Gravitational force between point masses

1. Overview

Gravitational force is a universal attractive force acting between all objects with mass. In the context of A-Level Physics, we utilize Newton’s Law of Gravitation to model the interactions between celestial bodies and satellites. The core principle is that gravity provides the centripetal force required for circular orbital motion. By treating uniform spherical masses as point masses, we can simplify complex three-dimensional interactions into precise mathematical models, allowing for the calculation of orbital periods, velocities, and the specific conditions required for geostationary orbits.

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Key Definitions

• 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: An idealized representation of an object where its entire mass is considered to be concentrated at a single point.
• Gravitational Field Strength ($g$): The gravitational force per unit mass exerted on a small "test" mass placed at a specific point within the field. It is a vector quantity directed towards the center of the mass creating the field.
• Geostationary Orbit: A specific circular orbit around the Earth in which a satellite appears stationary to an observer on the Earth's surface. It must have a period of 24 hours, travel from West to East, and be positioned directly over the Equator.
• Gravitational Constant ($G$): An empirical physical constant involved in the calculation of gravitational effects. Its value is approximately $6.67 \times 10^{-11} , \text{N},\text{m}^2,\text{kg}^{-2}$.

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Content

3.1 The Point Mass Approximation

A fundamental requirement of the syllabus is understanding how to treat large, spherical objects. For any point outside a uniform sphere of mass $M$, the gravitational field produced is identical to the field that would be produced if all the mass $M$ were concentrated at the sphere's geometric center.

• Radial Fields: The field lines around a uniform sphere are directed radially inwards. As you move further from the center, the lines spread out, representing the decrease in field strength according to the inverse square law.
• Application: When calculating the force between the Earth and a satellite, the distance $r$ used in equations must be the distance from the center of the Earth to the satellite, not the height above the surface.

3.2 Newton’s Law of Gravitation

Newton’s Law provides the magnitude of the attractive force between two masses.

$$\mathbf{F = \frac{Gm_1m_2}{r^2}}$$ (This equation is provided on the Formulae Sheet)

Variables:

• $F$: Gravitational force (N)
• $G$: Gravitational constant ($6.67 \times 10^{-11} , \text{N},\text{m}^2,\text{kg}^{-2}$)
• $m_1, m_2$: The two interacting masses (kg)
• $r$: The separation between the centers of the masses (m)

The Inverse Square Law ($F \propto \frac{1}{r^2}$): The force follows an inverse square relationship with distance. If the separation $r$ increases by a factor of $x$, the force $F$ decreases by a factor of $x^2$.

• If distance doubles ($2r$), force becomes $\frac{1}{4}F$.
• If distance triples ($3r$), force becomes $\frac{1}{9}F$.
• If distance is halved ($\frac{1}{2}r$), force becomes $4F$.

3.3 Analysis of Circular Orbits

For a planet or satellite to maintain a stable circular orbit, a centripetal force must act on it. In space, this centripetal force is provided entirely by the gravitational attraction between the orbiting body ($m$) and the central body ($M$).

Derivation of Orbital Relationships: To analyze any orbit, we start by equating the gravitational force to the centripetal force:

$$F_{grav} = F_{centripetal}$$

$$\frac{GMm}{r^2} = \frac{mv^2}{r} \quad \text{or} \quad \frac{GMm}{r^2} = mr\omega^2$$

1. Finding Orbital Velocity ($v$): From $\frac{GMm}{r^2} = \frac{mv^2}{r}$, we cancel $m$ and one $r$: $$\frac{GM}{r} = v^2 \implies \mathbf{v = \sqrt{\frac{GM}{r}}}$$ Note: The orbital speed is independent of the mass of the orbiting satellite.

2. Deriving Kepler’s Third Law ($T^2 \propto r^3$): Substitute $v = \frac{2\pi r}{T}$ into the orbital velocity equation: $$\frac{GM}{r} = \left(\frac{2\pi r}{T}\right)^2$$ $$\frac{GM}{r} = \frac{4\pi^2 r^2}{T^2}$$ Rearranging for $T^2$: $$\mathbf{T^2 = \left( \frac{4\pi^2}{GM} \right) r^3}$$ Since the term in the brackets is constant for a given central mass $M$, we conclude that $T^2 \propto r^3$.

3.4 Geostationary Orbits

A geostationary satellite is essential for global communications and weather monitoring. For a satellite to remain above the same point on the Earth's surface, it must satisfy these four conditions:

1. Period: The orbital period must be exactly 24 hours (86,400 seconds), matching the Earth's rotation on its axis.
2. Direction: It must rotate from West to East, following the Earth's own rotation.
3. Plane: It must be in an equatorial orbit (directly above the Earth's equator). If it were in a polar orbit, it would move north and south relative to the ground.
4. Radius: There is only one specific radius (and therefore one specific altitude) that satisfies the 24-hour period requirement.

Calculating the Geostationary Radius: Using $T = 86,400 , \text{s}$, $M_E = 5.97 \times 10^{24} , \text{kg}$, and $G = 6.67 \times 10^{-11}$: $$r^3 = \frac{GMT^2}{4\pi^2}$$ $$r = \sqrt[3]{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(86400)^2}{4\pi^2}} \approx 4.22 \times 10^7 , \text{m}$$ To find the altitude ($h$): $$h = r - R_{Earth} = 4.22 \times 10^7 - 6.37 \times 10^6 \approx 3.58 \times 10^7 , \text{m} , (35,800 , \text{km})$$

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Worked Example 1 — Force between Point Masses

Two uniform spheres, A and B, have masses of $4.0 , \text{kg}$ and $9.0 , \text{kg}$ respectively. Their centers are separated by a distance of $0.50 , \text{m}$. Calculate the magnitude of the gravitational force exerted by sphere B on sphere A.

Step 1: Identify the relevant equation $$F = \frac{Gm_1m_2}{r^2}$$

Step 2: Substitute the values $$F = \frac{(6.67 \times 10^{-11}) \times 4.0 \times 9.0}{(0.50)^2}$$

Step 3: Calculate intermediate steps $$F = \frac{2.4012 \times 10^{-9}}{0.25}$$

Step 4: Final Answer $$F = 9.6 \times 10^{-9} , \text{N}$$ (Note: The force exerted by A on B is identical in magnitude but opposite in direction, according to Newton's Third Law.)

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Worked Example 2 — Orbital Speed of the ISS

The International Space Station (ISS) orbits the Earth at an altitude of approximately $400 , \text{km}$. Calculate the orbital speed of the ISS. (Data: $M_E = 5.97 \times 10^{24} , \text{kg}$; $R_E = 6.37 \times 10^6 , \text{m}$)

Step 1: Calculate the total orbital radius ($r$) The radius is the sum of the Earth's radius and the altitude. $$r = R_E + h = (6.37 \times 10^6) + (400 \times 10^3) = 6.77 \times 10^6 , \text{m}$$

Step 2: State the orbital speed equation $$v = \sqrt{\frac{GM}{r}}$$

Step 3: Substitute and solve $$v = \sqrt{\frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{6.77 \times 10^6}}$$ $$v = \sqrt{5.88 \times 10^7}$$ $$v = 7670 , \text{m},\text{s}^{-1}$$

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Worked Example 3 — Comparing Orbits (Ratio Method)

Satellite X orbits a planet at a distance $R$ from the center with a period $T$. Satellite Y orbits the same planet at a distance $4R$. Determine the orbital period of Satellite Y in terms of $T$.

Step 1: Use Kepler's Third Law relationship $$T^2 \propto r^3 \implies \frac{T_Y^2}{T_X^2} = \frac{r_Y^3}{r_X^3}$$

Step 2: Substitute the ratios $$\frac{T_Y^2}{T^2} = \frac{(4R)^3}{R^3}$$ $$\frac{T_Y^2}{T^2} = \frac{64R^3}{R^3} = 64$$

Step 3: Solve for $T_Y$ $$T_Y^2 = 64T^2$$ $$T_Y = \sqrt{64T^2} = 8T$$ Answer: The period of Satellite Y is $8T$.

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Key Equations

Equation	Description	Status
$F = \frac{Gm_1m_2}{r^2}$	Newton's Law of Gravitation	Data Sheet
$g = \frac{F}{m}$	Definition of Gravitational Field Strength	Data Sheet
$g = \frac{GM}{r^2}$	Field strength for a point mass	Data Sheet
$v = \sqrt{\frac{GM}{r}}$	Orbital velocity for a circular orbit	Derive/Memorize
$T^2 = \left(\frac{4\pi^2}{GM}\right)r^3$	Kepler's Third Law (Period vs Radius)	Derive/Memorize
$\omega = \sqrt{\frac{GM}{r^3}}$	Orbital angular velocity	Derive

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Common Mistakes to Avoid

• ❌ Confusing Altitude and Radius: Using the height above the surface ($h$) instead of the distance from the center ($r = R + h$) in the denominator of the gravitational formula.
• ❌ Squaring Errors: Forgetting to square the distance $r$ in $F = \frac{GMm}{r^2}$ or forgetting to square the period $T$ and cube the radius $r$ in Kepler’s Law.
• ❌ Unit Mismatch: Using kilometers (km) instead of meters (m). Astronomical data is almost always given in km; you must convert to m ($10^3$) before calculating.
• ❌ Mass Confusion: In the orbital velocity formula $v = \sqrt{\frac{GM}{r}}$, $M$ is the mass of the central body (e.g., the Earth), not the mass of the satellite. The satellite's mass cancels out.
• ❌ Geostationary Misconceptions: Thinking a geostationary satellite can be placed over any city (e.g., London or New York). It must be over the Equator to remain stationary relative to the ground.

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Exam Tips

1. The "First Line" Rule: When asked to derive orbital equations, your very first line of working should always be:
   • "Gravitational force provides the centripetal force"
   • $\frac{GMm}{r^2} = \frac{mv^2}{r}$ or $\frac{GMm}{r^2} = mr\omega^2$
   • Failing to state this equality often results in the loss of the first method mark.
2. Significant Figures: Cambridge 9702 papers usually provide $G$ to 3 significant figures. Ensure your final answer is rounded to 2 or 3 s.f. as appropriate to the data provided in the question.
3. Standard Form: Get comfortable entering large exponents into your calculator. A single mistake in the power of 10 (e.g., $10^{24}$ vs $10^{22}$) will lead to an answer that is orders of magnitude incorrect.
4. Direction Matters: If a question asks for the "Gravitational Field Strength" as a vector, remember it is always directed towards the center of the mass creating the field.
5. Geostationary Explanations: If asked why a satellite must be geostationary, mention continuous communication. Because it is stationary relative to the Earth, ground-based satellite dishes do not need to move to track it.