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Gravitational field · Planetary motion

The Laws of Planetary Motion

When NASA's Mission Control at Johnson Space Center in Houston plans a Mars mission, how does it know how long the trip will take? By analyzing decades of observations made by Tycho Brahe, Johannes Kepler found three regularities in the motion of the planets (the first two published in 1609, the third in 1619), without yet knowing why they held. Seventy years later, Newton showed that all three laws are a direct consequence of his law of gravitation. Here you review the three laws, learn to derive the third law from gravitation, and use it to calculate the masses of celestial bodies and the length of trips like that one.

First law: elliptical orbits

For centuries it was assumed that celestial orbits were perfect circles traversed at uniform speed, but the meticulous observations of Mars made by Tycho Brahe showed a persistent deviation of $8'$ of arc from any circular model: which geometric figure describes the true trajectory of the planets with complete accuracy?

Elliptical orbit of a planet with the Sun at one of the foci: the perihelion (r_p), the aphelion (r_a), the center of the ellipse and the semi-major axis a are marked. center Sun (focus) perihelion rp aphelion ra a (semi-major axis)
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Johannes Kepler deduced three laws describing the motion of planets around the Sun from the observations of Tycho Brahe. a) State Kepler's first law (law of orbits). b)…

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 farthest 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.

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The Earth passes through perihelion at a distance of $r_p = 1.471\times10^{11}\ \text{m}$ from the Sun and through aphelion at $r_a = 1.521\times10^{11}\ \text{m}$. The…

The Earth passes through perihelion at a distance of $r_p = 1.471\times10^{11}\ \text{m}$ from the Sun and through aphelion at $r_a = 1.521\times10^{11}\ \text{m}$. The figure shows the orbit with both distances (not to scale):

Sun perihelion aphelion 1.471·10¹¹ m 1.521·10¹¹ m a = ?
  1. a) Calculate the semi-major axis of the Earth's orbit, $a = \frac{r_p + r_a}{2}$.
  2. b) Given that $1\ \text{UA} = 1.496\times10^{11}\ \text{m}$ (astronomical unit: mean Earth–Sun distance), express the result in astronomical units and comment on what you observe.
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The eccentricity of an ellipse is $e = \frac{c}{a}$, where $a$ is the semi-major axis and $c$ is the distance from the center of the ellipse to one of the foci. For the…

The eccentricity of an ellipse is $e = \frac{c}{a}$, where $a$ is the semi-major axis and $c$ is the distance from the center of the ellipse to one of the foci. For the Earth: $r_p = 1.471\times10^{11}\ \text{m}$ (perihelion) and $r_a = 1.521\times10^{11}\ \text{m}$ (aphelion). Observe the geometry of the orbit in the figure (not to scale):

Sun (focus) O perihelion aphelion 1.471·10¹¹ m 1.521·10¹¹ m a = ? c = ?
  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.
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Second law: the law of areas

When the Earth's orbital motion is measured over the course of a year, it turns out that it does not travel at constant speed: in January it moves at $30.3\ \text{km/s}$ and in July at $29.3\ \text{km/s}$. What geometric principle governs how a planet's speed changes at each point of its orbit? (Keep this in mind: we come back to it in the question below.)

Animation of Kepler's second law: two planets travel at the same time, over the same period, a long, fast arc near perihelion and a short, slow arc near aphelion, while the radius vector sweeps out two sectors of equal area. area (perihelion) area (aphelion)
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a) State Kepler's second law (law of areas). b) A planet moves along its elliptical orbit, passing through perihelion and aphelion. Apply the law of areas to reason in…
  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.
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Draw an elongated elliptical orbit in your notebook with the Sun located at one of its foci. a) Mark two equal time intervals, one centered at perihelion and the other…

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 centered 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.

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Third law: the law of periods

Observations of the Solar System show that the more distant planets take much longer to complete their orbit (Mercury: $a = 0.39\ \text{AU}$, $T = 0.24\ \text{years}$; Earth: $a = 1\ \text{AU}$, $T = 1\ \text{year}$; Jupiter: $a = 5.20\ \text{AU}$, $T = 11.86\ \text{years}$): what exact mathematical law relates the size of the orbit to the length of each planet's year?

Graph of T² against a³: the planets Mercury, Earth and Jupiter line up along a straight line through the origin, according to Kepler's third law. a³ T² 0 T² = C·a³ Mercury Earth Jupiter Gradient of the line = C
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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 $a$. b) The constant is equal…
  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 $C = \frac{4\times\pi^2}{G\times M}$, where $M$ is the mass of the central body. Explain whether $C$ is the same for all planets in the Solar System and whether it would also be the same for Jupiter's moons.
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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". a) Explain why…

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) Explain why this statement is false, taking into account that $C = \frac{4\times\pi^2}{G\times M}$, 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.
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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 $T^2 = a^3$.…

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).
Solve
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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…

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 to solve for the semi-major axis $a$.
  2. b) Calculate its value, in $\text{UA}$.
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Mars orbits the Sun with a semi-major axis $a = 1.524\ \text{UA}$. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units. a) Calculate…

Mars orbits the Sun with a semi-major axis $a = 1.524\ \text{UA}$. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units.

  1. a) Calculate the length of the Martian year, in Earth years.
  2. b) Express the result in days ($1$ year $= 365.25$ days).
Solve
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Kepler verified his third law with data from several planets. Earth has $a = 1\ \text{UA}$ and $T = 1$ year; Jupiter has $a = 5.20\ \text{UA}$ and $T = 11.86$ years. a)…

Kepler verified his third law with data from several planets. Earth has $a = 1\ \text{UA}$ and $T = 1$ year; Jupiter has $a = 5.20\ \text{UA}$ and $T = 11.86$ years.

  1. a) Calculate the ratio $\frac{T^2}{a^3}$ for Earth, in $\text{years}^2/\text{AU}^3$.
  2. b) Calculate the same ratio for Jupiter and check that it matches the previous one (allowing for rounding).
  3. c) Explain why this ratio is only identical for bodies orbiting the same central body.
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From gravitation to Kepler: deriving the third law

Circular orbit: the gravitational force F points from the body of mass m towards the central mass M and acts as the centripetal force; the velocity v is tangent to the orbit. r M F = G·M·m/r² v (tangent) m F acts as the centripetal force

Applications: comets and exoplanets

Animation of a Hohmann transfer orbit: the Earth and Mars move in the same direction along their orbits until Mars is 45° ahead of the Earth; then the ellipse appears and the spacecraft travels along it for half a period, faster near the Sun, arriving at Mars just as it reaches aphelion. transfer Earth Mars
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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…

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; $1\ \text{UA} = 1.496\times10^{11}\ \text{m}$.

  1. a) Calculate the semi-major axis of Ceres's orbit, in $\text{UA}$.
  2. b) Express its mean distance to the Sun in meters, using scientific notation.
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A comet has a semi-major axis twice that of another comet in the same system. a) Show, using $T^2 = a^3$, that if $a' = 2\times a$ then…

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 $a' = 2\times a$ then $T' = \sqrt{8}\times T \approx 2.83\times T$.
  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.
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A team of astronomers discovers an object whose orbital period is $T = 27$ years. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical…

A team of astronomers discovers an object whose orbital period is $T = 27$ years. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units.

  1. a) Calculate the semi-major axis of its orbit, in $\text{UA}$.
  2. b) Compare it with Neptune's orbit ($a \approx 30\ \text{UA}$): does the object orbit closer to or farther from the Sun than Neptune?
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