². This is the same as the gravitational field strength at that radius.

由于向心加速度 a = v² / r，我们也可得到 a = GM / r²。这与该半径处的引力场强度相同。

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3. Orbital Speed | 轨道速度

Rearranging GMm / r² = mv² / r gives the orbital speed for a circular orbit:

将 GMm / r² = mv² / r 重新整理，可得到圆周轨道的轨道速度：

v = √(GM / r)

This result shows that orbital speed decreases as the orbital radius increases. A satellite closer to Earth must travel faster than one farther away.

这一结果表明，轨道速度随轨道半径增大而减小。离地球较近的卫星必须比较远的卫星运动得更快。

Orbital speed does not depend on the satellite’s mass. All objects at the same radius around the same central body have the same orbital speed, regardless of their mass or size.

轨道速度不依赖于卫星的质量。绕同一中心天体、处于同一半径的所有物体，无论质量或大小如何，都具有相同的轨道速度。

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4. Orbital Period and Kepler’s Third Law | 轨道周期与开普勒第三定律

The orbital period T is the time for one complete revolution. Since the circumference is 2πr and speed is constant, T = 2πr / v.

轨道周期 T 是完成一整圈所需的时间。由于圆周长为 2πr 且速率恒定，T = 2πr / v。

Substituting v = √(GM / r) and squaring both sides gives the relation known as Kepler’s third law for circular orbits:

代入 v = √(GM / r) 并将两边平方，可得到圆周轨道上的开普勒第三定律关系式：

T² = 4π²r³ / GM

For any set of satellites orbiting the same central mass M, the ratio T² / r³ is constant. This means a satellite in a larger orbit has a longer period.

对于绕同一中心天体 M 运行的一组卫星，比值 T² / r³ 是恒定的。这意味着轨道半径越大的卫星周期越长。

Kepler’s original laws apply to elliptical orbits, but circular orbits are a special case. CIE questions often ask you to derive this equation by equating gravitational and centripetal forces.

开普勒原始定律适用于椭圆轨道，但圆周轨道是一个特例。CIE 考题经常要求你通过令引力和向心力相等来推导这个方程。

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5. Energy in Orbits | 轨道能量

A satellite in orbit has both kinetic energy and gravitational potential energy. For a circular orbit, the kinetic energy is K = ½mv² = GMm / 2r.

轨道上的卫星同时具有动能和引力势能。对于圆周轨道，动能 K = ½mv² = GMm / 2r。

The gravitational potential energy is taken as zero at infinity, so at a distance r it is negative: U = -GMm / r.

引力势能在无穷远处取为零，因此在距离 r 处为负值：U = -GMm / r。

E_total = K + U = -GMm / 2r

The total energy is negative, meaning the satellite is bound to the central body. To move to a higher orbit, energy must be supplied; the total energy becomes less negative.

总能量为负，说明卫星被中心天体束缚。要移动到更高轨道，必须提供能量；总能量会变得不那么负。

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6. Geostationary Orbits | 地球同步轨道

A geostationary satellite appears fixed above one point on the Earth’s equator. It must have a period of

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