Lesson Explainer: Orbital Mechanics Physics • 9th Grade

In this explainer, we will learn how to use the formula ๐‘‡=2๐œ‹๐‘Ÿ๐‘ฃ to calculate the orbital characteristics of a planet, moon, or man-made satellite in a circular orbit.

Recall that there are two types of orbits. In a circular orbit, the object in orbit remains a constant distance away from the other object it is orbiting. Some objects are also in elliptical orbits, where the distance between the two objects is constantly changing. Here, we are going to focus on circular orbits.

The diagram above shows an object in a circular orbit around another larger object. The orbital path indicated as a circle is the path traced out by the center of mass of the smaller object as it moves along its orbit. The orbital radius, ๐‘Ÿ, is the distance from the center of mass of the larger object to the orbital path.

If we know the value of ๐‘Ÿ, we can also work out the total length of the orbital path, or the circumference of the orbit. If we call the total circumference ๐‘, then ๐‘=2๐œ‹๐‘Ÿ, or the circumference of a circle of radius ๐‘Ÿ.

Letโ€™s look at at example of working out the distance traveled along a circular orbit.

Example 1: Finding the Distance Traveled along a Circular Orbit

Mars orbits the Sun at a distance of about 228โ€Žโ€‰โ€Ž000โ€Žโ€‰โ€Ž000 km. Assuming that Mars has a circular orbit, what is the distance traveled by the planet as it makes 1 full orbit around the Sun? Give your answer in scientific notation to one decimal place.

Answer

In this example, we have the planet Mars in a circular orbit around the Sun, as in the diagram below.

We are given the distance, ๐‘Ÿ, between Mars and the Sun, as ๐‘Ÿ=228000000km. We need to calculate the distance traveled by the planet in 1 full orbit, that is, the circumference of the orbit.

Recall that the circumference, ๐‘, of a circle is related to the radius, ๐‘Ÿ, by ๐‘=2๐œ‹๐‘Ÿ.

Notice that ๐‘Ÿ was given in units of kilometres (km), so our answer for ๐‘ will also be in kilometres. We have ๐‘=2๐œ‹๐‘Ÿ=2ร—3.14โ€ฆร—228000000=1432566250kmkm to the nearest kilometre.

Now, we need to state the answer in scientific notation to one decimal place, so the final answer for the distance, ๐‘, traveled by Mars in 1 orbit around the Sun is ๐‘=1.4ร—10.๏Šฏkm

In the following example, we will do this in reverse, finding the orbital radius from the circumference.

Example 2: Finding the Radius of a Circular Orbit

Mercury travels 364 million kilometres as it makes 1 full orbit around the Sun. Assuming that Mercury has a circular orbit, at what distance away from the Sun does Mercury orbit? Give your answer in scientific notation to two decimal places.

Answer

In this example, we are given the total distance that Mercury travels over the course of one orbit, which is equivalent to the circumference of the orbit. The quantity we need to find is the distance between Mercury and Sun, or the orbital radius.

Recall that the radius, ๐‘Ÿ, and circumference, ๐‘, of a circle are related by ๐‘=2๐œ‹๐‘Ÿ.

We need to find ๐‘Ÿ, so we first need to divide both sides of this equation by 2๐œ‹ so that we have ๐‘2๐œ‹=๐‘Ÿ.

We are given the circumference, ๐‘, as 364 million kilometres, which is equivalent to 364โ€Žโ€‰โ€Ž000โ€Žโ€‰โ€Ž000 km. Substituting this into the equation above, we have ๐‘Ÿ=๐‘2๐œ‹=3640000002ร—3.14โ€ฆ=57932399kmkm to the nearest kilometre. Note that the radius is in kilometres because the circumference we used was in kilometres.

The distance away from the Sun that Mercury orbits is, therefore, 5.79ร—10๏Šญ km, written in scientific notation to two decimal places.

Letโ€™s now consider the velocity of an object in a circular orbit.

The diagram below shows the same object in two positions along its circular orbit. In the first position at the top of the diagram, it has a velocity โƒ‘๐‘ฃ๏Šง, and in the second position to the right, it has a velocity โƒ‘๐‘ฃ๏Šจ.

We can see in the diagram that the directions of โƒ‘๐‘ฃ๏Šง and โƒ‘๐‘ฃ๏Šจ are different. However, the magnitudes are the same. This is true for any position along the orbit.

For any object in a circular orbit, the direction of its velocity changes with time, but the magnitude of the velocity, also known as the speed, remains constant.

Another way of saying this is that the orbital speed is constant. If we call the orbital speed ๐‘ , we can say that ๐‘  is the same at all points around the orbit.

The final quantity we want to work with relating to circular orbits is the time taken to complete one orbit, also known as the orbital period. We will denote the orbital period as ๐‘‡. This is the time taken, from any starting position along the orbit, for the object to travel around the entire circumference and return to the starting point.

For a planet moving in orbit around the Sun, the orbital period defines one year for that planet. For example, one year on Earth is about 365 days, because it takes approximately 365 days for Earth to complete one orbit around the Sun.

In the next example, we will work on converting between years for different planets.

Example 3: Understanding Orbital Periods

Venus takes 225 days to orbit the Sun.

How long is this period in Earth years? Use a value of 365 for the number of days in an Earth year. Give your answer to 2 decimal places.

How long does Earth take to orbit the Sun in Venus years ? Give your answer to 2 decimal places.

Answer

This example requires us to relate the orbital periods, or years, of two planets: Earth and Venus.

We approach the first part of the question the same way we would convert between different units of any quantity: here, we are given the period of Venusโ€™s orbit in days, and we need to express it in years. We do that by multiplying it by a fraction equal to 1, so that we are not changing the value, that converts the units from days to Earth years. We are given the length of an Earth year as 365 days, so we can use that 1=365yeardays. If we call the orbital period of Venus ๐‘‡V, then we have ๐‘‡=225ร—1365=0.62Vdaysyeardaysyears to two decimal places.

For the second part, we need to determine how long Earth takes to orbit the Sun in Venus years. Here, we need to recall that the time taken to orbit the Sun is the orbital period, which is equivalent to one year for each planet.

We are given the length of an Earth year as 365 days, so this is the time it takes for Earth to orbit the Sun in days, which we will call ๐‘‡E. We now need to convert this to the unit of Venus years, or the orbital period of Venus, which is given as 225 days. To convert the unit from days to a Venus year, we need to multiply the number of days in an Earth year by 1225Venusyeardays, so ๐‘‡=365ร—1225=1.62EdaysVenusyeardaysVenusyears to 2 decimal places.

Now we know the distance the object travels in the course of one orbit, ๐‘, and we know that it is moving with a constant speed, ๐‘ . Recall that speed and distance are related by speeddistancetime=.

Using the formula above, we can relate the orbital period, ๐‘‡, to the orbital speed, ๐‘ , and the total distance traveled, which is 2๐œ‹๐‘Ÿ, where ๐‘Ÿ is the radius of the orbital path, as follows: ๐‘ =2๐œ‹๐‘Ÿ๐‘‡.

Letโ€™s get some practice using this equation in some examples.

Example 4: Finding the Orbital Speed from the Radius and Period for Circular Orbits

A satellite orbits Earth at an orbital radius of 10โ€Žโ€‰โ€Ž000 km. Its orbital period is 2.8 hours. How fast is the satellite moving? Give your answer to the nearest kilometre per second.

Answer

In this example we have a satellite in orbit around Earth. We are given the orbital radius, which is the distance of the satellite from Earthโ€™s center of mass, and the orbital period, which is the time taken for the satellite to complete one orbit. We need to find the speed of the satellite.

Recall that the orbital speed, ๐‘ , is given by ๐‘ =2๐œ‹๐‘Ÿ๐‘‡, where ๐‘Ÿ is the orbital radius and ๐‘‡ is the orbital period. These quantities are both given in the question: ๐‘Ÿ=10000km and ๐‘‡=2.8hours.

We need to find ๐‘  in units of kilometres per second (km/s) which means we need to provide the distance in kilometres and the time in seconds. We already have ๐‘Ÿ in kilometres, but ๐‘‡ is given in hours, so we first need to convert that to seconds. Recall that 1=60hourminutes, and 1=60minuteseconds, so 1=60ร—60=3600hours. Therefore, ๐‘‡=2.8ร—36001=10080.hoursshours

Now, we can substitute ๐‘Ÿ=10000km and ๐‘‡=10080s into the equation for ๐‘ , which gives ๐‘ =2๐œ‹๐‘Ÿ๐‘‡=2ร—3.14โ€ฆร—1000010080=6/kmskms to the nearest kilometre per second.

Example 5: Finding the Orbital Period from the Radius and Speed for Circular Orbits

The table shows some data about Saturnโ€™s moon, Titan. Using this data, calculate the orbital period of Titan around Saturn. Assume that Titan has a circular orbit. Give your answer in days to the nearest day.

Titan (Moon of Saturn)
Diameter5โ€Žโ€‰โ€Ž150 km
Orbital Radius1โ€Žโ€‰โ€Ž220โ€Žโ€‰โ€Ž000 km
Orbital Speed5.57 km/s
Mass1.35ร—10๏Šจ๏Šฉ kg

Answer

Here, we are asked to calculate the orbital period of the moon Titan in a circular orbit around the planet Saturn. To help us, we are given Titanโ€™s diameter, orbital radius, orbital speed, and mass.

Recall that orbital period, ๐‘‡, is related to orbital speed, ๐‘ , and orbital radius, ๐‘Ÿ, by the equation ๐‘ =2๐œ‹๐‘Ÿ๐‘‡, so these are the only quantities we need to use; we can ignore the mass and diameter of Titan, as these have no effect on the orbital period.

Since the orbital period, ๐‘‡, is the quantity we need to find, we need to rearrange this equation in terms of ๐‘‡. We do this by multiplying both sides of the equation by ๐‘‡๐‘  so that ๐‘‡=2๐œ‹๐‘Ÿ๐‘ .

Before we substitute in the values, ๐‘Ÿ=1220000km and ๐‘ =5.57/kms, we should check that they are in the correct units. Both of the distance units (the radius in kilometres and the distance element of the speed in kilometres per second) are given in kilometres, so those units will cancel out correctly. The time is given in seconds, so we will obtain a result for the orbital period in seconds. Substituting in these values, we have ๐‘‡=2๐œ‹๐‘Ÿ๐‘ =2ร—3.14โ€ฆร—12200005.57/=1376209kmkmss to the nearest second.

We are asked to give this in days to the nearest day, so we need to convert this value to units of days. Recall that 1=24dayhours, 1=60hourminutes, and 1=60minutes. Therefore, ๐‘‡=1376209ร—124ร—60ร—60=16sdaysdays to the nearest day.

Example 6: Finding the Orbital Period from the Radius and Speed for Circular Orbits

A satellite orbits Earth at an orbital radius of 7โ€Žโ€‰โ€Ž720 km and moves at a speed of 7.2 km/s. The satellite has a circular orbit.

What is the circumference of the satelliteโ€™s orbit? Give your answer in kilometres, using scientific notation, to one decimal place.

What is the period of the satelliteโ€™s orbit? Give your answer in minutes to the nearest minute.

Answer

In this example, we are considering an artificial satellite in a circular orbit around Earth. We are given the orbital radius and speed, and we first need to find the circumference of the satelliteโ€™s orbit.

Recall that the circumference, ๐‘, of a circle is related to the radius, ๐‘Ÿ, by ๐‘=2๐œ‹๐‘Ÿ.

We can therefore find ๐‘ from the orbital radius, ๐‘Ÿ=7720km, as follows: ๐‘=2๐œ‹๐‘Ÿ=2ร—3.14โ€ฆร—7720=48506kmkm to the nearest kilometre. We are asked to give this in scientific notation to one decimal place, so this becomes ๐‘=4.9ร—10๏Šชkm.

Next, we need to find the period of the satelliteโ€™s orbit. Recall that orbital period, ๐‘‡, is related to orbital radius, ๐‘Ÿ, and speed, ๐‘ , by ๐‘ =2๐œ‹๐‘Ÿ๐‘‡, and we have already found ๐‘=2๐œ‹๐‘Ÿ, so we can substitute that in so that we have ๐‘ =๐‘๐‘‡.

We need to find ๐‘‡, so we can rearrange this equation in terms of ๐‘‡ by multiplying both sides by ๐‘‡๐‘ , which gives ๐‘‡=๐‘๐‘ .

We can now substitute in the numbers ๐‘=48506km and ๐‘ =7.2/kms. Note that we use the value for ๐‘ before rounding so that we do not accumulate rounding errors. Both distance units are kilometres, and the time is in seconds, so using these values will result in an orbital period in seconds. We now have ๐‘‡=485067.2/=6736kmkmss to the nearest second.

We are asked for ๐‘‡ in minutes to the nearest minute, so next we need to convert this value to units of minutes. Recall that 1=60minutes, so ๐‘‡=6736ร—160=112sminutesminutes to the nearest minute.

Example 7: Understanding Orbital Speed and Period

Two planets, A and B, orbit a star. Both planets have circular orbits. Planet A orbits the star at a distance of 1.5ร—10๏Šฎ km and at a speed of 30 km/s. Planet B orbits the star at a distance of 4.8ร—10๏Šฎ km and at a speed of 17 km/s.

How many times is the length of planet Bโ€™s orbit greater than planet Aโ€™s?

How many times longer does it take for planet B to orbit the star than planet A? Give your answer to 1 decimal place.

Answer

Here, we have a system consisting of a star orbited by two planets, A and B. The setup is shown in the diagram below.

The orbital radius of planet A is labeled ๐‘ŸA, and its speed ๐‘ A. The orbital radius of planet B is shown as ๐‘ŸB, and its orbital speed as ๐‘ B.

Planet B is located further from the star than planet A, so its orbital radius is larger, and the length of its orbit will also be greater. We are asked to find how many times longer the orbital path of planet B is than the orbital path of planet A. Another way of stating this is that if we call the length of planet Aโ€™s orbital path ๐‘A, and the length of planet Bโ€™s orbital path ๐‘B, we need to find ๐‘๐‘BA.

Recall that the circumference of a circle, ๐‘, is related to the radius, ๐‘Ÿ, by ๐‘=2๐œ‹๐‘Ÿ. We can, therefore, say that ๐‘๐‘=2๐œ‹๐‘Ÿ2๐œ‹๐‘Ÿ.BABA

On the right-hand side, we have the factor 2๐œ‹ on both the top and the bottom of the fraction, so those cancel each other out, leaving us with ๐‘๐‘=๐‘Ÿ๐‘Ÿ.BABA

We can now substitute in the values given in the question, ๐‘Ÿ=1.5ร—10A๏Šฎkm and ๐‘Ÿ=4.8ร—10B๏Šฎkm, so we have ๐‘๐‘=๐‘Ÿ๐‘Ÿ=4.8ร—101.5ร—10=3.2,BABA๏Šฎ๏Šฎ so the length of planet Bโ€™s orbit is 3.2 times longer than the orbit of planet A.

In the second part of the question, we need to find how many times longer it takes planet B to orbit the star than planet A. In other words, we need to find ๐‘‡๐‘‡BA, where ๐‘‡B is the orbital period of planet B and ๐‘‡A is the orbital period of planet A.

Recall that orbital period, ๐‘‡, is related to orbital speed, ๐‘ , and radius, ๐‘Ÿ, by ๐‘ =2๐œ‹๐‘Ÿ๐‘‡.

We can rearrange this in terms of ๐‘‡ by multiplying both sides of the equation by ๐‘‡๐‘ , which gives ๐‘‡=2๐œ‹๐‘Ÿ๐‘ .

We can write ๐‘‡๐‘‡BA as ๐‘‡ร—1๐‘‡BA. Substituting in the equation for ๐‘‡, we have ๐‘‡๐‘‡=2๐œ‹๐‘Ÿ๐‘ ๐‘ 2๐œ‹๐‘Ÿ.BABBAA

If we collect like terms together, we have ๐‘‡๐‘‡=2๐œ‹2๐œ‹๐‘Ÿ๐‘Ÿ๐‘ ๐‘ .BABAAB

We have 2๐œ‹ on both the top and the bottom of the fraction, so they cancel out, and we have already found ๐‘Ÿ๐‘Ÿ=3.2BA above. If we substitute in this and the orbital speeds ๐‘ A and ๐‘ B, we have ๐‘‡๐‘‡=๐‘Ÿ๐‘Ÿ๐‘ ๐‘ =3.2ร—30/17/=5.6BABAABkmskms to 1 decimal place. Therefore, it takes planet B approximately 5.6 times longer to orbit the star than planet A.

Key Points

  • Objects in circular orbits have a constant speed known as the orbital speed.
  • In a circular orbit, the circumference of the orbit, ๐‘, is related to the radius of the orbit, ๐‘Ÿ, as ๐‘=2๐œ‹๐‘Ÿ.
  • The time taken to complete one orbit is called the orbital period and is denoted as ๐‘‡.
  • Orbital speed, ๐‘ , orbital period, ๐‘‡, and the radius of the orbital path, ๐‘Ÿ, are related by ๐‘ =2๐œ‹๐‘Ÿ๐‘‡.

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