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Gravitational field · Planetary motion

The Laws of Planetary Motion

From the Sierra Negra, where the Large Millimeter Telescope is located, or from any rooftop in Mexico City on a clear night, the planets look like simple points that move slowly among the stars. Johannes Kepler analyzed Tycho Brahe's observations for years and found three regularities in that motion, without yet knowing why they held. Seventy years later, Newton showed that they are a consequence of the law of universal gravitation. On this page you will review the three laws, learn how to derive the third one, and use it to "weigh" the Sun and calculate periods.

First Law: Elliptical Orbits

For centuries it was believed that the heavenly bodies traced perfect circles. But Tycho Brahe's observations of Mars deviated by $8'$ of arc from any circular model. What shape exactly describes the real path of the planets?

Elliptical orbit of a planet with the Sun at one of the foci: the perihelion, the aphelion, the center of the ellipse, and the semi-major axis a are marked. center Sun (focus) perihelion aphelion 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 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.

Solve
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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.
Solve
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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.
Solve

Second Law: The Law of Areas

Earth 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 principle governs how a planet's speed changes along its orbit?

Two sectors of equal area swept out in equal times: one wide with a long arc near perihelion, the other narrow with a short arc near aphelion. Sun long arc, short r short arc, long r same area in the same time
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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.
Solve
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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 centred at perihelion and the other at…

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.

Solve
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By conservation of angular momentum, at perihelion and aphelion $r_p\times v_p = r_a\times v_a$ holds. The Earth passes through perihelion (…

By conservation of angular momentum, at perihelion and aphelion $r_p\times v_p = r_a\times v_a$ holds. The Earth passes through perihelion ($r_p = 1.471\times10^{11}\ \text{m}$) with a speed of $30.3\ \text{km/s}$, and aphelion is at $r_a = 1.521\times10^{11}\ \text{m}$. The diagram shows the orbit and the speeds at both points (not to scale):

Sun perihelion v = 30.3 km/s aphelion v = ? 1.471·10¹¹ m 1.521·10¹¹ m
  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.
Solve

Third Law: The Law of Periods

The farthest planets take much longer to complete one revolution. Mercury: $a = 0.39\ \text{AU}$, $T = 0.24$ years. Earth: $a = 1\ \text{AU}$, $T = 1$ year. Jupiter: $a = 5.20\ \text{AU}$, $T = 11.86$ years. What exact relationship links the size of the orbit to the length of the year?

Graph of T squared versus a cubed: Mercury and Jupiter line up on a straight line through the origin, whose slope is the constant C. a³ T² 0 T² = C·a³ Mercury Jupiter Slope 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. 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.
Solve
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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 for the semi-major axis $a$.
  2. b) Calculate its value, in $\text{UA}$.
Solve
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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) Reason 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) Reason 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.
Solve
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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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Venus completes its orbit around the Sun in $T = 0.615$ years. For bodies orbiting the Sun, $T^2 = a^3$ with $T$ in years and $a$ in astronomical units;…

Venus completes its orbit around the Sun in $T = 0.615$ 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 Venus's orbit, in $\text{UA}$.
  2. b) Express it in kilometers, using scientific notation.
Solve
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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?
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 coincides with the previous one (allowing for rounding).
  3. c) Explain why this ratio is only identical for bodies orbiting the same central body.
Solve

From Gravitation to Kepler: Deriving the Third Law

Circular orbit: the gravitational force F points from mass m toward the central mass M and acts as the centripetal force; the velocity v is tangent to the orbit. r M F = GMm/r² v (tangent) m F acts as the centripetal force
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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…

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:

r M m v F
  1. a) Equate the gravitational force, $F = \frac{G\times M\times m}{r^2}$, with the centripetal force of circular motion, $F_c = \frac{m\times v^2}{r}$, and obtain the orbital speed, $v = \sqrt{\frac{G\times M}{r}}$.
  2. b) Substitute $v = \frac{2\times\pi\times r}{T}$ into the previous result and deduce that $T^2 = \frac{4\times\pi^2}{G\times M}\times r^3$.
  3. c) Identify the constant of Kepler's third law in the expression obtained and explain what quantities it depends on.
Solve
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Kepler's third law allows us to "weigh" the Sun. Data: mean Earth–Sun distance $a = 1.496\times10^{11}\ \text{m}$, Earth's orbital period $T = 365.25$ days and…

Kepler's third law allows us to "weigh" the Sun. Data: mean Earth–Sun distance $a = 1.496\times10^{11}\ \text{m}$, Earth's orbital period $T = 365.25$ days and $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$. The figure represents Earth's orbit with the known data:

M = ? a = 1.496·10¹¹ m Earth T = 365.25 days
  1. a) Express the orbital period in seconds.
  2. b) Rearrange for the mass of the Sun from the expression $T^2 = \frac{4\times\pi^2\times a^3}{G\times M_S}$.
  3. c) Calculate $M_S$ and compare the result with the accepted value, $1.99\times10^{30}\ \text{kg}$.
Solve
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The Juno probe has studied Jupiter's system. Its moon Io orbits the planet with a semi-major axis $a = 4.22\times10^{8}\ \text{m}$ and a period of $T = 1.77$ days. Data:…

The Juno probe has studied Jupiter's system. Its moon Io orbits the planet with a semi-major axis $a = 4.22\times10^{8}\ \text{m}$ and a period of $T = 1.77$ days. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$.

  1. a) Express Io's period in seconds.
  2. b) Rearrange for the mass of Jupiter from Kepler's third law, $T^2 = \frac{4\times\pi^2\times a^3}{G\times M_J}$.
  3. c) Calculate $M_J$ and compare it with the accepted value, $1.90\times10^{27}\ \text{kg}$.
Solve
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The constant of Kepler's third law is $C = \frac{4\times\pi^2}{G\times M}$, where $M$ is the mass of the central body. Data:…

The constant of Kepler's third law is $C = \frac{4\times\pi^2}{G\times M}$, where $M$ is the mass of the central body. Data: $G = 6.67\times10^{-11}\ \text{N}\times\text{m}^2/\text{kg}^2$, mass of the Sun $M_S = 1.99\times10^{30}\ \text{kg}$, mass of the Earth $M_T = 5.97\times10^{24}\ \text{kg}$ and semi-major axis of the Moon's orbit $a = 3.84\times10^{8}\ \text{m}$.

  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.
Solve

Applications: Comets, Exoplanets, and Interplanetary Travel

Hohmann transfer orbit: an ellipse tangent to Earth's orbit at perihelion and to Mars's orbit at aphelion, with the Sun at a focus. Sun Earth Mars transfer
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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.
Solve
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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.
Solve
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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 $T$…

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):

Sun perihelion 0.586 AU aphelion ? T = 76 years
  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, $r_p\times v_p = r_a\times v_a$, calculate the ratio $\frac{v_p}{v_a}$ and explain why Halley is only observable from Earth for a few months every $76$ years.
Solve
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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…

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 $r_p = a\times(1 - e)$ and $r_a = a\times(1 + e)$.

  1. a) Calculate $r_p$ and $r_a$, in $\text{UA}$.
  2. b) Applying conservation of angular momentum, $r_p\times v_p = r_a\times v_a$, 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.
Solve
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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…

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:

transfer Sun Earth Mars 1 AU 1.524 AU a = ?
  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.
Solve
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An exoplanet is discovered orbiting a star of mass $M_E = 4\times M_S$ (four times the mass of the Sun) with a semi-major axis $a = 1\ \text{UA}$, equal to Earth's. a)…

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

  1. a) Starting from $T^2 = \frac{4\times\pi^2\times a^3}{G\times M}$, show that $\frac{T^2}{T_{Tierra}^2} = \frac{a^3}{a_{Tierra}^3}\times\frac{M_S}{M_E}$.
  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.
Solve

Back to the beginning: the planets you see as points in the sky trace ellipses with the Sun at a focus, sweep out equal areas in equal times, and satisfy $T^2/a^3 = C$.

Summary

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