Gravitation and Kepler's Laws
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You should be comfortable with the concept of an effective potential and the Binet orbit equation derived in the previous lecture, as well as the definitions of gravitational field and potential energy.
Why this matters
Newton's law of universal gravitation provides the specific force law that turns the general theory of central forces into the precise, testable predictions of planetary motion.
While the previous lecture established the mathematical machinery for any central force, this lecture applies it to the inverse-square law. This specific form of force is unique in classical mechanics because it yields closed, stable orbits described by conic sections. Understanding how the abstract parameters of the orbit equation map onto physical observables like semi-major axis and period is the key to connecting theoretical dynamics with astronomical data.
The Geometry of Bound Orbits
Imagine a planet moving around the Sun. In the previous lecture, we learned that any central force produces a planar orbit, and the shape of that orbit is determined by the total energy and angular momentum of the system. For the gravitational force, which weakens with the square of the distance, the mathematics simplifies beautifully. If the total energy of the system is negative, meaning the planet is "bound" to the Sun and cannot escape to infinity, the resulting path is not a circle, but an ellipse. Crucially, the Sun does not sit at the center of this ellipse; it sits at one of the two foci. This is the geometric essence of Kepler's First Law. The "eccentricity" of the orbit, a number between zero and one, tells us how stretched out this ellipse is. A perfectly circular orbit has an eccentricity of zero, while a highly elongated orbit has an eccentricity close to one. This geometric result is not an assumption; it is a direct consequence of the inverse-square nature of gravity.
From Binet's Equation to the Conic Section
We begin with the Binet orbit equation derived in D-029. For a gravitational force \( F(r) = -\dfrac{GMm}{r^2} \), substituting this into the general Binet form yields a specific differential equation for \( u(\theta) = 1/r(\theta) \). The equation becomes linear and has a well-known general solution. By applying the boundary conditions of the orbit, we find that the solution is a cosine function, which geometrically represents a conic section.
The general solution to this differential equation is \( u(\theta) = A \cos(\theta - \theta_0) + \frac{GM\mu}{L^2} \). By choosing our coordinate system such that the periapsis (closest approach) occurs at \( \theta = 0 \), we can rewrite this in the standard form of a conic section. The parameter \( p \), known as the semi-latus rectum, is determined by the angular momentum \( L \) and the reduced mass \( \mu \). The eccentricity \( e \) is fixed by the total energy \( E \) together with \( L \). For bound orbits where \( E < 0 \), the eccentricity \( e \) is strictly less than 1, confirming the elliptical shape.
Here, \( p = \dfrac{L^2}{\mu GM} \). This equation is the master key for orbital shape. It tells us that the distance \( r \) varies periodically with the angle \( \theta \), reaching a minimum at \( \theta = 0 \) and a maximum at \( \theta = \pi \).
Areal Velocity and the Second Law
Kepler's Second Law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. In the language of mechanics, this is a direct statement of the conservation of angular momentum. Since the gravitational force is central, the torque is zero, and the angular momentum \( L \) is constant. The rate at which area is swept out, \( \dfrac{dA}{dt} \), is directly proportional to this constant angular momentum.
This result is profound because it implies that the planet moves faster when it is closer to the Sun (smaller \( r \)) and slower when it is farther away (larger \( r \)), such that the product \( r^2 \dot{\theta} \) remains constant. This is not an empirical rule but a necessary consequence of the rotational symmetry of the gravitational potential.
The Period and the Third Law
Kepler's Third Law relates the orbital period \( T \) to the size of the orbit. For an elliptical orbit, the "size" is characterized by the semi-major axis \( a \). By integrating the areal velocity over one full period, we can derive the relationship between \( T \) and \( a \). The area of an ellipse is \( \pi a b \), where \( b \) is the semi-minor axis. Using the geometric relation \( b = a\sqrt{1-e^2} \) and the definition of the semi-latus rectum \( p = a(1-e^2) \), we can express the total area in terms of \( a \) and \( e \). Combining this with the constant areal velocity yields the final result.
In the case where the central mass \( M \) is much larger than the orbiting mass \( m \) (such as the Sun and Earth), the term \( M+m \) is approximately \( M \). This shows that the square of the period is proportional to the cube of the semi-major axis. This law allows astronomers to determine the mass of a central body simply by observing the period and distance of a satellite.
The derivations
Kepler's Three Laws from Gravity
For a body of mass \(m\) bound to a spherically symmetric mass \(M\) by the Newtonian inverse-square force \(\vec{F} = -\dfrac{GMm}{r^2}\,\hat{r}\), the central-orbit shape equation integrates exactly to the conic \(r(\theta)=\dfrac{p}{1+e\cos\theta}\). For total energy \(E<0\) (equivalently \(0\le e<1\)) this is an ellipse with the centre of force at one focus (First Law). Conservation of angular momentum makes the areal velocity constant, \(\dfrac{dA}{dt}=\dfrac{L}{2\mu}\) (Second Law). Integrating that constant rate over one complete ellipse gives \(T^2=\dfrac{4\pi^2 a^3}{G(M+m)}\) (Third Law).
Effective Potential and the Orbit Equation
For a particle of (reduced) mass \(\mu\) moving under a central force \(\mathbf{F}=F(r)\,\hat{\mathbf{r}}\), conservation of angular momentum \(L=\mu r^2\dot\theta\) reduces the planar two-dimensional problem to one-dimensional radial motion in an effective potential \(U_{\text{eff}}(r)=U(r)+\dfrac{L^2}{2\mu r^2}\), governed by \(\tfrac12\mu\dot r^2+U_{\text{eff}}(r)=E\); the same substitution, with \(u\equiv 1/r\) taken as a function of the polar angle \(\theta\), yields the Binet orbit equation \(\dfrac{d^2u}{d\theta^2}+u=-\dfrac{\mu}{L^2u^2}\,F(1/u)\), whose solution is the geometric orbit \(r(\theta)\).
Gravitational Field, Potential, Circular Orbits and Escape Velocity
For a spherically symmetric body of mass \(M\) and radius \(R\), the gravitational field at a distance \(r \ge R\) from the centre is \(\vec{g} = -\frac{GM}{r^2}\,\hat{\vec{r}}\), the gravitational potential is \(\phi(r) = -\frac{GM}{r}\) (with \(\phi(\infty)=0\)), the potential energy of a test mass \(m\) is \(U(r) = -\frac{GMm}{r}\), the circular orbital speed is \(v = \sqrt{GM/r}\), and the escape velocity from the surface is \(v_{\text{esc}} = \sqrt{2GM/R}\).
A common misconception is that the semi-major axis \( a \) is the maximum distance from the focus. In fact, the maximum distance (apoapsis) is \( r_{\text{max}} = a(1+e) \), while the minimum distance (periapsis) is \( r_{\text{min}} = a(1-e) \). The semi-major axis is the average of these two distances, \( a = \dfrac{r_{\text{max}} + r_{\text{min}}}{2} \). Confusing \( a \) with the maximum radius leads to significant errors in calculating orbital periods and energies. C
Worked example
Problem. A satellite of mass \( m = 500\ \mathrm{kg} \) orbits the Earth (mass \( M = 5.97 \times 10^{24}\ \mathrm{kg} \), radius \( R = 6.37 \times 10^6\ \mathrm{m} \)) in a circular orbit at an altitude of \( h = 35,786\ \mathrm{km} \). Calculate the orbital period \( T \) and the total mechanical energy \( E \) of the satellite. Use \( G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2} \).
- Determine the orbital radius \( r \): \( r = R + h = 6.37 \times 10^6\ \mathrm{m} + 3.5786 \times 10^7\ \mathrm{m} = 4.2156 \times 10^7\ \mathrm{m} \).
- For a circular orbit, the semi-major axis \( a \) is equal to the radius \( r \). Thus, \( a = 4.2156 \times 10^7\ \mathrm{m} \).
- Calculate the period \( T \) using Kepler's Third Law: \( T = \sqrt{\dfrac{4\pi^2 a^3}{G M}} \). Since \( m \ll M \), we neglect \( m \) in the denominator.
- Substitute the values: \( T = \sqrt{\dfrac{4\pi^2 (4.2156 \times 10^7)^3}{(6.674 \times 10^{-11})(5.97 \times 10^{24})}} \).
- Calculate the numerator: \( 4\pi^2 (4.2156 \times 10^7)^3 \approx 39.478 \times 7.492 \times 10^{22} \approx 2.958 \times 10^{24} \).
- Calculate the denominator: \( (6.674 \times 10^{-11})(5.97 \times 10^{24}) \approx 3.984 \times 10^{14} \).
- Compute \( T^2 \): \( \dfrac{2.958 \times 10^{24}}{3.984 \times 10^{14}} \approx 7.425 \times 10^9 \ \mathrm{s^2} \).
- Compute \( T \): \( \sqrt{7.425 \times 10^9} \approx 8.616\times10^{4}\ \mathrm{s} \). Converting to hours: \( 86\,160 / 3600 \approx 23.93\ \mathrm{h} \).
- Calculate the total energy \( E \) for a circular orbit: \( E = -\dfrac{GMm}{2a} \).
- Substitute values: \( E = -\dfrac{(6.674 \times 10^{-11})(5.97 \times 10^{24})(500)}{2(4.2156 \times 10^7)} \).
- Numerator: \( 6.674 \times 5.97 \times 500 \times 10^{13} \approx 1.992 \times 10^{17} \).
- Denominator: \( 8.4312 \times 10^7 \).
- Result: \( E \approx -2.363 \times 10^9\ \mathrm{J} \).
Units check. The period is in seconds, consistent with time. The energy is in Joules, consistent with work. The negative sign correctly indicates a bound system.
Three quick tests
- If the eccentricity of an orbit is \( e = 0.5 \), what is the ratio of the apoapsis distance to the periapsis distance?
Answer
The ratio is \( \dfrac{1+e}{1-e} = \dfrac{1.5}{0.5} = 3 \). - Why does the total energy of a bound orbit have to be negative?
Answer
Because the potential energy is defined as zero at infinity and negative at finite distances, and the kinetic energy is insufficient to reach infinity, resulting in a net negative total energy. - How does the orbital period change if the semi-major axis is doubled?
Answer
According to \( T^2 \propto a^3 \), if \( a \) doubles, \( T^2 \) increases by a factor of 8, so \( T \) increases by a factor of \( \sqrt{8} \approx 2.83 \).
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-030 · D-029 · D-423 · Threads: force, energy, matter