Circular Motion and Gravitation | 圆周运动与引力

📚 Circular Motion and Gravitation | 圆周运动与引力

Circular motion and gravitation are fundamental topics in IB Physics, bridging kinematics, dynamics and celestial mechanics. From the path of a car rounding a bend to the orbits of planets and satellites, the same essential principles—centripetal force and Newton's law of gravitation—underpin a vast range of phenomena. This article unpacks the key concepts, equations and applications that you need to master for your Cambridge IB Physics examinations.

圆周运动与引力是 IB 物理中的基础课题，将运动学、动力学与天体力学联系起来。从汽车转弯的轨迹到行星和卫星的轨道，同样的核心原理——向心力和牛顿万有引力定律——支撑着极为广泛的现象。本文拆解了你在剑桥 IB 物理考试中必须掌握的关键概念、方程和应用。

1. Uniform Circular Motion – The Basics | 匀速圆周运动的基本概念

Δθ / Δt, measured in rad s⁻¹. The time for one complete revolution is the period T, and the frequency f is the number of revolutions per second, with f = 1/T. The linear speed v is related to the angular velocity by v = ωr, where r is the radius of the circle.

物体以恒定速率做圆周运动称为匀速圆周运动。尽管速率不变，但由于运动方向持续改变，速度矢量不断变化。物体扫过角度的速率就是角速度，定义为 ω = Δθ / Δt，单位为 rad s⁻¹。完成一整圈的时间是周期 T，频率 f 是每秒转动的圈数，且 f = 1/T。线速度 v 与角速度的关系为 v = ωr，其中 r 是圆周半径。

v = ωr ; ω = 2πf = 2π/T

2. Centripetal Acceleration and Force | 向心加速度与向心力

Since the velocity vector is always tangent to the circle and its direction changes, there must be an acceleration directed toward the centre. This centripetal acceleration has magnitude a_c = v² / r = r ω². By Newton's second law, a net force toward the centre—the centripetal force—is required to produce this acceleration. The magnitude of the centripetal force is F_c = m a_c tand that centripetal force is not a new type of force; it is simply the name for the net radial force that causes circular motion.

由于速度矢量始终与圆周相切且方向不断改变，必定存在一个指向圆心的加速度。这个向心加速度的大小为 a_c = v² / r = r ω²。根据牛顿第二定律，需要一个指向圆心的净力——向心力——来产生这个加速度。向心力的大小为 F_c = m v² / r = m r ω²。关键要理解，向心力并非一种新型的力，它只是导致圆周运动的净径向力的统称。

a_c = v² / r = r ω²
F_c = m v² / r = m r ω²

3. Sources of Centripetal Force | 向心力的来源

Centripetal force can be provided by tension (e.g., a ball on a string), friction (a car turning a flat bend), the normal reaction (a roller coaster loop), or a component of gravity (a conical pendulum). In every case, it is the resultant force acting radially inward. For an object moving in a horizontal circle, you identify all real forces, resolve them radially, and set the net inward component equal to m v²/r.

向心力可以由张力（如绳端小球）、摩擦力（汽车在水平弯道转弯）、法向反力（过山车回环）或重力的分量（锥摆）来提供。在任何情况下，它都是指向圆心的合力。对于在水平面内做圆周运动的物体，需找出所有真实力，沿径向分解，并令净内向分量等于 m v²/r。

• Tension in a string: T provides the whole centripetal force if the circle is horizontal and gravity is balanced by another force. | 绳中张力：若圆周在水平面内且重力被另一力平衡，则 T 提供全部向心力。
• Friction on a flat road: f = μ N provides the centripetal force, limiting safe speed. | 平坦路面摩擦力：f = μ N 提供向心力，限制了安全车速。
• Conical pendulum: the horizontal component of tension acts as the centripetal force. | 锥摆：张力的水平分量充当向心力。

4. Horizontal Circular Motion | 水平面内的圆周运动

A classic example is a car turning on a flat, horizontal road. The friction force between the tyres and the road points toward the centre of the turn. The maximum static friction is f_max = μ_s N, and since N = mg on a level road, the maximum speed without skidding is v_max = √μ_s g r friction. The ideal banking angle for a frictionless turn is given by tan θ = v² / (r g).

经典的例子是汽车在平坦水平道路上转弯。轮胎与路面之间的摩擦力指向转弯圆心。最大静摩擦力为 f_max = μ_s N，由于平路 N = mg，则车辆不打滑的最大速度为 v_max = √μ_s g r。为实现更高速度，道路会按角度 θ 倾斜，使得法向反力的一个分量提供向心力，从而减少对摩擦的依赖。无摩擦转弯的理想倾斜角满足 tan θ = v² / (r g)。

tan θ = v² / (r g)

tally: T sin θ = m v² / r. Combining these gives tan θ = v² / (r g), and the period T = 2π √L cos θ / g, where L is the string length.

锥摆由系在绳端在水平面内旋转的质量块构成，绳子扫出一个圆锥面。竖直分解：T cos θ = mg；水平分解：T sin θ = m v² / r。两式联立得 tan θ = v² / (r g)，周期 T = 2π √L cos θ / g，其中 L 为绳长。

5. Vertical Circular Motion | 竖直面内的圆周运动

d: T + mg = m v² / r. For the mass to just maintain circular motion at the top, the minimum speed must satisfy mg = m v_min² / r, giving v_min = √g r. Below this speed the string goes slack.

物体在竖直面内做圆周运动时，重力起变化作用。考虑系在绳上的质点在竖直面内摆动。在圆的最低点，张力与重力方向相反：T − mg = m v² / r，故张力最大。在最高点，张力和重力均向下：T + mg = m v² / r。质点恰好能维持圆周运动时，最高点最小速度满足 mg = m v_min² / r，得 v_min = √g r。低于此速度绳子会松弛。

Top: T + mg = m v² / r ; Bottom: T − mg = m v² / r

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

Newton' ² kg⁻² is the universal gravitational constant. This force is always attractive and acts along the line joining the two masses. The law applies strictly to point masses but also holds for spherical bodies with radially symmetric mass distributions when r is measured between centres.

牛顿万有引力定律指出，任何两个质点都相互吸引，引力大小与两者质量的乘积成正比，与它们中心之间距离的平方成反比。大小为 F = G M m / r²，其中 G = 6.67 × 10⁻¹¹ N m² kg⁻² 是万有引力常量。该力永远是吸引力，沿两质点的连线方向。该定律严格适用于质点，但也适用于具有径向对称质量分布的球体，此时 r 取球心间距。

F = G M m / r²

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7. Gravitational Field Strength | 引力场强度

force per unit mass experienced by a small test mass placed at that point. Near the Earth' t g decreases with the square of the distance. On the surface, r = R (planet radius), so g_surface = G M / R². This relation allows calculation of the mass of a planet if g and R are known.

引力场强度是放置在该点的小检验质量所受的每单位质量的力。地球表面附近记为 g，数值约为 9.81 N kg⁻¹。对于质量为 M 的球形行星，距离其中心 r 处的场强为 g = G M / r²。这表明 g 随距离的平方减小。在行星表面，r = R（行星半径），故 g_surface = G M / R²。该关系可用于在已知 g 和 R 时计算行星的质量。

g = G M / r²

8. Orbital Motion and Satellite Mechanics | 轨道运动与卫星力学

al force: G M m / r² = m v² / r. Cancelling m and rearranging yields the orbital speed v = √G M / r ² / G M) r³. This shows that T² is proportional to r³, consistent with Kepler’s third law.

对于在稳定圆轨道上的卫星，万有引力提供所需的向心力：G M m / r² = m v² / r。消去 m 并整理得到轨道速度 v = √G M / r。注意卫星的质量被消去；轨道速度仅取决于中心天体的质量和轨道半径。轨道周期 T 可由 v = 2π r / T 求得，导出 T² = (4π² / G M) r³。这表明 T² 与 r³ 成正比，与开普勒第三定律一致。

v = √G M / r
T² = (4π² / G M) r³

e period relation gives an orbital radius of about 42 200 km, or an altitude of roughly 35 800 km above the Earth’s surface.

地球同步卫星在地球赤道上空运行，周期恰好为 24 小时。将 T=24 h 代入周期关系可得轨道半径约 42 200 km，即地表以上约 35 800 km。

9. Kepler’s Laws | 开普勒定律

ocus. Second law: a line joining a planet to the Sun sweeps out equal areas in equal times – f the orbital period is proportional to the cube of the semi-major axis of the ellipse, T² ∝ proof: T² = (4π²/GM) r³. Thus Kepler’s laws are fully consistent with Newtonian mechanics.

开普勒提出了行星运动的三条经验定律。第一定律：行星沿椭圆轨道运动，太阳位于其中一个焦点。第二定律：行星与太阳的连线在相等时间内扫过相等面积——这是角动量守恒的结果。第三定律：轨道周期的平方与椭圆半长轴的立方成正比，即 T² ∝ a³。对于圆轨道，a = r，而牛顿万有引力定律提供了理论证明：T² = (4π²/GM) r³。因此开普勒定律与牛顿力学完全一致。

T² ∝ r³ (circular orbits: T² = 4π² r³ / (G M))

10. Gravitational Potential Energy | 引力势能

nfinite separation. The negative sign indicates that work must be done to separate the masses against their mutual attraction. The change in potential energy when moving a mass from r₁ to r₂ is ΔU = −G M m (1/r₂ − 1/r₁).

在地表附近的均匀场中，势能为 mgh。对于大尺度距离，引力场并不均匀。两质点相距 r 时引力势能的一般表达式为 U = − G M m / r，势能零点选在无穷远处。负号表示使两质量克服相互吸引而分离，必须做功。将质量从 r₁ 移至 r₂ 引起的势能变化为 ΔU = −G M m (1/r₂ − 1/r₁)。

U = − G M m / r

11. Escape Velocity | 逃逸速度

The escape velocity is the minimum speed needed for an object to completely escape a planet’s gravitational field, reaching infinity with zero final kinetic energy. Using conservation of energy, the sum of kinetic and potential energy at the surface must equal zero at infinity: ½ m v_esc² − G M m / R = 0. Solving gives v_esc = √2 G M / R. This speed is independent of the escaping object’s mass. For Earth, v_esc ≈ 11.2 km s⁻¹. Note that v_esc is √2 times the circular orbital speed at the surface.

逃逸速度是物体完全脱离行星引力场到达无穷远且末动能为零所需的最小速度。利用能量守恒，物体在行星表面的动能与势能之和必须等于无穷远处的零：½ m v_esc² − G M m / R = 0。解得 v_esc = √2 G M / R。该速度与逃逸物体的质量无关。地球的逃逸速度约为 11.2 km s⁻¹。注意，v_esc 是表面圆轨道速度的 √2 倍。

v_esc = √2 G M / R

12. Apparent Weight and Geostationary Orbits | 视重与地球同步轨道

slightly less than the true gravitational attraction mg. At the equator, mg’ = mg − m ω² R_E. This difference is about 0.34% munications. The orbital radius is found from T = 24 h and the equation T² = 4π² r³ / (G M_E), giving r ≈ 4.22 × 10⁷ m.

地球表面的物体（两极除外）因地球自转而做圆周运动。所需的向心力来自法向接触力略微减小，因此视重 mg’ 略小于真实的引力 mg。在赤道，mg’ = mg − m ω² R_E。这一差值约为 g 的 0.34%。地球同步轨道是一个特殊的赤道轨道，其周期等于地球自转周期（24 小时）。这样的卫星在天空中似乎静止不动，广泛用于通信。由 T = 24 h 及方程 T² = 4π² r³ / (G M_E) 可求出轨道半径 r ≈ 4.22 × 10⁷ m。

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