📚 Orbiting the Earth | 绕地球轨道运动
An Earth satellite stays in orbit because the gravitational pull of the Earth continuously changes the direction of the satellite's velocity. For a circular orbit, this gravitational force acts as the centripetal force, so the satellite is in a state of free fall around the Earth rather than falling straight down. This single idea - gravity as the centripetal force - unlocks almost every orbital calculation you will meet at A-Level, from the speed of the International Space Station to the period of a geostationary satellite.
地球卫星之所以能保持在轨道上，是因为地球的引力不断改变卫星速度的方向。对于圆轨道，这个引力充当向心力，因此卫星处于绕地球自由下落的状态，而不是竖直掉落。就是这一个核心思想——引力充当向心力——几乎可以解释 A-Level 中所有轨道计算问题，从国际空间站的速度到地球同步卫星的周期。
This article develops that idea step by step: the force equation, orbital speed, period and Kepler's third law, geostationary and polar orbits, gravitational field strength at altitude, weightlessness, orbital energy, escape velocity, worked examples and the most common exam traps.
本文将逐步展开这一思想：受力方程、轨道速度、周期与开普勒第三定律、地球同步轨道与极轨道、高空的引力场强度、失重、轨道能量、逃逸速度、例题以及最常见的考试陷阱。
1. Gravity as the centripetal force | 引力作为向心力
For a satellite of mass m moving in a circular orbit of radius r around the Earth (mass M), Newton's law of gravitation gives the attractive force between the Earth and the satellite. The distance r is measured from the centre of the Earth to the centre of the satellite, not from the Earth's surface.
对于质量为 m 的卫星绕地球（质量为 M）在半径为 r 的圆轨道上运动，牛顿万有引力定律给出地球与卫星之间的吸引力。这里的距离 r 是从地心量到卫星质心，而不是从地面量起。
F = GMm / r²
For a circular path at constant speed v, the resultant force toward the centre of the circle must equal the centripetal force mv²/r. Since gravity is the only force providing this centripetal requirement, we can equate the two expressions.
对于以恒定速率 v 沿圆轨道运动的天体，指向圆心的合力必须等于向心力 mv²/r。由于引力是提供这一向心需求的唯一力，我们可以令两个表达式相等。
GMm / r² = mv² / r
The two facts that matter are therefore: (i) gravity supplies the whole centripetal force, and (ii) the satellite is continuously accelerating towards the Earth even though its speed is constant. Acceleration changes velocity as a vector, so a change of direction alone is an acceleration.
因此有两点最重要：第一，引力提供全部的向心力；第二，卫星虽然速率不变，却一直在向地球方向加速。加速度改变的是作为矢量的速度，所以仅仅方向改变就已经是加速运动。
- G = 6.67 × 10⁻¹¹ N m² kg⁻² (gravitational constant / 万有引力常量)
- M = 5.97 × 10²⁴ kg (mass of the Earth / 地球质量)
- R = 6.37 × 10⁶ m (mean radius of the Earth / 地球平均半径)
2. Orbital speed | 轨道速度
ellite does not appear in the result. A 1 kg CubeSat and a 400-tonne space station at the same orbital radius must move at exactly the same speed.
把 GMm/r² = mv²/r 两边消去 m 后，卫星质量不再出现在结果中。在同一轨道半径上，1 kg 的立方星和 400 吨的空间站必须以完全相同的速度运动。
v = √GM / r
This is the orbital speed for a circular orbit of radius r. It depends only on the mass of the central body and the orbital radius. A larger radius means a smaller speed, so satellites in low Earth orbit move faster than the Moon.
这就是半径为 r 的圆轨道的轨道速度。它只取决于中心天体的质量和轨道半径。半径越大速度越小，因此近地轨道卫星比月球运动得更快。
Combining v = √GM/r with the definition of angular speed ω = v/r gives ω = √GM/r³. The whole orbit therefore has one fixed speed, one fixed angular speed and one fixed period - this is what 'circular orbit' means.
把 v = √GM/r 与角速度的定义 ω = v/r 结合，得 ω = √GM/r³。因此整条轨道只有一个固定速率、一个固定角速度和一个固定周期——这正是"圆轨道"的含义。
A useful alternative form replaces GM with gR², since at the Earth's surface g = GM/R². Then v = R√g/r, which is handy when the data sheet gives g but you cannot remember G and M.
一个有用的替代形式是用 gR² 代入 GM，因为在地球表面 g = GM/R²。于是 v = R√g/r，当数据表给出 g 而你记不住 G 和 M 时非常方便。
3. Orbital period and Kepler's third law | 轨道周期与开普勒第三定律
nce the orbit is circular, the satellite travels a distance 2πr in time T, so v = 2πr / T.
周期 T 是完成一整圈轨道所需的时间。由于轨道是圆的，卫星在时间 T 内经过的距离是 2πr，因此 v = 2πr / T。
Substituting this into v = √GM/r and rearranging gives the period of a circular orbit:
把它代入 v = √GM/r 并整理，得到圆轨道的周期公式：
T = 2π √r³ / GM
Squaring both sides produces the relation known as Kepler's third law for circular orbits:
两边平方，就得到圆轨道情形下的开普勒第三定律：
T² / r³ = 4π² / GM = constant
body, T² / r³ has the same value. Doubling the orbital radius multiplies the period by 2³ᐟ² ≈ 2.83, so higher orbits are much slower as well as longer.
对于绕同一中心天体运动的所有卫星和行星，T² / r³ 的值相同。轨道半径变为两倍，周期就变为 2³ᐟ² ≈ 2.83 倍，所以高轨道不仅更长，而且慢得多。
| Quantity | 物理量 | Expression | 表达式 | Depends on | 依赖于 |
| Orbital speed 轨道速度 | v = √GM/r | r only (not m) 只与 r 有关 |
| Period 周期 | T = 2π√r³/GM | r only (not m) 只与 r 有关 |
| Angular speed 角速度 | ω = √GM/r³ | r only 只与 r 有关 |
| Centripetal acceleration 向心加速度 | a = GM/r² | r only 只与 r 有关 |
4. Low Earth orbit and the International Space Station | 近地轨道与国际空间站
about 400 km, so its orbital radius is r = R + h = 6.37 × 10⁶ + 4.0 × 10⁵ ≈ 6.77 × 10⁶ m.
近地轨道（LEO）指距地面大约 200 km 到 2000 km 的高度。国际空间站在约 400 km 高度运行，因此其轨道半径 r = R + h = 6.37 × 10⁶ + 4.0 × 10⁵ ≈ 6.77 × 10⁶ m。
Using v = √GM/r ³ m s⁻¹, that is about 7.7 km s⁻¹ or roughly 27 600 km h⁻¹. The corresponding period is close to 92 minutes.
用 v = √GM/r 并代入 M = 5.97 × 10²⁴ kg，得 v ≈ 7.67 × 10³ m s⁻¹，即约 7.7 km s⁻¹，约合 27 600 km h⁻¹。对应周期约为 92 分钟。
That is why astronauts see about 16 sunrises every day. It is also why LEO satellites need occasional re-boosting: a very thin residual atmosphere still produces a small drag force, which removes energy and slowly lowers the orbit.
这就是为什么宇航员每天能看到约 16 次日出。这也是近地轨道卫星偶尔需要"抬轨"的原因：极其稀薄的残余大气仍会产生微小阻力，消耗能量并缓慢降低轨道。
5. Geostationary and polar orbits | 地球同步轨道与极轨道
A geostationary orbit is a circular equatorial orbit whose period is exactly one sidereal day (about 86 164 s, often approximated as 24 hours) and whose direction matches the Earth's rotation from west to east. The satellite therefore appears fixed above one point on the equator.
地球同步轨道是一条圆形的赤道轨道，其周期恰好等于一个恒星日（约 86 164 s，常近似取 24 小时），且运动方向与地球自西向东的自转一致。因此卫星看起来固定于赤道上空的某一点。
Rearranging T² = 4π²r³/GM gives r = (GMT² / 4π²)1/3 tude is r − R ≈ 3.58 × 10⁷ m, about 36 000 km. The orbital speed is only about 3.1 km s⁻¹.
由 T² = 4π²r³/GM 整理得 r = (GMT² / 4π²)1/3。取 T = 86 164 s，得 r ≈ 4.22 × 10⁷ m，故高度为 r − R ≈ 3.58 × 10⁷ m，约 36 000 km。轨道速度只有约 3.1 km s⁻¹。
- Geostationary: equatorial, period 24 h, moves west to east, used for TV, weather and communications. 地球同步：赤道平面、周期 24 h、自西向东，用于电视、气象和通信。
- Polar orbit: passes over the poles, so the Earth rotates underneath it and the satellite can scan the whole surface. 极轨道：经过两极上空，地球在其下方自转，卫星可扫描整个地表。
- Sun-synchronous polar orbit: crosses the equator at the same local time each pass, ideal for imaging and remote sensing. 太阳同步极轨道：每次经过赤道时地方时相同，适合成像与遥感。
A geostationary orbit must be equatorial, because any inclined orbit would make the satellite drift north and south in the sky. It must also have exactly the right radius, since the period is fixed by r.
地球同步轨道必须位于赤道平面，因为任何倾斜轨道都会使卫星在天空中南北漂移。其半径也必须恰好合适，因为周期完全由 r 决定。
6. Gravitational field strength at altitude | 高空处的引力场强度
e force per unit mass, g = F/m = GM/r². At the surface, r = R, so g = GM/R² = 9.81 N kg⁻¹.
距地心 r 处的引力场强度 g 定义为单位质量所受的力，g = F/m = GM/r²。在地表 r = R，故 g = GM/R² = 9.81 N kg⁻¹。
ength falls as 1/r². At the ISS altitude of 400 km, g is still about 8.7 N kg⁻¹, roughly 89 % of its surface value. Gravity has certainly not disappeared.
在高度 h 处，r = R + h，场强按 1/r² 减小。在国际空间站 400 km 的高度，g 仍约为 8.7 N kg⁻¹，大约是地表值的 89 %。引力绝对没有"消失"。
e, the approximation Δg/g ≈ −2h/R is often used. A 1 km climb reduces g by only about 0.03 %.
对于地表附近较小的升高量，常用近似 Δg/g ≈ −2h/R。上升 1 km 只会让 g 减小约 0.03 %。
7. Weightlessness and free fall | 失重与自由落体
Astronauts on the ISS are described as weightless, yet the gravitational field there is nearly as strong as on the ground. The correct explanation is that the astronaut, the station and everything inside are all falling towards the Earth with the same acceleration, so there is no contact force between the astronaut and the floor.
国际空间站上的宇航员被称为"失重"，但那里的引力场几乎和地面一样强。正确的解释是：宇航员、空间站以及舱内一切物体都以相同的加速度向地球下落，因此宇航员与地板之间没有接触力。
act force N = 0, so the apparent weight is zero even though the true weight GMm/r² is not.
视重是支持力的大小，而不是引力本身。在轨道空间站内部，法向接触力 N = 0，因此视重为零，尽管真实引力 GMm/r² 并不为零。
The same physics explains the feeling of lightness in a lift whose cable has snapped, in a parabolic 'vomit comet' flight, and in a freely falling experiment package. All are examples of free fall, not of the absence of gravity.
同样的物理也解释了电梯缆绳断裂时的轻盈感、抛物线飞行的"失重飞机"，以及自由下落的实验舱。它们全都是自由落体的例子，而不是引力不存在的例子。
8. Energy in a circular orbit | 圆轨道中的能量
The gravitational potential energy of a satellite of mass m at distance r from the centre of the Earth is defined as zero at infinity, so it is negative for a bound orbit:
质量 m 的卫星在距地心 r 处的引力势能规定在无穷远处为零，因此在束缚轨道中它为负值：
Ep = − GMm / r
The kinetic energy follows from v² = GM/r:
动能由 v² = GM/r 得到：
Ek = ½ m v² = GMm / 2r
Adding the two gives the total energy of a circular orbit:
两者相加即得圆轨道的总能量：
Etotal = Ek + Ep = − GMm / 2r
The total energy is negative, as it must be for a bound orbit, and Ek = −½ Ep. A striking consequence is that drag makes satellites speed up: losing energy lowers the orbit, which reduces r and increases v.
总能量为负，这是束缚轨道必须满足的条件，并且 Ek = −½ Ep。一个令人惊讶的推论是：空气阻力反而会让卫星加速——损失能量使轨道降低，r 减小而 v 增大。
9. Escape velocity | 逃逸速度
its total energy to zero at infinity. Setting ½mv² − GMm/R = 0 gives the escape velocity.
要完全逃离地球，从地面发射的物体必须有足够的动能，使其在无穷远处的总能量为零。令 ½mv² − GMm/R = 0，就得到逃逸速度。
vesc = √2GM / R = √2gR ≈ 11.2 km s⁻¹
Escape velocity is independent of the mass and the launch direction of the projectile, and it is √2 times the speed of a circular orbit at the same radius, that is about 41 % larger.
逃逸速度与物体质量和发射方向无关，它是同一半径处圆轨道速度的 √2 倍，即大约大 41 %。
Escape velocity is not the only condition for reaching space; a rocket can escape at any speed if it keeps thrusting, but a ballistic projectile needs 11.2 km s⁻¹ to coast away without further propulsion.
逃逸速度并不是进入太空的唯一条件；火箭只要持续提供推力，任何速度都能脱离地球，但弹道式抛体需要 11.2 km s⁻¹ 才能在不继续推进的情况下滑行离开。
10. Worked examples | 典型例题
Example 1. 0 km altitude. Use r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m and GM = 3.98 × 10¹⁴ m³ s⁻².
例 1. 求在 400 km 高度圆轨道上卫星的轨道速度。取 r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m，GM = 3.98 × 10¹⁴ m³ s⁻²。
v = √GM/r = √3.98 × 10¹⁴ / 6.77 × 10⁶ = √5.88 × 10⁷ ≈ 7.7 × 10³ m s⁻¹.
v = √GM/r = √3.98 × 10¹⁴ / 6.77 × 10⁶ = √5.88 × 10⁷ ≈ 7.7 × 10³ m s⁻¹。
Example 2. Show that a geostationary orbit has radius about 4.2 × 10⁷ m. Use T = 8.62 × 10⁴ s.
例 2. 证明地球同步轨道的半径约为 4.2 × 10⁷ m。取 T = 8.62 × 10⁴ s。
× 10⁴)²) / (4π²) = (3.98 × 10¹⁴ × 7.43 × 10⁹) / 39.5 ≈ 7.49 × 10²² m³, so r ≈ 4.2 × 10⁷ m.
(3.98 × 10¹⁴ × (8.62 × 10⁴)²) / (4π²) = (3.98 × 10¹⁴ × 7.43 × 10⁹) / 39.5 ≈ 7.49 × 10²² m³，故 r ≈ 4.2 × 10⁷ m。
Example 3. ellite has the less negative (higher) total energy and the larger potential energy, but the smaller kinetic energy.
例 3. 比较两个质量相同、半径分别为 r 和 2r 的卫星的总能量。由于 E = −GMm/2r，较远的卫星总能量负得更少（更高），势能更大，但动能更小。
11. Common misconceptions and exam traps | 常见误区与考试陷阱
Misconception 1: there is no gravity in space. Gravity extends to infinity. Satellites stay up because they move sideways fast enough, not because gravity is switched off.
误区 1：太空中没有引力。 引力延伸到无穷远。卫星能"悬"在轨道上是因为它横向速度足够快，而不是因为引力被关闭了。
- Forgetting to add the Earth's radius: orbital radius r = R + h, never just h. 忘记加上地球半径：轨道半径 r = R + h，不能只写 h。
- Using 24 h instead of the sidereal day for a geostationary period - usually accepted, but state your approximation. 用 24 h 代替恒星日计算同步轨道——通常可以接受，但要说明你的近似。
- Writing 'centrifugal force' as a real force in the inertial frame. In the inertial frame there is only gravity, acting inward. 在惯性系中把"离心力"当作真实力。惯性系中只有指向内侧的引力。
- Assuming the heavier satellite needs a higher speed. The mass cancels, so it does not. 误以为更重的卫星需要更大速度。质量被消去，所以并不需要。
- Mixing up escape velocity √2GM/R with orbital speed √GM/r - remember the factor √2. 混淆逃逸速度 √2GM/R 与轨道速度 √GM/r——记住因子 √2。
- Quoting GM with the wrong power of ten: GM ≈ 3.98 × 10¹⁴ m³ s⁻² for the Earth. GM 的幂次写错：地球 GM ≈ 3.98 × 10¹⁴ m³ s⁻²。
Exam technique. earrange symbolically before substituting numbers. This usually earns method marks even if the arithmetic slips.
答题技巧。 先写出受力方程 GMm/r² = mv²/r，再消去 m，然后先用符号整理、最后代入数值。这样即使算错数字，通常也能拿到方法分。
12. Summary and revision checklist | 总结与复习清单
is the centripetal force. From GMm/r² = mv²/r you can derive the speed, the period, Kepler' rd law, the orbital energy and the relationship between escape velocity and orbital speed.
本主题的所有结论都来自同一个起点：对于圆轨道，引力就是向心力。由 GMm/r² = mv²/r 可以推出速度、周期、开普勒第三定律、轨道能量，以及逃逸速度与轨道速度之间的关系。
GMm / r² = mv² / r → v = √GM/r → T = 2π√r³/GM → T²/r³ = 4π²/GM
Ek = GMm/2r, Ep = −GMm/r, Etotal = −GMm/2r, vesc = √2GM/R
- Can you explain why the satellite's speed is constant but its velocity is not? 你能解释为什么卫星速率不变而速度在变吗？
- Can you derive T² = 4π²r³/GM from first principles in under two minutes? 你能在两分钟内从头推导 T² = 4π²r³/GM 吗？
- Can you list the three conditions for a geostationary orbit? 你能列出地球同步轨道的三个条件吗？
- Can you explain weightlessness in terms of the normal contact force? 你能用接触力解释失重吗？
- Can you state why a satellite slows down when its orbit is raised? 你能说明为什么轨道升高时卫星反而变慢吗？
Master these ideas and orbital mechanics becomes one of the most reliable sources of marks in the A-Level physics paper - the physics is elegant, the algebra is short, and the same few equations keep reappearing.
掌握这些内容之后，轨道力学将成为 A-Level 物理考卷中最稳定的得分点之一——物理图像优美、代数简洁，而且总是那几个公式反复出现。
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