Kepler's third law of planetary motion says that the average distance of a
planet from the Sun *cubed* is directly proportional to the orbital
period *squared.* Newton found that his gravity force law could explain Kepler's
laws. Since Newton's law of gravity applies to any object with mass, Kepler's
laws can be used for any object orbiting another object. Let's look at
satellites orbiting a planet.

If you have two satellites (#1 and #2) orbiting a planet, Kepler's third law says:

If you have measured the orbital period of one satellite around a planet, you can then easily find how long it would take any other satellite to orbit the planet in any size oribt. Kepler's third law can be simplified down to

period #1 | = | period #2 × Sqrt[(distance #1/distance #2)^{3}] |

OR | ||

period #1 | = | period #2 × (distance #1/distance #2)^{3/2}. |

Those of you with a scientific calculator (one that does powers, trig functions, scientific notation, etc.) will want to use the formula on the last line (remember that 3/2 = 1.5). Those with a calculator that just has a square root button will want to use the formula on the second-to-last line.

If the satellite is orbiting the Sun,
then the relation can be greatly simplified with an appropriate choice of
units: the unit of *years* for the orbit period
and the distance unit of *astronomical units.* In this case, the
reference ``satellite'' is the Earth and Kepler's third law becomes
period = distance^{3/2}. Let's use this to find out how long it takes
to explore the solar system.

Let's
go to Mars! The relative positions of Earth and
Mars must be just right
at launch so that Mars will be at the right position to greet the spacecraft
when it arrives several months later. These good positionings happen once every
780 days (the synodic period of Mars). The spacecraft must be launched within a
time interval called the ``launch window'' that is just few of weeks long to use a
Hohmann orbit for the spacecraft's path. The Earth is at the
**perihelion** (point closest to the Sun) of the spacecraft orbit (here,
1.0 A.U.) and Mars is at the **aphelion** (point farthest from the Sun---here,
1.52 A.U.).

Kepler's third law relates the semi-major axis of the orbit to its sidereal
period. The major axis is the total length of the long axis of the elliptical
orbit (from perihelion to aphelion). For the Mars journey, the major axis =
1.52 + 1.0 A.U. = 2.52 A.U. *The semi-major axis is one-half of the major
axis,* so divide the major axis by two: 2.52/2 = 1.26 A.U. Now apply Kepler's
third law to find the orbital period of the spacecraft = 1.26^{3/2} =
1.41 years. This is the period for a full orbit (Earth to Mars and back to Earth),
but you want to go only half-way (just Earth to Mars). Travelling from Earth to
Mars along this path will take (1.41 / 2) years = 0.71 years or about 8.5 months.

When the craft is launched, it already has the Earth's orbital velocity of
about 30 kilometers/second. Since this is the speed for a circular orbit around the Sun
at 1.0 A.U., a reduction in the spacecraft's speed would make it fall closer
to the Sun and the Hohmann orbit would be *inside* the Earth's orbit. Since you
want to go beyond the Earth's orbit, the spacecraft needs an increase in its
speed to put it in an orbit that is *outside* the Earth's orbit. It will slow
down gradually as it nears aphelion.

At aphelion the spacecraft will not be
travelling fast enough to be in a circular orbit at Mars' distance (1.52 A.U.)
so it will need to arrive at aphelion *slightly* before Mars does. Mars
will then catch up to it. But the spacecraft will be moving much too fast to be in
a circular orbit *around Mars,* so it will need to slow down to go in orbit
around Mars.

On its journey to Mars, the spacecraft's distance from the Sun is continuously monitored to be sure the craft is on the correct orbit. Though the spacecraft responds mostly to the Sun's gravity, the nine planets' gravitational pulls on the spacecraft can affect the spacecraft's path as it travels to Mars, so occasional minor firings of on-board thrusters may be required to keep the craft exactly on track.

aphelion | perihelion |
---|

- Kepler's third law: period #1 = period #2 ×
**Sqrt**[(distance #1/distance #2)^{3}] - Kepler's third law: period #1 = period #2 ×
(distance #1/distance #2)
^{3/2} - If considering objects orbiting the Sun, measure the orbit period in
*years*and the distance in*A.U.*With these units, Kepler's third law is simply: period = distance^{3/2}.

- How can you predict the orbital period of Jupiter's satellite Europa from observations of the other jovian moon Io?
- If Io takes 1.8 days to orbit Jupiter at a distance of 422,000 kilometers from its center, find out how long it would take Europa to orbit Jupiter at 671,000 kilometers from its center.
- If the Moon were twice as far from the Earth as it is now, how long could a solar eclipse last? (Solar eclipses currently last up to about two hours from the start of the cover-up to when the Moon no longer blocks the Sun at all.)
- The Hubble Space Telescope orbits the Earth 220 kilometers above the surface and takes about 1.5 hours to complete one orbit. How can you find out how far up to put a communication satellite, so that it takes 24 hours to circle the Earth? (Such an orbit is called a ``geosychronous orbit'' because the satellite remains above a fixed point on the Earth.)
- Why does NASA not launch interplanetary spacecraft when the planets are at opposition (closest to the Earth)?
- Find out how long it will take the Cassini spacecraft to travel to Saturn 9.5 A.U. from the Sun.

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last updated: 12 May 2001

Author of original content: Nick Strobel