The Laws of Planetary Motion

Johannes Kepler deduced three laws describing the motion of planets around the Sun from the observations of Tycho Brahe.

1. a) State Kepler's first law (law of orbits).
2. b) Draw an elliptical orbit in your notebook and indicate on it the two foci of the ellipse, the position of the Sun, perihelion (the point closest to the Sun) and aphelion (the point furthest away).
3. c) State what the semi-major axis $a$ of the orbit is, and write its expression in terms of the distances at perihelion, $r_p$, and aphelion, $r_a$.

You can do it on paper and send a photo with your answer.

The Earth passes through perihelion at a distance of $r_p = 1,471\cdot10^{11}\ \text{m}$ from the Sun and through aphelion at $r_a = 1,521\cdot10^{11}\ \text{m}$. The figure shows the orbit with both distances (not to scale):

1. a) Calculate the semi-major axis of the Earth's orbit, $a = \frac{r_p + r_a}{2}$.
2. b) Given that $1\ \text{UA} = 1,496\cdot10^{11}\ \text{m}$ (astronomical unit: mean Earth–Sun distance), express the result in astronomical units and comment on what you observe.

The eccentricity of an ellipse is $e = \frac{c}{a}$, where $a$ is the semi-major axis and $c$ is the distance from the centre of the ellipse to one of the foci. For the Earth: $r_p = 1,471\cdot10^{11}\ \text{m}$ (perihelion) and $r_a = 1,521\cdot10^{11}\ \text{m}$ (aphelion). Observe the geometry of the orbit in the figure (not to scale):

1. a) Calculate the semi-major axis, $a = \frac{r_p + r_a}{2}$.
2. b) Calculate the focal distance, $c = \frac{r_a - r_p}{2}$, and the eccentricity $e$ of the Earth's orbit.
3. c) A value of $e = 0$ corresponds to a circle and a value close to $1$ to a highly elongated ellipse. Comment on the shape of the Earth's orbit according to the value obtained.

1. a) State Kepler's second law (law of areas).
2. b) A planet moves along its elliptical orbit, passing through perihelion and aphelion. Apply the law of areas to reason in which of these two points it moves faster.
3. c) The areal velocity (area swept out by the radius vector per unit time) is constant because the planet's angular momentum is conserved. Explain qualitatively why a shorter radius vector at perihelion implies a higher velocity.

Draw an elongated elliptical orbit in your notebook with the Sun located at one of its foci.

1. a) Mark two equal time intervals, one centred at perihelion and the other at aphelion, and shade the areas swept out by the radius vector in both sections.
2. b) According to the law of areas, what relationship exists between the two shaded areas?
3. c) Reason, from the diagram, in which section the planet covers a longer arc in the same time, and what this implies about its velocity.

You can do it on paper and send a photo with your answer.

By conservation of angular momentum, at perihelion and aphelion $r_p\cdot v_p = r_a\cdot v_a$ holds. The Earth passes through perihelion ($r_p = 1,471\cdot10^{11}\ \text{m}$) with a speed of $30,3\ \text{km/s}$, and aphelion is at $r_a = 1,521\cdot10^{11}\ \text{m}$. The diagram shows the orbit and the speeds at both points (not to scale):

1. a) Rearrange for the speed at aphelion, $v_a$, from the above expression.
2. b) Calculate its value, in $\text{km/s}$.
3. c) Perihelion occurs in early January and aphelion in early July. State in which of these months the Earth moves faster along its orbit and relate this to Kepler's second law.

1. a) State Kepler's third law (law of periods) and write its mathematical expression, $\frac{T^2}{a^3} = C$, stating the meaning of $T$ and $a$.
2. b) The constant is equal to $C = \frac{4\cdot\pi^2}{G\cdot M}$, where $M$ is the mass of the central body. Reason whether $C$ is the same for all planets in the Solar System and whether it would also be the same for Jupiter's moons.

For bodies orbiting the Sun, if the period $T$ is expressed in years and the semi-major axis $a$ in astronomical units, Kepler's third law reduces to $T^2 = a^3$. Mercury orbits with a semi-major axis $a = 0,387\ \text{UA}$.

1. a) Calculate Mercury's orbital period, in years.
2. b) Express it in Earth days ($1$ year $= 365,25$ days).

An asteroid in the main belt takes $T = 8$ years to complete its orbit around the Sun. For bodies orbiting the Sun, $T^2 = a^3$ holds, with $T$ in years and $a$ in astronomical units.

1. a) Rearrange the above expression for the semi-major axis $a$.
2. b) Calculate its value, in $\text{UA}$.

A student claims: "as $\frac{T^2}{a^3} = C$ is a universal constant, its value is the same for the Earth around the Sun as for the Moon around the Earth".

1. a) Reason why this statement is false, taking into account that $C = \frac{4\cdot\pi^2}{G\cdot M}$, where $M$ is the mass of the central body.
2. b) State a set of bodies for which the constant would coincide with that of Jupiter's moons.

For a circular orbit of radius $r$, Kepler's third law can be derived from Newton's law of universal gravitation. The diagram shows a planet in a circular orbit with the force acting on it:

1. a) Equate the gravitational force, $F = \frac{G\cdot M\cdot m}{r^2}$, with the centripetal force of circular motion, $F_c = \frac{m\cdot v^2}{r}$, and obtain the orbital speed, $v = \sqrt{\frac{G\cdot M}{r}}$.
2. b) Substitute $v = \frac{2\cdot\pi\cdot r}{T}$ into the previous result and deduce that $T^2 = \frac{4\cdot\pi^2}{G\cdot M}\cdot r^3$.
3. c) Identify the constant of Kepler's third law in the expression obtained and explain what quantities it depends on.

Kepler's third law allows us to "weigh" the Sun. Data: mean Earth–Sun distance $a = 1,496\cdot10^{11}\ \text{m}$, Earth's orbital period $T = 365,25$ days and $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. The figure represents Earth's orbit with the known data:

1. a) Express the orbital period in seconds.
2. b) Rearrange for the mass of the Sun from the expression $T^2 = \frac{4\cdot\pi^2\cdot a^3}{G\cdot M_S}$.
3. c) Calculate $M_S$ and compare the result with the accepted value, $1,99\cdot10^{30}\ \text{kg}$.

The Juno probe has studied Jupiter's system. Its moon Io orbits the planet with a semi-major axis $a = 4,22\cdot10^{8}\ \text{m}$ and a period of $T = 1,77$ days. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$.

1. a) Express Io's period in seconds.
2. b) Rearrange for the mass of Jupiter from Kepler's third law, $T^2 = \frac{4\cdot\pi^2\cdot a^3}{G\cdot M_J}$.
3. c) Calculate $M_J$ and compare it with the accepted value, $1,90\cdot10^{27}\ \text{kg}$.

The constant of Kepler's third law is $C = \frac{4\cdot\pi^2}{G\cdot M}$, where $M$ is the mass of the central body. Data: $G = 6,67\cdot10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$, mass of the Sun $M_S = 1,99\cdot10^{30}\ \text{kg}$, mass of the Earth $M_T = 5,97\cdot10^{24}\ \text{kg}$ and semi-major axis of the Moon's orbit $a = 3,84\cdot10^{8}\ \text{m}$.

1. a) Calculate, in SI units, the constant $C$ for the Solar System and for the Earth–Moon system.
2. b) Reason what causes the huge difference between both values.
3. c) Use the constant for the Earth–Moon system to predict the Moon's orbital period and express it in days.

Ceres, the largest body in the asteroid belt, has an orbital period of $T = 4,60$ years. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units; $1\ \text{UA} = 1,496\cdot10^{11}\ \text{m}$.

1. a) Calculate the semi-major axis of Ceres's orbit, in $\text{UA}$.
2. b) Express its mean distance to the Sun in metres, using standard form.

A comet has a semi-major axis twice that of another comet in the same system.

1. a) Show, using $T^2 = a^3$, that if $a' = 2\cdot a$ then $T' = \sqrt{8}\cdot T \approx 2,83\cdot T$.
2. b) An asteroid orbits the Sun with $a = 2\ \text{UA}$. Calculate its period, in years.
3. c) Reason, without carrying out new calculations, whether a planet located at half the distance from the Sun as the Earth would have a year longer or shorter than Earth's.

Halley's Comet has a period of $T = 76$ years and passes through perihelion at $r_p = 0,586\ \text{UA}$ from the Sun. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units. The figure shows its elongated orbit (not to scale):

1. a) Calculate the semi-major axis of its orbit, in $\text{UA}$.
2. b) Taking into account that $a = \frac{r_p + r_a}{2}$, calculate the distance at aphelion, $r_a$.
3. c) Applying conservation of angular momentum, $r_p\cdot v_p = r_a\cdot v_a$, calculate the ratio $\frac{v_p}{v_a}$ and explain why Halley is only observable from Earth for a few months every $76$ years.

Mercury has the most eccentric planetary orbit in the Solar System: $a = 0,387\ \text{UA}$ and $e = 0,206$. The distances to the Sun at perihelion and aphelion are $r_p = a\cdot(1 - e)$ and $r_a = a\cdot(1 + e)$.

1. a) Calculate $r_p$ and $r_a$, in $\text{UA}$.
2. b) Applying conservation of angular momentum, $r_p\cdot v_p = r_a\cdot v_a$, calculate the ratio $\frac{v_p}{v_a}$.
3. c) For the Earth, this ratio is equal to $1,034$. Compare both results and explain what role eccentricity plays in the variation of orbital speed.

To travel from Earth to Mars with minimum fuel consumption, a Hohmann transfer orbit is used: an ellipse with perihelion at Earth's orbit ($1\ \text{UA}$) and aphelion at Mars's orbit ($1,524\ \text{UA}$). For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units. The diagram shows the transfer orbit:

1. a) Calculate the semi-major axis of the transfer ellipse.
2. b) Calculate the complete period of that ellipse, in years.
3. c) The outbound journey takes half a period. Calculate its duration in days ($1$ year $= 365,25$ days) and reason why missions to Mars can only be launched in certain windows that repeat every $26$ months.

An exoplanet is discovered orbiting a star of mass $M_E = 4\cdot M_S$ (four times the mass of the Sun) with a semi-major axis $a = 1\ \text{UA}$, equal to Earth's.

1. a) Starting from $T^2 = \frac{4\cdot\pi^2\cdot a^3}{G\cdot M}$, show that $\frac{T^2}{T_{Tierra}^2} = \frac{a^3}{a_{Tierra}^3}\cdot\frac{M_S}{M_E}$.
2. b) Calculate the period of the exoplanet, in Earth years.
3. c) Explain why, in this case, the simplified relationship $T^2 = a^3$ valid for the Solar System cannot be applied directly.