Gravitational field of a point mass

1. Overview

The gravitational field of a point mass is a fundamental concept in classical mechanics, describing how a single point of matter influences the space around it. In physics, a gravitational field is a region of space where a mass experiences a gravitational force. For a point mass—or any object that can be modeled as one, such as a uniform spherical planet—the field is radial. This means the field lines converge at the center of the mass, and the strength of the field depends solely on the distance from that center.

The behavior of this field is governed by an inverse-square law. As you move further away from the source of the field, the gravitational field strength ($g$) decreases in proportion to the square of the distance. This relationship is central to understanding the motion of satellites, the orbits of planets, and the weight of objects at different altitudes. While we often treat $g$ as a constant ($9.81 \text{ m s}^{-2}$) for everyday calculations on Earth, this is merely a local approximation. This topic explores the precise mathematical derivation and application of the gravitational field strength equation, $g = \frac{GM}{r^2}$.

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Key Definitions

• Gravitational field strength ($g$): The gravitational force exerted per unit mass on a small test mass placed at a specific point in the field. It is a vector quantity, always directed towards the center of the mass creating the field.
• Newton’s Law of Gravitation: The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation.
• Point Mass: A theoretical model where the entire mass of an object is considered to be concentrated at a single point in space. This model is valid for calculating the field outside any uniform spherical object.
• Inverse-Square Law: A physical law stating that a specified physical quantity is inversely proportional to the square of the distance from the source of that physical quantity.
• Radial Field: A field in which the field lines are straight and converge at a single point (the center of the mass). The field strength is not uniform, as the lines become more spread out as distance from the center increases.

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Content

3.1 Derivation of $g = \frac{GM}{r^2}$

The syllabus requires you to derive the expression for the gravitational field strength ($g$) due to a point mass by combining the definition of field strength with Newton's Law of Gravitation.

Step 1: Define Gravitational Field Strength By definition, the gravitational field strength $g$ at a point is the force $F$ acting on a small test mass $m$ placed at that point: $$g = \frac{F}{m}$$

Step 2: State Newton's Law of Gravitation The magnitude of the gravitational force $F$ between a source mass $M$ and a test mass $m$, where their centers are separated by a distance $r$, is given by: $$F = \frac{GMm}{r^2}$$ (Where $G$ is the universal gravitational constant, $6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$)

Step 3: Substitute and Simplify Substitute the expression for $F$ from Newton's Law into the definition of $g$: $$g = \frac{\left( \frac{GMm}{r^2} \right)}{m}$$

The test mass $m$ appears in both the numerator and the denominator, so it cancels out: $$g = \frac{GM}{r^2}$$

Conclusion of Derivation: This result proves that the gravitational field strength at a point depends only on the mass of the source ($M$) and the distance from its center ($r$). It is entirely independent of the mass of the object experiencing the field.

3.2 The Inverse-Square Relationship and Graphical Representation

The equation $g = \frac{GM}{r^2}$ demonstrates that $g \propto \frac{1}{r^2}$. This has significant implications for how gravity behaves over large distances:

• If the distance $r$ from the center of a planet is doubled ($2r$), the field strength becomes $\frac{1}{2^2} = \frac{1}{4}$ of its original value.
• If the distance $r$ is tripled ($3r$), the field strength becomes $\frac{1}{3^2} = \frac{1}{9}$ of its original value.

The $g$ vs. $r$ Graph: When plotting gravitational field strength ($g$) against distance from the center ($r$):

1. Outside the mass ($r \geq R$): The curve follows an inverse-square decay ($g \propto 1/r^2$). The maximum value of $g$ occurs at the surface of the mass ($r = R$).
2. Inside the mass ($r < R$): For a uniform sphere, the field strength actually decreases linearly as you move toward the center ($g \propto r$), reaching zero at the very center. However, the point mass derivation specifically describes the field for $r \geq R$.

3.3 Why $g$ is Approximately Constant Near the Earth's Surface

In many mechanics problems, we use $g = 9.81 \text{ m s}^{-2}$ as a constant. We can justify this using the point mass equation.

Let $R$ be the radius of the Earth ($\approx 6.4 \times 10^6 \text{ m}$) and $h$ be the height above the surface. The distance from the center is $r = R + h$. $$g = \frac{GM}{(R+h)^2}$$

If the change in height $h$ is very small compared to the radius of the Earth ($h \ll R$), then: $$R + h \approx R$$ Therefore: $$g \approx \frac{GM}{R^2} \approx \text{constant}$$

Numerical Evidence:

• At the surface ($h = 0$): $g = \frac{GM}{R^2} \approx 9.81 \text{ N kg}^{-1}$.
• At the top of Mount Everest ($h \approx 8.8 \text{ km}$): The ratio of the new field strength to the surface field strength is $\left( \frac{R}{R+h} \right)^2$. $$\left( \frac{6400}{6400 + 8.8} \right)^2 \approx 0.997$$ This represents a change of only 0.3%, which is negligible for most practical purposes. Thus, for small-scale motion (like a ball being thrown), the field is treated as uniform.

3.4 Resultant Gravitational Fields

Gravitational field strength is a vector. If a point is affected by two or more masses (e.g., a spacecraft between the Earth and the Moon), the resultant field strength is the vector sum of the individual fields.

If two masses $M_1$ and $M_2$ are separated by a distance $d$, the field strength at a point between them is: $$g_{\text{net}} = g_1 - g_2$$ (Assuming we define the direction towards $M_1$ as positive). There is always a "null point" between two masses where the two fields cancel out perfectly, resulting in $g_{\text{net}} = 0$.

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4. Worked Examples

Worked Example 1 — Field Strength on a Different Planet

A hypothetical planet has a mass of $4.80 \times 10^{24} \text{ kg}$ and a radius of $5.50 \times 10^6 \text{ m}$. Calculate the gravitational field strength at a height of $1.50 \times 10^6 \text{ m}$ above the planet's surface.

Step 1: Identify the total distance $r$ from the center $$r = R + h$$ $$r = (5.50 \times 10^6) + (1.50 \times 10^6) = 7.00 \times 10^6 \text{ m}$$

Step 2: Use the gravitational field strength equation $$g = \frac{GM}{r^2}$$

Step 3: Substitute the values $$g = \frac{(6.67 \times 10^{-11}) \times (4.80 \times 10^{24})}{(7.00 \times 10^6)^2}$$

Step 4: Calculate intermediate steps Numerator: $3.2016 \times 10^{14}$ Denominator: $4.90 \times 10^{13}$

Step 5: Final Answer $$g = 6.53 \text{ N kg}^{-1} \text{ (or m s}^{-2})$$

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Worked Example 2 — The Ratio Method

Planet X has twice the mass of Earth and three times the radius of Earth. If the gravitational field strength on Earth's surface is $g_E$, calculate the field strength on the surface of Planet X in terms of $g_E$.

Step 1: Write the expression for both planets $$g_E = \frac{GM_E}{R_E^2}$$ $$g_X = \frac{G(2M_E)}{(3R_E)^2}$$

Step 2: Simplify the expression for Planet X $$g_X = \frac{2GM_E}{9R_E^2}$$

Step 3: Relate $g_X$ to $g_E$ $$g_X = \frac{2}{9} \left( \frac{GM_E}{R_E^2} \right)$$ $$g_X = \frac{2}{9} g_E \approx 0.22 g_E$$

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Worked Example 3 — Finding the Null Point

The Earth ($M_E = 6.0 \times 10^{24} \text{ kg}$) and the Moon ($M_M = 7.4 \times 10^{22} \text{ kg}$) are separated by a center-to-center distance of $3.8 \times 10^8 \text{ m}$. Find the distance $x$ from the center of the Earth where the resultant gravitational field strength is zero.

Step 1: Set the field strengths equal in magnitude At the null point, $g_{\text{Earth}} = g_{\text{Moon}}$. $$\frac{GM_E}{x^2} = \frac{GM_M}{(d - x)^2}$$

Step 2: Cancel $G$ and take the square root of both sides $$\frac{M_E}{x^2} = \frac{M_M}{(d - x)^2}$$ $$\frac{\sqrt{M_E}}{x} = \frac{\sqrt{M_M}}{d - x}$$

Step 3: Substitute the mass values and solve for $x$ $$\frac{\sqrt{6.0 \times 10^{24}}}{x} = \frac{\sqrt{7.4 \times 10^{22}}}{3.8 \times 10^8 - x}$$ $$\frac{2.45 \times 10^{12}}{x} = \frac{2.72 \times 10^{11}}{3.8 \times 10^8 - x}$$

Step 4: Cross-multiply and rearrange $$2.45 \times 10^{12} (3.8 \times 10^8 - x) = 2.72 \times 10^{11} x$$ $$9.31 \times 10^{20} - 2.45 \times 10^{12} x = 2.72 \times 10^{11} x$$ $$9.31 \times 10^{20} = 2.722 \times 10^{12} x$$ $$x = 3.4 \times 10^8 \text{ m}$$

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Key Equations

Equation	Description	Data Sheet?
$g = \frac{F}{m}$	Definition of gravitational field strength	Yes
$F = \frac{GMm}{r^2}$	Newton's Law of Gravitation	Yes
$g = \frac{GM}{r^2}$	Field strength of a point mass	No (Must Derive)
$r = R + h$	Total distance from center	No

Constants:

• $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
• $g_{\text{surface}} = 9.81 \text{ m s}^{-2}$

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Common Mistakes to Avoid

• ❌ Using Height instead of Radius: Using the altitude $h$ as $r$ in the equation.
  • ✅ Right: Always use $r = R + h$. Gravity acts from the center of the mass, not the surface.
• ❌ Forgetting the Square: Calculating $g = \frac{GM}{r}$ instead of $g = \frac{GM}{r^2}$.
  • ✅ Right: The inverse-square law is strict. Always square the distance.
• ❌ Incorrect Unit Conversions: Using kilometers instead of meters.
  • ✅ Right: $G$ is defined in terms of meters. Convert all distances (km, Mm) to meters (m) before calculating.
• ❌ Confusing $g$ and $G$: Treating the gravitational constant and field strength as the same thing.
  • ✅ Right: $G$ is a universal constant ($6.67 \times 10^{-11}$); $g$ is a local property that changes depending on where you are.
• ❌ Calculator Errors with Standard Form: Entering $6.67 \times 10^{-11}$ as 6.67 * 10^-11 in a way that the calculator misinterprets the order of operations.
  • ✅ Right: Use the EXP or x10^x key to ensure the exponent is tied to the coefficient.

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Exam Tips

1. Show the Derivation: If a question asks you to "show that $g = GM/r^2$", you must start from $g = F/m$ and $F = GMm/r^2$. Simply writing the final equation will earn zero marks.
2. The "Ratio" Shortcut: For multiple-choice questions comparing two planets, don't calculate the full value of $g$. Use the ratio $\frac{g_1}{g_2} = \frac{M_1}{M_2} \times \left( \frac{r_2}{r_1} \right)^2$. It is much faster and prevents power-of-ten errors.
3. Significant Figures: In Paper 2 and Paper 4, look at the data provided. If the mass is $6.0 \times 10^{24}$ (2 s.f.), your answer should be to 2 or 3 s.f. Never give 5 or 6 significant figures.
4. Vector Direction: If a question asks for the "field strength" at a point between two planets, remember it is a vector. You must subtract the magnitudes if the fields are in opposite directions.
5. Graph Interpretation: Be prepared to identify the $1/r^2$ shape. If you are asked to sketch it, ensure the curve never touches the x-axis (it is an asymptote) and starts at a clear $(R, g_s)$ point.