IB Physics: Gravitational Fields (HL Included) | IB 物理:引力场考点精讲(含HL)

📚 IB Physics: Gravitational Fields (HL Included) | IB 物理:引力场考点精讲(含HL)

Gravitational fields are a fundamental topic in the IB Physics syllabus, appearing in both SL and HL papers. This guide breaks down every key concept—from Newton’s law to orbital mechanics—with exam-style explanations to help you secure full marks.

引力场是 IB 物理教学大纲中的基础核心专题,在 SL 和 HL 试卷中均会出现。本文将从牛顿定律到轨道力学逐项拆解全部考点,配合考试风格解析,助你稳拿满分。


1. Gravitational Field Concept | 引力场基本概念

A gravitational field is a region of space where a mass experiences a gravitational force. Every object with mass creates a gravitational field around itself.

引力场是空间中任何有质量的物体周围存在的区域,在该区域内,其他有质量的物体会受到引力作用。任何具有质量的物体都会在自身周围产生引力场。

  • Field direction: always towards the centre of the mass | 场方向:始终指向质量中心

  • Field strength is measured in N kg⁻¹ | 场强的单位为 N kg⁻¹(牛顿每千克)

  • Gravitational forces are always attractive | 引力始终表现为吸引力

  • The field extends infinitely, but strength decreases with distance | 场可延伸至无限远,但强度随距离增大而减弱


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

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

牛顿万有引力定律指出:任意两个质点之间相互吸引的力,大小与两物体质量的乘积成正比,与两物体间距离的平方成反比。

F = G·m₁m₂ / r²

Where G = 6.674 × 10⁻¹¹ N m² kg⁻² (gravitational constant), m₁ and m₂ are the masses (kg), and r is the centre-to-centre distance (m).

其中 G = 6.674 × 10⁻¹¹ N·m²·kg⁻²(万有引力常量),m₁、m₂ 为两物体的质量(kg),r 为两物体质心间的距离(m)。

Key exam points:

考查要点:

  • r is always measured from centre to centre | r 始终为质心到质心的距离

  • The law applies to point masses and uniform spheres | 该定律适用于质点和均匀球体

  • For spherical objects, r is measured from the centre | 对球体而言,r 从球心量起


3. Gravitational Field Strength | 引力场强度

Gravitational field strength at a point is defined as the gravitational force per unit mass acting on a small test mass placed at that point.

引力场强度定义为放置在电场中某点的单位质量小测试物体所受的引力大小,即该点处每单位质量所受的引力。

g = F / m

For a point mass or uniform sphere, this becomes:

对于质点或均匀球体,上式可化为:

g = GM / r²

  • At the surface of Earth: g ≈ 9.81 m s⁻² | 在地球表面:g ≈ 9.81 m/s²

  • g decreases with altitude: g = GM/(R+h)² | g 随高度增加而减小:g = GM/(R+h)²

  • g is a vector quantity pointing toward the centre | g 为矢量,方向指向球心


4. Gravitational Potential Energy | 引力势能

In the IB syllabus, gravitational potential energy is defined as the work done in bringing a mass from infinity to a given point. At infinity, gravitational potential energy is taken as zero.

在 IB 教学大纲中,引力势能定义为将物体从无穷远处移动到某点所做的功。在无穷远处,引力势能取为零。

Eₚ = -GMm / r

The negative sign is crucial: it indicates that gravitational potential energy is lower (more negative) closer to Earth. You must use this formula for exam solutions rather than the simpler mgh, which only applies near Earth’s surface.

负号至关重要:它表明物体距离地球越近,引力势能越低(负值越大)。解题时必须使用此公式,而非简化式 mgh——后者仅适用于地球表面附近。

ΔEₚ = GMm(1/r₁ – 1/r₂),物体从 r₁ 移到 r₂ 时的势能变化


5. Gravitational Potential | 引力势(重力势)

Gravitational potential V at a point is the gravitational potential energy per unit mass at that point.

引力势 V 定义为某点处单位质量所具有的引力势能,即该点的引力势能除以质量。

V = Eₚ / m = -GM / r

  • V is measured in J kg⁻¹ | V 的单位为 J/kg(焦耳每千克)

  • V is a scalar quantity | V 为标量

  • Equipotential surfaces for a point mass are concentric spheres | 质点引力场的等势面为同心球面

  • At Earth’s surface: V = -GM/R ≈ -6.25 × 10⁷ J kg⁻¹ | 地球表面:V = -GM/R ≈ -6.25 × 10⁷ J/kg

HL: Potential gradient

HL 拓展:势梯度

g = -ΔV / Δr

The field strength is the negative gradient of potential. On a V–r graph, the slope at any point gives the field strength at that point.

引力场强度等于引力势的负梯度。在 V-r 图像上,任意点的斜率即为该点的场强大小。


6. Kepler’s Laws | 开普勒定律

Kepler’s laws describe planetary motion around the Sun and are directly examinable in IB Physics.

开普勒定律描述行星绕太阳运动的规律,是 IB 物理的直接考查内容。

First Law (Elliptical Orbits) | 第一定律(椭圆轨道):Each planet moves in an ellipse with the Sun at one focus. | 每颗行星沿椭圆轨道运行,太阳位于其中一个焦点上。

Second Law (Equal Areas) | 第二定律(等面积定律):A line joining a planet to the Sun sweeps out equal areas in equal times. This implies planets move faster when closer to the Sun. | 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时运动速度较快。

Third Law (Harmonic Law) | 第三定律(和谐定律):The square of a planet’s period is proportional to the cube of the semi-major axis of its orbit. | 行星公转周期的平方与其轨道半长轴的立方成正比。

T² ∝ r³

For circular orbits, combining Newton’s law with circular motion gives:

对于圆轨道,结合牛顿定律与圆周运动可得:

T² = 4π²r³ / GM


7. Satellite Motion and Orbital Speed | 卫星运动与轨道速度

For a satellite in a circular orbit, the gravitational force provides the required centripetal force.

对于在圆轨道上运行的卫星,引力提供所需的向心力。

GMm / r² = mv² / r

Solving for orbital speed:

由此解出轨道速度:

v = √(GM / r)

  • Orbital speed is independent of satellite mass | 轨道速度与卫星质量无关

  • Higher orbits → slower speed | 轨道越高,速度越慢

  • Period increases with orbital radius | 轨道半径越大,周期越长

  • Geostationary satellites have T = 24 h and orbit above the equator | 地球同步卫星周期为 24 小时,且轨道位于赤道正上方


8. Escape Velocity | 逃逸速度

Escape velocity is the minimum speed an object needs to escape a body’s gravitational field without further propulsion, starting from a given radius.

逃逸速度是指物体从某一半径处出发、无需额外推进即可摆脱天体引力场所需的最小速度。

½mv² = GMm / R ⇒ vₑₛ꜀ = √(2GM / R)

For Earth: vₑₛ꜀ ≈ 11.2 km s⁻¹.

地球的逃逸速度约为 11.2 km/s。

HL connection: Total energy of an orbiting satellite is E = -GMm/2r. When E ≥ 0, the satellite is unbound and can escape.

HL 关联:轨道卫星的总能量为 E = -GMm/2r。当 E ≥ 0 时,卫星不受约束,可以逃逸。


9. Energy in Orbital Motion (HL) | 轨道运动能量(HL)

For a satellite in a stable circular orbit, the total mechanical energy is constant and negative.

对于处于稳定圆轨道的卫星,其总机械能为常数且为负值。

Eₜₒₜₐₗ = KE + Eₚ = ½mv² – GMm/r = -GMm/2r

Notice that: KE = -Eₜₒₜₐₗ (positive), and Eₚ = 2Eₜₒₜₐₗ (negative). To move a satellite to a higher orbit, energy must be supplied; the satellite slows down, yet its total energy increases.

注意:动能 KE = -Eₜₒₜₐₗ(正值),势能 Eₚ = 2Eₜₒₜₐₗ(负值)。要将卫星送入更高轨道,需要提供额外能量;卫星速度降低,但总能量增大。

Exam tip: When a satellite spirals inward due to air drag, it speeds up while losing total energy—a common trick question!

考试技巧:当卫星因空气阻力逐渐向内螺旋时,其速度反而增大,但总能量减少——这是常见的易错题!


10. Gravitational Field vs Electric Field | 引力场与电场的对比

Comparing gravitational and electrostatic fields helps consolidate understanding, especially for Paper 2 questions.

对比引力场与静电场有助于加深理解,尤其在 Paper 2 的综合性试题中十分有效。

Property | 性质 Gravitational | 引力场 Electric | 电场
Force law | 力定律 F = GMm/r² (attractive only) F = kq₁q₂/r² (attractive/repulsive)
Field strength | 场强 g = GM/r² E = kq/r²
Potential | 势 V = -GM/r V = kq/r (±)
Nature | 性质 Always attractive | 始终吸引 Can attract/repel | 可吸引或排斥

11. Common Mistakes and How to Avoid Them | 常见错误与规避方法

Here are the most frequently lost marks in IB exams on this topic:

以下是 IB 考试中该专题最常见的丢分点:

  • Using mgh for large distances: mgh only works when h ≪ R. Use ΔEₚ = GMm(1/r₁ – 1/r₂) instead.

  • 忽略负号:用 mgh 处理大距离问题。mgh 仅在 h ≪ R 时适用。应改用 ΔEₚ = GMm(1/r₁ – 1/r₂)。

  • Confusing g and V: g is a vector (force per unit mass); V is a scalar (energy per unit mass).

  • 混淆 g 和 V:g 是矢量(单位质量的力),V 是标量(单位质量的能量)。

  • Forgetting r is centre-to-centre: When a satellite is at altitude h, r = R + h.

  • 忘记 r 为质心距:当卫星距地面高度为 h 时,r = R + h。

  • Misapplying Kepler’s third law: T² ∝ r³ only for orbits around the same central mass.

  • 错误套用开普勒第三定律:T² ∝ r³ 仅适用于绕同一中心天体运行的情形。


12. Exam Worked Example | 真题例题精解

Question (HL-style): A satellite of mass 2000 kg orbits Earth at an altitude of 600 km. Calculate (a) the orbital speed, (b) the orbital period, and (c) the total energy. (Take G = 6.67 × 10⁻¹¹ N m² kg⁻², Mₑ = 5.97 × 10²⁴ kg, Rₑ = 6.37 × 10⁶ m)

题目(HL 风格):一颗质量为 2000 kg 的卫星在距地面 600 km 的高度绕地球运行。计算:(a) 轨道速度;(b) 轨道周期;(c) 总能量。(取 G = 6.67 × 10⁻¹¹ N·m²/kg²,Mₑ = 5.97 × 10²⁴ kg,Rₑ = 6.37 × 10⁶ m)

Solution | 解答:

r = Rₑ + h = 6.37 × 10⁶ + 0.6 × 10⁶ = 6.97 × 10⁶ m

(a) v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.97 × 10⁶) ≈ √(5.71 × 10⁷) ≈ 7.56 × 10³ m s⁻¹

(a) v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.97 × 10⁶) ≈ √(5.71 × 10⁷) ≈ 7.56 × 10³ m/s

(b) T = 2πr/v = 2π(6.97 × 10⁶)/(7.56 × 10³) ≈ 5790 s ≈ 96.5 min

(b) T = 2πr/v = 2π × 6.97 × 10⁶ / 7.56 × 10³ ≈ 5790 s ≈ 96.5 分钟

(c) Eₜₒₜₐₗ = -GMm/2r = -(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 2000) / (2 × 6.97 × 10⁶) ≈ -5.71 × 10¹⁰ J

(c) Eₜₒₜₐₗ = -GMm/2r = -(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 2000) / (2 × 6.97 × 10⁶) ≈ -5.71 × 10¹⁰ J

Scoring tip: Always state the formula before substitution, and show units in every line. | 得分技巧:先写公式再代入数据,并在每一行中标明单位。


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