Orbits and The Wider Universe

AS UNIT 4 Fields and Options 4.3 Orbits and The Wider Universe Learners should be able to demonstrate and apply their knowledge and understanding of:
a)	Kepler’s three laws of planetary motion
b)	Newton’s law of gravitation [math]F = G \frac{M_1 M_2}{r^2}[/math] in simple examples, including the motion of planets and satellites
c)	How to derive Kepler’s 3rd law, for the case of a circular orbit from Newton’s law of gravity and the formula for centripetal acceleration
d)	How to use data on orbital motion, such as period or orbital speed, to calculate the mass of the central object
e)	How the orbital speeds of objects in spiral galaxies implies the existence of dark matter
f)	How the recently discovered Higgs boson may be related to dark matter
g)	How to determine the position of the centre of mass of two spherically symmetric objects, given their masses and separation, and calculate their mutual orbital period in the case of circular orbits
h)	The Doppler relationship in the form [math]\frac{∆λ}{λ} = \frac{v}{c}[/math]
i)	How to determine a star’s radial velocity (i.e. the component of its velocity along the line joining it and an observer on the Earth) from data about the Doppler shift of spectral lines
j)	The use of data on the variation of the radial velocities of the bodies in a double system (for example, a star and orbiting exo-planet) and their orbital period to determine the masses of the bodies for the case of a circular orbit edge on as viewed from the Earth
k)	How the Hubble constant ([math]H_o[/math] ) relates galactic radial velocity (v) to distance (D) and it is defined by [math]v = H_0 D[/math]
l)	Why [math]\frac{1}{H_0} [/math] approximates the age of the universe
m)	How the equation [math]\rho_c = \frac{3H_0^2}{8\pi G}[/math] for the critical density of a ‘flat’ universe can be derived very simply using conservation of energy

a) Kepler’s Three Laws of Planetary Motion & Newton’s Law of Gravitation
1. Kepler’s Three Laws of Planetary Motion
Johannes Kepler formulated three fundamental laws describing planetary motion based on observational data collected by Tycho Brahe. These laws apply to any object orbiting another under the influence of gravity, such as planets around the Sun or satellites around Earth.
⇒ Kepler’s First Law (Law of Ellipses)
A planet moves in an elliptical orbit with the Sun at one focus.
– Ellipse: A stretched-out circle with two foci.
– The Sun is located at one focus of the elliptical orbit.
– Some planets have nearly circular orbits, but technically all follow ellipses.
Figure 1 Kepler’s first law
⇒ Example:
Earth’s orbit is almost circular, but comets like Halley’s follow highly elliptical paths.
⇒ Kepler’s Second Law (Law of Equal Areas)
A line drawn from the Sun to a planet sweeps out equal areas in equal time intervals.
– Planets move faster when they are closer to the Sun (perihelion) and slower when they are farther away (aphelion).
– This happens because gravity pulls the planet in more strongly when it is closer to the Sun.
⇒ Example:
Earth moves faster in January (when it is closest to the Sun) and slower in July (when it is farther).
Figure 2 Kepler’s second Law
⇒ Kepler’s Third Law (Law of Periods)
The square of a planet’s orbital period is proportional to the cube of its average distance from the Sun.
Mathematically:
[math]T^2 \propto r^3[/math]
or
[math]\frac{T^2}{r^3} = Constant[/math]
Where:
– T = orbital period (time taken for one complete orbit).
– r = average distance from the Sun (semi-major axis of the ellipse).
⇒ Example:
Mars is about 1.52 times farther from the Sun than Earth.
Since
[math]T^2 \propto r^3[/math]
its orbital period is about 1.88 Earth years.
Figure 3 Kepler’s third Law
b) Newton’s Law of Gravitation
Newton extended Kepler’s laws by proposing that the gravitational force between two masses follows the equation:
[math]F = G \frac{M_1 M_2}{r^2}[/math]
Where:
– F = gravitational force (N),
– G = gravitational constant ([math]6.674 \times 10^{-11} \,\text{Nm}^2/\text{kg}^2[/math] ),
– [math]M_1 \ and \ M_2[/math] masses of the two objects (kg),
– r = distance between their centers (m).
⇒ Example 1: Earth and Moon
– Earth’s mass = [math]5.972 \times 10^{24} \text{kg}[/math]
– Moon’s mass = [math]7.348 \times 10^{22} \text{kg}[/math]
– Distance = [math]3.84 \times 10^8 \text{m}[/math]
Using Newton’s formula, the gravitational force between Earth and the Moon can be calculated.
⇒ Example 2: Satellites
– A satellite orbiting Earth is pulled inward by gravity, preventing it from flying off into space.
– The satellite’s velocity balances gravity, keeping it in orbit.
c) Deriving Kepler’s 3rd Law from Newton’s Laws
To derive Kepler’s Third Law, we consider a circular orbit where the gravitational force provides the necessary centripetal force.
⇒ Centripetal force formula:
[math]F_{\text{centripetal}} = \frac{mv^2}{r}[/math]
⇒ Gravitational force formula:
[math]F_{\text{gravity}} = \frac{GMm}{r^2}[/math]
Since gravity provides the centripetal force, we set these equal:
[math]\frac{mv^2}{r} = \frac{GMm}{r^2}[/math]
Cancel m from both sides:
[math]\frac{v^2}{r} = \frac{GM}{r^2} \\ v = \sqrt{\frac{GM}{r}}[/math]
We know orbital velocity is related to period T by:
[math]T = \frac{2\pi r}{T}[/math]
Substituting this into the equation:
[math]\frac{GM}{r^2} = \frac{(2\pi r)^2}{T^2 r}[/math]
Simplifying:
[math]\frac{GM}{r} = \frac{4\pi^2 r^2}{T^2}[/math]
Rearranging for [math]T^2 [/math]:
[math]T^2 = \frac{4\pi^2 r^3}{GM}[/math]
Since [math] \frac{4\pi^2}{GM}[/math] is a constant, we get:
[math]T^2 \propto r^3[/math]
This is Kepler’s Third Law derived from Newton’s laws!
d) Calculating the Mass of a Central Object Using Orbital Motion Data
We can use the equation derived above:
[math]T^2 = \frac{4\pi^2 r^3}{GM}[/math]
Rearranging for M:
[math]M = \frac{4\pi^2 r^3}{GT^2}[/math]
⇒ Example:
Finding the Mass of the Sun
– Earth’s orbital period T=1 year = [math]3.154 \times 10^7 \,\text{s}[/math]
– Earth’s average distance from the Sun r= [math]1.496 \times 10^{11} \text{m}[/math]
– [math]G = 6.674 \times 10^{-11} \,\text{Nm}^2/\text{kg}^2[/math]
[math]M = \frac{4\pi^2 r^3}{G T^2} \\
M = \frac{4\pi^2 (1.496 \times 10^{11})^3}{(6.674 \times 10^{-11}) (3.154 \times 10^7)^2} \\
M = 1.989 \times 10^{30} \,\text{kg}[/math]
Solving gives [math]M = 1.989 \times 10^{30} \,\text{kg}[/math], which is the mass of the Sun.
Finding the mass of Earth
The orbital period, T and the orbital radius r, we can calculate the mass of an astronomical body by rearranging out formula for M
[math]M = \frac{4\pi^2 r^3}{G T^2}[/math]
can estimate the mass of earth by measuring the orbital period and radius of the Moon. It takes 27 days (2.3 million seconds) for the Moon to orbit the Earth once and the radius of orbit is approximately 385000 km. Thus
[math]M = \frac{4\pi^2 (r_{\text{moon}})^3}{G (T_{\text{Moon}})^2} \\
M \approx 6.38 \times 10^{24} \,\text{kg}[/math]

e) Orbital Speeds in Spiral Galaxies & Dark Matter:
⇒ Observations of Orbital Speeds in Galaxies:
In a spiral galaxy, stars orbit around the galactic center under the influence of gravity. Based on Newton’s laws and Kepler’s third law, we expect that:
– Closer stars (near the galaxy’s center) should orbit faster due to the strong gravitational pull from the dense mass concentration.
– Farther stars (in the outer regions) should orbit slower because the visible mass decreases, reducing gravity.
However, observations show something unexpected:
– Stars in the outer parts of galaxies move at nearly constant speeds, much faster than expected.
– This contradicts the prediction that velocity should decrease at large distances.
⇒ Dark Matter Hypothesis
Since visible mass (stars, gas, dust) cannot account for the high speeds, scientists propose the existence of dark matter, an invisible form of matter that provides extra gravitational pull.
– Dark matter must be spread out in a halo around galaxies, exerting additional gravitational force to keep the outer stars moving faster.
– The rotation curves of galaxies (graphs of orbital speed vs. distance) remain flat instead of decreasing, supporting the dark matter theory.
Figure 4 Dark matter hypothesis
⇒ Example:
The Milky Way’s rotation curve suggests that over 80% of its mass is dark matter.
f) The Higgs Boson and Dark Matter
The Higgs boson was discovered in 2012 at the Large Hadron Collider (LHC). It plays a key role in the Standard Model of particle physics, giving particles their mass via the Higgs field.
Possible Connection Between the Higgs Boson and Dark Matter
1. Dark Matter Particles Could Interact with the Higgs Field
– If dark matter particles have mass, they may acquire it through the Higgs mechanism, just like ordinary particles.
2. The Higgs Boson Could Decay Into Dark Matter
– Some theories suggest the Higgs boson might decay into unknown particles, which could be dark matter.
– Physicists are searching for invisible decays of the Higgs boson at the LHC to test this idea.
Figure 5 Higgs Boson
3. Extensions of the Standard Model
– Some models propose additional Higgs-like particles or new forces related to dark matter.
– Candidates for dark matter include Weakly Interacting Massive Particles (WIMPs) and axions, which might interact with the Higgs field.
Although the Higgs boson’s exact connection to dark matter is still unknown, ongoing experiments aim to uncover new physics beyond the Standard Model.
g) Centre of Mass of Two Spherical Objects & Mutual Orbital Period
When two objects (like planets, stars, or binary systems) orbit each other, they actually orbit around their common center of mass (COM), also called the barycenter.
⇒ Finding the Centre of Mass
For two objects with masses [math]M_1[/math] and [math]M_2[/math] separated by distance r, the centre of mass is located at distance [math]r_1[/math] from [math]M_1[/math], given by:
[math]r_1 = \frac{M_2}{M_1 + M_2} \times r[/math]
Similarly, the distance of the center of mass from [math]M_2[/math] is:
[math]r_2 = \frac{M_1}{M_2 + M_1} \times r[/math]
The two objects orbit this center of mass in circular orbits.
⇒ Example:
– Earth and Moon system: The barycenter is inside the Earth, but not at its center.
– Sun and Jupiter system: The barycenter is slightly outside the Sun, meaning the Sun also wobbles!
Finding the Orbital Period in a Circular Orbit
Using Kepler’s Third Law for a binary system, the mutual orbital period T is given by:
[math]T^2 = \frac{4\pi^2 r^3}{G (M_1 + M_2)} \\
T = 2\pi \sqrt{\frac{d^3}{G (M_1 + M_2)}}[/math]
Where:
– r is the total separation between the two objects,
– G is the gravitational constant,
– [math]M_1[/math] and [math]M_2[/math] are the masses of the two objects.
⇒ Example Calculation:
Earth-Moon system: Given [math]M_1 = 5.972 \times 10^{24} \,\text{kg} \quad (\text{Earth}) M_2 = 7.348 \times 10^{22} \,\text{kg} \quad (\text{Moon}) r = 3.84 \times 10^8 \,\text{m}[/math], we can calculate T, which is about 27.3 days.
h) Doppler Effect & Redshift
The Doppler effect describes how the wavelength of light or sound changes when an object moves relative to an observer.
Figure 6 Doppler effect
For light, it is given by:
[math]\frac{\Delta \lambda}{\lambda} = \frac{v}{c}[/math]
Where:
– [math]\Delta \lambda = \lambda_{\text{observed}} – \lambda_{\text{rest}}[/math](Change in wavelength),
– [math]\lambda[/math] is the original wavelength,
– v is the velocity of the source relative to the observer,
– c is the speed of light ([math]3.00 \times 10^8 m/s[/math]).
⇒ Redshift and Blueshift
If an object is moving away, the wavelength increases → Redshift ([math]z = \frac{\Delta \lambda}{\lambda} > 0[/math]).
If an object is moving toward us, the wavelength decreases → Blueshift ([math]z<0[/math]).
Figure 7 Redshift and blueshift
⇒ Example: Expanding Universe
Light from distant galaxies is redshifted, meaning they are moving away from us.
This supports the Big Bang theory, showing that the universe is expanding.
The Hubble Law relates redshift to distance:
[math]v=H_0d[/math]
Where [math]H_0[/math] is Hubble’s constant.
⇒ Example: Exoplanet Detection
The wobble of a star due to an orbiting planet causes periodic Doppler shifts in its light.
Scientists measure these shifts to detect exoplanets
Determining a Star’s Radial Velocity Using the Doppler Shift
The radial velocity of a star is the component of its velocity along the line of sight between the star and an observer on Earth. We determine it using the Doppler effect, which describes how wavelengths of light change due to motion.
The Doppler shift equation for light is:
[math]\frac{\Delta \lambda}{\lambda} = \frac{v}{c}[/math]
Where:
– [math]\Delta \lambda = \lambda_{\text{observed}} – \lambda_{\text{rest}}[/math]is the shift in wavelength,
– [math]\lambda[/math]is the rest (original) wavelength,
– v is the radial velocity (velocity along the line of sight),
– c is the speed of light ([math]3.00 \times 10^8 m/s[/math]).
Figure 8 Radial velocity
⇒ Use This Equation
– If the star moves away, its spectral lines shift to longer (redshifted) wavelengths → v>0.
– If the star moves toward us, its spectral lines shift to shorter (blue shifted) wavelengths → v<0
– Measuring Δλ for known spectral lines allows us to calculate the radial velocity v.
⇒ Example Calculation
A hydrogen spectral line has a rest wavelength of 656.3 nm, but is observed at 656.8 nm. The shift is:
[math]\Delta \lambda = 656.8 – 656.3 = 0.5 \,\text{nm}[/math]
Using the Doppler equation:
[math]\frac{0.5}{656.3} = \frac{v}{3 \times 10^8}[/math]
Solving for v:
[math]v = \frac{0.5}{656.3} \times 3 \times 10^8 \\
v = 2.28 \times 10^5 \,\text{m/s}[/math]
So, the star is moving away from Earth at 228 km/s.
j) Determining the Masses of a Star and Exoplanet from Radial Velocity Variations
⇒ Detecting Exoplanets Using Radial Velocity
In a star-planet system, both the star and planet orbit their common center of mass (barycenter).
The star’s motion around this barycenter causes periodic changes in its radial velocity (detected as Doppler shifts).
By analyzing these shifts, we can determine:
– Orbital period T
– Star’s velocity v∗ (from spectral line shifts)
– Planet’s velocity [math]v_p[/math](via momentum conservation)
⇒ Finding the Mass of the Exoplanet
For a circular orbit, the planet and star obey:
[math]M_* v_* = M_p v_p[/math]
Where:
– [math]M_*[/math] is the mass of the star,
– [math]M_p[/math] is the mass of the planet,
– [math]v_*[/math] and [math]v_p[/math] are their respective orbital speeds.
We measure [math]v_*[/math] from Doppler shifts and estimate [math]v_p[/math] using Kepler’s third law:
[math]T^2 = \frac{4\pi^2 r^3}{G (M_* + M_p)}[/math]
From this, we can determine [math]M_p[/math], the planet’s mass.
k) Hubble’s Law and the Expansion of the Universe
⇒ Hubble’s law states:
[math]v = H_0D[/math]
Where:
– v is the recession velocity of a galaxy (measured via redshift),
– D is its distance from Earth
– [math]H_0[/math] is the Hubble constant (current best estimate [math]67 – 74 \,\text{km/s/Mpc}[/math]).
⇒ Implications of Hubble’s Law
Galaxies farther away move faster, meaning the universe is expanding.
This supports the Big Bang theory: all galaxies were closer together in the past.
Figure 9 Hubble’s Law
i) Approximating the Age of the Universe Using [math]\frac{1}{H_0}[/math]
Since [math]H_0[/math] relates velocity and distance, we can estimate the time since the universe started expanding by considering:
[math]\text{Time} = \frac{\text{Distance}}{\text{Velocity}} \\
\text{Time} = \frac{D}{H_0 D} \\
\text{Time} = \frac{1}{H_0}[/math]
Thus, the age of the universe is approximately:
[math]t_{\text{universe}} \approx \frac{1}{H_0}[/math]
⇒ Example Calculation
Using [math]H_0 \approx 70 \,\text{km/s/Mpc}[/math] ,first convert:
[math]H_0 = \frac{70 \times 10^3 \,\text{m/s}}{3.09 \times 10^{22} \,\text{m}} \\
H_0 \approx 2.27 \times 10^{-18} \,\text{s}^{-1} \\
t_{\text{universe}} \approx \frac{1}{H_0} \\
t_{\text{universe}} \approx \frac{1}{2.27 \times 10^{-18}} \\
t_{\text{universe}} \approx 4.4 \times 10^{17} \,\text{s}[/math]
Converting to years:
[math]t_{\text{universe}} \approx 14 \text{billion years}[/math]
This agrees with cosmic microwave background measurements
m) Deriving the Critical Density Equation:[math]\rho_c = \frac{3H_0^2}{8\pi G}[/math]
The critical density [math]\rho_c [/math] is the density required for the universe to be flat (neither collapsing nor expanding forever).
We use conservation of energy in an expanding universe:
Total Energy=Kinetic Energy+Gravitational Potential Energy=0
For a mass m in a spherical universe of radius R:
– Kinetic Energy:
[math]\text{KE} = \frac{1}{2} m v^2[/math]
– Gravitational Potential Energy:
[math]\text{PE} = -\frac{GMm}{R}[/math]
Where M is the total mass of the universe:
[math]M = \frac{4}{3} \pi R^3 \rho [/math]
Setting total energy to zero:
[math]\frac{1}{2} m v^2 – \frac{G \frac{4}{3} \pi R^3 \rho m}{R} = 0[/math]
Using Hubble’s law [math]v = H_0 R[/math], substituting [math]v^2[/math]:
[math]\frac{1}{2} m H_0^2 R^2 = \frac{4}{3} \pi G R^2 \rho m [/math]
Canceling mmm and [math]R^2[/math], solving for ρ:
[math]\rho_c = \frac{3H_0^2}{8\pi G}[/math]
This is the critical density required for a flat universe. If:
– [math]\rho > \rho_c[/math] → Universe collapses (closed universe).
– [math]\rho < \rho_c[/math]→ Universe expands forever (open universe).
– [math]\rho = \rho_c[/math] → Universe is flat, expanding at a balanced rate.
⇒ Example Calculation
For [math]H_0 = 70 \,\text{km/s/Mpc}[/math]:
[math]\rho_c = \frac{3H_0^2}{8\pi G} \\
\rho_c = \frac{3(2.27 \times 10^{-18})^2}{8\pi (6.674 \times 10^{-11})} \\
\rho_c = 9.2 \times 10^{-27} \,\text{kg/m}^3[/math]
This is extremely low—only a few hydrogen atoms per cubic meter.