Gravitational Field Strength g (N kg⁻¹) | 重力场强度 g（N kg⁻¹）

📚 Gravitational Field Strength g (N kg⁻¹) | 重力场强度 g（N kg⁻¹）

The quantity g, typically expressed in newtons per kilogram (N kg⁻¹), is one of the most fundamental concepts in A-Level Physics. It appears in mechanics, gravitational fields, and even thermal physics. Yet many students confuse it with the acceleration due to gravity, or fail to distinguish it from the gravitational constant G. This article unpacks everything you need to know about g for your AQA A-Level examination.

物理量 g 通常以牛顿每千克（N kg⁻¹）为单位，是 A-Level 物理中最基础的概念之一。它出现在力学、引力场乃至热学中。然而许多学生将其与重力加速度混淆，或无法将其与万有引力常量 G 区分开来。本文将为你全面解析 AQA A-Level 考试中关于 g 的所有考点。

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1. Definition of Gravitational Field Strength | 重力场强度的定义

Gravitational field strength at a point is defined as the gravitational force acting per unit mass placed at that point. Mathematically:

重力场强度定义为：放置在某一位置的单位质量所受到的重力。数学表达式为：

g = F / m

where F is the gravitational force (in newtons) acting on a mass m (in kilograms). The unit of g is therefore N kg⁻¹. It is a vector quantity, always directed towards the centre of the mass that produces the field.

其中 F 是作用在质量 m（千克）上的重力（牛顿）。因此 g 的单位是 N kg⁻¹。它是一个矢量，方向始终指向产生该场的质量中心。

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2. N kg⁻¹ and m s⁻²: Are They the Same? | N kg⁻¹ 与 m s⁻²：它们相同吗？

An essential insight is that 1 N kg⁻¹ is equivalent to 1 m s⁻². This follows from the definition of the newton: 1 N = 1 kg × 1 m s⁻². Therefore, dividing by 1 kg gives:

一个关键的认识是：1 N kg⁻¹ 等价于 1 m s⁻²。这源于牛顿的定义：1 N = 1 kg × 1 m s⁻²。因此，除以 1 kg 可得：

1 N kg⁻¹ = 1 (kg × m s⁻²) / kg = 1 m s⁻²

This means that gravitational field strength and acceleration due to gravity are numerically identical on Earth — approximately 9.81 N kg⁻¹ or 9.81 m s⁻². Physically, however, they are interpreted differently: N kg⁻¹ emphasises the field concept (force per unit mass), while m s⁻² emphasises the kinematic consequence (acceleration of a free-falling object).

这意味着重力场强度和重力加速度在地球表面数值相同——约为 9.81 N kg⁻¹ 或 9.81 m s⁻²。但物理意义上，两者的侧重点不同：N kg⁻¹ 强调场的概念（单位质量所受的力），而 m s⁻² 强调运动学后果（自由落体物体的加速度）。

Perspective | 视角	Quantity | 物理量	Unit | 单位
Field viewpoint | 场的观点	Gravitational field strength | 重力场强度	N kg⁻¹
Kinematic viewpoint | 运动学观点	Acceleration of free fall | 自由落体加速度	m s⁻²

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3. Weight: W = mg | 重量：W = mg

The most direct application of gravitational field strength is calculating weight. The weight of an object is the gravitational force exerted on it, given by:

重力场强度最直接的应用是计算重量。物体的重量是作用在它上面的重力，公式为：

W = mg

where W is weight in newtons, m is mass in kilograms, and g is gravitational field strength in N kg⁻¹. Crucially, mass is invariant — it does not change with location — whereas weight depends on g. An astronaut of mass 80 kg weighs approximately 80 × 9.81 = 785 N on Earth, but only 80 × 1.62 ≈ 130 N on the Moon, where g ≈ 1.62 N kg⁻¹.

其中 W 为重量（牛顿），m 为质量（千克），g 为重力场强度（N kg⁻¹）。关键在于：质量不随位置变化，而重量取决于 g。一个质量为 80 kg 的宇航员在地球上重约 80 × 9.81 = 785 N，但在月球上（g ≈ 1.62 N kg⁻¹）仅重约 130 N。

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4. g as a Radial Field: The Inverse Square Law | g 作为径向场：平方反比定律

For a point mass or a spherically symmetric body such as the Earth, the gravitational field strength at a distance r from the centre is given by:

对于质点或球形对称物体（如地球），距离中心 r 处的重力场强度为：

g = GM / r²

where G is the gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²) and M is the mass of the body producing the field. This equation shows that g decreases with the square of the distance from the centre of the Earth.

其中 G 为万有引力常量（6.67 × 10⁻¹¹ N m² kg⁻²），M 为产生场的物体的质量。该方程表明 g 随距地球中心距离的平方而减小。

At the Earth’s surface, r = Rₑ (Earth’s radius ≈ 6.37 × 10⁶ m), giving:

在地球表面，r = Rₑ（地球半径 ≈ 6.37 × 10⁶ m），可得：

g₀ = GMₑ / Rₑ² ≈ 9.81 N kg⁻¹

At a height h above the surface, r = Rₑ + h. For example, at an altitude of 300 km (the approximate height of the ISS), g ≈ 9.81 × (6370/6670)² ≈ 8.9 N kg⁻¹ — noticeably less than at the surface.

在距地面高度 h 处，r = Rₑ + h。例如，在 300 km 的高空（国际空间站的大致高度），g ≈ 9.81 × (6370/6670)² ≈ 8.9 N kg⁻¹——明显小于地面值。

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5. Distinguishing g from G | 区分 g 与 G

A common source of confusion in A-Level exams is the difference between g (gravitational field strength) and G (the universal gravitational constant).

A-Level 考试中常见的困惑来源是 g（重力场强度）与 G（万有引力常量）之间的区别。

g	G
Gravitational field strength | 重力场强度	Universal gravitational constant | 万有引力常量
Unit: N kg⁻¹ | 单位：N kg⁻¹	Unit: N m² kg⁻² | 单位：N m² kg⁻²
Varies with location and mass distribution | 随位置和质量分布变化	Universal constant: same everywhere | 普适常量：处处相同
A field quantity | 场的物理量	A proportionality constant in Newton’s law | 牛顿定律中的比例常数

On Earth, g ≈ 9.81 N kg⁻¹; G = 6.67 × 10⁻¹¹ N m² kg⁻². Notice the enormous difference in magnitude — G is a universal constant that does not depend on any particular planet or star.

在地球上，g ≈ 9.81 N kg⁻¹；G = 6.67 × 10⁻¹¹ N m² kg⁻²。注意它们在数值上的巨大差异——G 是不依赖于任何特定行星或恒星的普适常量。

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6. Experimental Measurement of g | g 的实验测量

AQA requires you to be familiar with methods for determining g. Two classic experiments are:

AQA 要求你熟悉测定 g 的方法。两个经典实验是：

• Free-fall method: A ball bearing is dropped through light gates connected to a timer. The time t to fall a measured distance s is recorded. Using s = ½gt², the value of g is calculated.

• 自由落体法：将钢球穿过连接计时器的光电门。记录下落已知距离 s 所需的时间 t。利用 s = ½gt² 计算 g 值。

• Pendulum method: For a simple pendulum, the period T relates to length L by T = 2π√(L/g). Averaging multiple oscillations reduces timing errors, and a graph of T² against L yields a straight line through the origin with gradient 4π²/g.

• 单摆法：对于单摆，周期 T 与摆长 L 的关系为 T = 2π√(L/g)。多次振荡取平均可减少计时误差，绘制 T² 对 L 的图像可得过原点的直线，斜率为 4π²/g。

Common sources of systematic error include neglecting air resistance and timing reaction errors; using light gates or a data logger greatly improves accuracy.

常见的系统误差来源包括忽略空气阻力和计时反应误差；使用光电门或数据记录器可显著提高精度。

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7. g Inside the Earth | 地球内部的 g

An interesting extension: inside a uniform spherical planet, the mass outside the radius r exerts no net gravitational effect (this is a consequence of Newton’s shell theorem). The effective mass is M(r) = M × (r/R)³, giving:

一个有趣的延伸：在均匀球形星球内部，半径 r 之外的质量不产生净引力效应（这是牛顿壳层定理的推论）。有效质量为 M(r) = M × (r/R)³，因此：

g_inside = (GM/R³) × r

This shows that g varies linearly with r inside the Earth: it is zero at the centre and reaches g₀ at the surface. Beyond the surface, it follows the inverse square law. This piecewise behaviour is a favourite exam question.

这表明在地球内部 g 随 r 线性变化：在地心为零，在地表达到最大值 g₀。在地表以外，遵循平方反比定律。这种分段行为是考试的热门考点。

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8. Gravitational Field Strength vs Gravitational Potential | 重力场强度与重力势

In AQA A-Level, you are also expected to relate g to gravitational potential V. The relationship is:

在 AQA A-Level 中，你还需掌握 g 与重力势 V 的关系：

g = −dV/dr

The negative sign indicates that g points in the direction of decreasing potential. Since V = −GM/r for a radial field, differentiating gives:

负号表示 g 指向势减小的方向。对于径向场，V = −GM/r，微分可得：

g = −d(−GM/r)/dr = GM/r²

This mathematical link is tested regularly, especially the graphical interpretation: the gradient of a potential–distance graph gives the field strength.

这一数学联系经常被考查，尤其是图像解释：势–距离图像的斜率给出场强度。

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9. Common Misconceptions and Exam Traps | 常见误解与考试陷阱

• “g = 9.81 N kg⁻¹ everywhere” — wrong. This value applies only at the Earth’s surface. At high altitudes, on other planets, or inside the Earth, g differs.

• “g = 9.81 N kg⁻¹ 适用于任何地方”——错误。该值仅适用于地球表面。在高空、其他行星或地球内部，g 的数值不同。

• “Mass and weight are the same” — wrong. Mass is a scalar quantity measured in kg; weight is a force measured in newtons, equal to mg.

• “质量和重量相同”——错误。质量是标量，单位为 kg；重量是力，单位为牛顿，等于 mg。

• Forgetting that g is a vector: On a field diagram, g is represented by arrows pointing towards the mass. The magnitude decreases with distance, but the direction is always radial.

• 忘记 g 是矢量：在场线图中，g 用指向质量的箭头表示。大小随距离减小，但方向始终是径向的。

• Unit confusion: When substituting into W = mg, ensure g is in N kg⁻¹. If using kinematic equations such as s = ½gt², g is numerically identical but interpreted as 9.81 m s⁻².

• 单位混淆：代入 W = mg 时，确保 g 以 N kg⁻¹ 为单位。若使用运动学方程如 s = ½gt²，g 数值相同但解释为 9.81 m s⁻²。

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10. Worked Example | 例题解析

Question: The mass of the Earth is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m. A satellite orbits at an altitude of 400 km. Calculate (a) g at the Earth’s surface, (b) g at the satellite’s position.

例题：地球质量为 5.97 × 10²⁴ kg，半径为 6.37 × 10⁶ m。一颗卫星在 400 km 高空轨道运行。计算 (a) 地球表面的 g，(b) 卫星位置处的 g。

Solution | 解答：

(a) g₀ = GMₑ/Rₑ² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶)²

= (3.98 × 10¹⁴) / (4.06 × 10¹³) ≈ 9.81 N kg⁻¹ ✓

(b) r = Rₑ + h = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m

g = GMₑ/r² = (3.98 × 10¹⁴) / (6.77 × 10⁶)² = (3.98 × 10¹⁴) / (4.58 × 10¹³) ≈ 8.69 N kg⁻¹

Alternatively, using proportionality: g = g₀ × (Rₑ/r)² = 9.81 × (6.37/6.77)² ≈ 8.69 N kg⁻¹. The satellite experiences a gravitational field about 11% weaker than at the surface.

或者用比例关系：g = g₀ × (Rₑ/r)² = 9.81 × (6.37/6.77)² ≈ 8.69 N kg⁻¹。卫星处所受的重力场比地面约弱 11%。

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11. g and Orbital Motion | g 与轨道运动

Gravitational field strength is central to satellite and planetary motion. For an object in a stable circular orbit, the gravitational force provides the required centripetal force:

重力场强度是卫星和行星运动的核心。对于稳定圆周轨道上的物体，引力提供所需的向心力：

mg = mv²/r ⇒ g = v²/r

This yields the orbital velocity v = √(gr). Note that g here is evaluated at the orbital radius, not at the Earth’s surface. Many students erroneously use g₀ = 9.81 N kg⁻¹ for satellites — always check the radius being used.

由此可得轨道速度 v = √(gr)。注意此处的 g 是在轨道半径处取值，而非地球表面。许多学生错误地使用 g₀ = 9.81 N kg⁻¹ 来计算卫星——务必检查所用的半径。

The period of a satellite can also be expressed in terms of g:

卫星的周期也可以用 g 表示：

T = 2π√(r³/GMₑ) = 2π√(r/g)

where r is the orbital radius measured from the Earth’s centre. This equation combines Newton’s laws with the definition of gravitational field strength.

其中 r 是从地球中心测量的轨道半径。该方程将牛顿定律与重力场强度的定义结合起来。

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12. Summary and Exam Strategy | 总结与应试策略

For AQA A-Level Physics, ensure you can:

针对 AQA A-Level 物理，确保你能：

• Define gravitational field strength in words and equations, using the correct unit N kg⁻¹.
• 用文字和方程定义重力场强度，并使用正确单位 N kg⁻¹。
• Convert between N kg⁻¹ and m s⁻² when required by the question context.
• 在题目需要时在 N kg⁻¹ 与 m s⁻² 之间换算。
• Apply g = GM/r² to calculate field strength at any distance, including inside and outside a planet.
• 运用 g = GM/r² 计算任意距离处的场强度，包括行星内部和外部。
• Distinguish clearly between g and G — a frequent 1-mark definition question.
• 清楚区分 g 与 G——这是常见的 1 分定义题。
• Describe and evaluate experiments for measuring g, identifying sources of error.
• 描述并评价测量 g 的实验，识别误差来源。
• Link g to gravitational potential via g = −dV/dr and interpret graphs.
• 通过 g = −dV/dr 将 g 与重力势联系起来，并解读图像。
• Use g correctly in orbital mechanics problems.
• 在轨道力学问题中正确使用 g。

In the exam, read carefully whether the question asks for field strength (N kg⁻¹), acceleration (m s⁻²), or gravitational potential (J kg⁻¹). These are related but distinct quantities. Units are your best guide — writing the correct unit tells the examiner you understand the physics.

考试中，请仔细阅读题目问的是场强度（N kg⁻¹）、加速度（m s⁻²）还是重力势（J kg⁻¹）。这些量相关但不同。单位是你最好的指南——写出正确的单位向考官表明你理解了物理。

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