Lesson Explainer: Orbital Speed Physics • 9th Grade

In this explainer, we will learn how to calculate the orbital speed of an object moving along a circular orbit given its orbital radius and the mass of the object it orbits.

To begin, let us recall some of the key properties of circular orbital motion.

Remember that, for circular orbits, any orbiting body has a velocity with a constant magnitude but an ever-changing direction. The diagram below shows Jupiter’s moon Europa orbiting Jupiter. The direction of the moon’s velocity always points along a tangent to its orbital path, indicated by the blue arrow. Jupiter’s gravity acts as a centripetal force, which we know must always point radially inward, indicated by the red arrow.

This relationship helps explain why the moon’s orbital speed is constant: the gravitational force has no component in the same direction as the moon’s velocity. Thus, the magnitude of the velocity does not change due to gravity, but the direction constantly changes because the gravitational force constantly redirects it along a circular path. At any point in the orbit, the directions of the two quantities always point at a right angle, or 90, to each other.

As illustrated above, the gravitational force from a massive object always points inward, toward its own center of mass. Remember that the acceleration due to gravity, 𝑎, at a point near a massive object, such as a planet, is given by 𝑎=𝐺𝑀𝑟, where 𝐺 is the universal gravitational constant, 𝑀 is the mass of the planet, and 𝑟 is the distance between the point and the planet.

In this case, the gravitational acceleration, which always points inward and keeps the moon in a circular orbit, acts as a centripetal acceleration. Thus, for an object, such as a moon, in circular orbit, the centripetal acceleration of the object is given by 𝑎=𝑣𝑟, where 𝑣 is the linear speed of the moon and 𝑟 is the radius of the orbit.

Since gravity is providing the centripetal acceleration, we can equate the values of 𝑎 and 𝑎: 𝑣𝑟=𝐺𝑀𝑟.

On both sides of the equation, 𝑟 appears in the denominator, so we can simplify the equation by multiplying both sides by 𝑟: 𝑟𝑣𝑟=𝑟𝐺𝑀𝑟𝑣=𝐺𝑀𝑟.

We can take the square root of both sides to make 𝑣 the subject of the equation: 𝑣=𝐺𝑀𝑟.

Now, we have an equation for the orbital speed of an object moving along a circular orbit, given its orbital radius and the mass of the object it orbits.

Notice that the equation does not depend on 𝑚, the mass of the object in orbit. This means that all objects—regardless of their masses—that move in circular orbits with the same orbital radius, around planets with the same mass, will have the same orbital speed. Remember that this relationship only applies in the case of circular orbits, where orbital speed is constant; noncircular orbits do not obey this equation.

Before we practice some calculations, it should be noted that the universal gravitational constant is found commonly throughout astronomy and other fields of physics and has an unchanging value of 𝐺=6.67×10/mkgs.

Definition: Orbital Speed Equation—Circular Orbit

In the special case of a circular orbit, an object’s orbital speed, 𝑣, is given by the equation 𝑣=𝐺𝑀𝑟, where 𝐺 is the universal gravitational constant, 𝑀 is the mass of the large object at the center of the orbit, and 𝑟 is the orbital radius.

Now that we have an equation for orbital speed, let us put it to use in some examples.

Example 1: Calculating Orbital Speed

For a satellite to follow a circular orbit around Earth at a radius of 10‎ ‎000 km, what orbital speed must it have? Use a value of 5.97×10 kg for the mass of Earth and 6.67×10 m3/kg⋅s2 for the value of the universal gravitational constant. Give your answer to the nearest metre per second.


Here, we are given values for 𝑀, 𝑟, and 𝐺. Notice that 𝑟 is given in kilometres, so we must convert to metres before we can use it in the equation: 𝑟=10000=1.0×10.kmm

We are now ready to substitute all of our values into the orbital speed equation: 𝑣=𝐺𝑀𝑟=(6.67×10/)(5.97×10)1.0×10=6310.3/.mkgskgmms

Rounding to the nearest metre per second, we have found that the satellite must orbit at a speed of 6‎ ‎310 m/s.

Notice that we did not need to know the mass of the satellite to find the answer. This is because the orbital speed equation does not depend on the mass of the object in orbit, 𝑚.

Sometimes, we will be given information about an orbital system other than just 𝑀 and 𝑟. Depending on what information is provided, we can rearrange the orbital speed equation to solve for another quantity, besides 𝑣. For instance, in the following example, we will calculate the orbital radius for a system.

Example 2: Calculating Orbital Radius

A planet has a circular orbit around a star. It orbits the star at a speed of 17.9 km/s, and the star has a mass of 2.18×10 kg. What is the radius of the planet’s orbit? Use a value of 6.67×10 m3/kg⋅s2 for the universal gravitational constant and 1.5×10 m for the length of 1 AU. Give your answer to the nearest astronomical unit.


Here, we are given values for 𝑣, 𝑀, and 𝐺 and we must solve for 𝑟. To do this, we can rearrange the orbital speed equation so that 𝑣=𝐺𝑀𝑟 becomes 𝑟=𝐺𝑀𝑣.

Before we can calculate, we must convert the value for 𝑣 into units of metres per second: 𝑣=17.9/=17900/.kmsms

Now, substituting our values, we have 𝑟=6.67×10/2.18×10(17900/)=4.54×10.mkgskgmsm

However, this is not our final answer, since 𝑟 is presently expressed in metres and we need to convert it into astronomical units: 𝑟=4.54×10×11.5×10=3.03.mAUmAU

Rounding to the nearest astronomical unit, we find that the planet has an orbital radius of 3 AU.

Let us now shift our focus from these computational examples to a couple examples that are more conceptual.

Example 3: Variable Dependence in the Orbital Speed Equation

Which line on the graph shows the relation between orbital speed and orbital radius for objects moving along circular orbits due to gravity?


We can begin by recalling the orbital speed equation, which relates orbital speed, 𝑣, and orbital radius, 𝑟: 𝑣=𝐺𝑀𝑟.

From this, we will devise a proportion to help us describe the relationship between orbital speed and radius. Since 𝐺 and 𝑀 are both constants in this context, they will not appear in our proportion. Using a 1 as a placeholder in the numerator, we have 𝑣1𝑟.

Thus, we can say that 𝑣 and 𝑟 are inversely proportional to each other, since an increase in one quantity indicates a decrease in the other. Because of this inverse relationship, we know that 𝑣 should decrease as 𝑟 increases, so the green line is incorrect.

Further, since 𝑟, the independent variable, appears in the denominator, we know that the relationship between 𝑣 and 𝑟 cannot be linear. Therefore, the red line is also incorrect.

The blue and orange lines have similar shapes, but they are quite different at the 𝑦-axis. The orange curve intersects the axis, where the blue curve has a vertical asymptote. To determine how the graph should look here, let us think about how the orbital speed equation behaves near 𝑟=0.

As mentioned before, 𝑟 appears in the denominator, and we know that we cannot meaningfully divide by zero. Thus, as 𝑟 goes to zero, the function goes to infinity and is undefined at the 𝑦-axis.

Therefore, the blue line correctly shows the relation between orbital speed and orbital radius for objects in circular orbit.

Example 4: Variable Dependence in the Orbital Speed Equation

A satellite follows a circular orbit around Earth at a radial distance 𝑅 and with an orbital speed 𝑣. If the satellite were moved closer to Earth, so that it followed a circular orbit with a radius of 𝑅9, at what speed, in terms of 𝑣, would it have to move in order to maintain its orbit?


In this example, we will consider how orbital speed and the radius relate to each other by devising a proportion. First, we need to identify an equation that includes both of these quantities, so we will use the orbital speed equation: 𝑣=𝐺𝑀𝑟.

Next, we identify the constant values in the equation so that we can remove them to form our proportion, since proportions explore how variables change with respect to one another. We know that 𝐺 represents a constant, so it will not appear in the proportion. And although 𝑀 appears as a variable in the orbital speed equation, within the context of this problem, 𝑀 represents the mass of Earth, which is essentially constant.

Now, we can write a proportion to show how 𝑣 and 𝑟 depend on each other using the symbol for proportionality, . We can no longer use an equals sign to relate 𝑣 and 𝑟, since the true equality of their values does depend on some constant quantities that are not relevant to our proportion. The number 1 will act as a placeholder in the numerator where the constant values were in the original equation: 𝑣1𝑟.

Thus, it can be stated that 𝑣 is inversely proportional to the square root of 𝑟. The relationship is described as inverse because an increase in one quantity causes a decrease in the other—remember that increasing a denominator corresponds to a decrease in the overall value and decreasing a denominator corresponds to an increase in the overall value. If the satellite is moved closer to Earth, we know that the orbital radius will decrease, so we can predict that the orbital speed will increase.

We can find the exact factor to describe the change in 𝑣 by substituting in the factor that describes the change in 𝑟—a proportion behaves similarly to an equation in this way. We want to change the satellite’s orbital radius from 𝑅 to 𝑅9, so 𝑅 is being multiplied by a factor of 19, which we substitute into the proportion: 𝑣1.

Simplifying the math, 1=1=3.

Therefore, multiplying 𝑅 by a factor of 19 means 𝑣 must be multiplied by a factor of 3.

Thus, moving the satellite to a new orbital radius of 𝑅9 will cause the satellite to move at a new orbital speed of 3𝑣.

In the previous example, we used the orbital speed equation without having any real, exact values to substitute in, yet we were still able to find a meaningful result by examining the relationship between the relevant variables.

As demonstrated in the following example, we are not limited to only using the orbital speed equation in our calculations. We may be given other information about a system, such as an orbital period, 𝑇. Remember that the orbital period equation is 𝑇=2𝜋𝑟𝑣.

We may need to use this in conjunction with the orbital speed equation to learn about a system.

Example 5: Calculating the Mass of a Body in Orbit

Io is one of the four Galilean moons of Jupiter. Io makes one complete orbit of Jupiter every 1.77 days. Assuming that Io’s orbit is circular with a radius of 422‎ ‎000 km, calculate the mass of Jupiter. Use a value of 6.67×10 m3/kg⋅s2 for the universal gravitational constant. Give your answer in scientific notation to two decimal places.


We want to solve for mass, which appears in our equation for orbital speed, 𝑣=𝐺𝑀𝑟.

Let us now solve the orbital speed equation for 𝑀. We start by squaring both sides to undo the radical on the right side: 𝑣=𝐺𝑀𝑟.

Now, we will multiply both sides of the equation by 𝑟𝐺: 𝑟𝐺𝑣=𝑟𝐺𝐺𝑀𝑟𝑀=𝑟𝑣𝐺.

We now have an equation we can use to solve for the mass of Jupiter, but notice that we do not yet have a value for each of the variables—we know the values of 𝑟 and 𝐺, but we do not know the orbital speed of Io, so we need a way to calculate 𝑣. We are given the orbital period of Io, 𝑇=1.77days, and we know an equation that relates orbital period and orbital speed: 𝑇=2𝜋𝑟𝑣.

We can rewrite the equation to solve for 𝑣: 𝑣=2𝜋𝑟𝑇.

We know 𝑟 and 𝑇, but notice that we are given a value for 𝑇 expressed in days, which is not an SI unit, so we must convert into seconds, the SI unit for time. The conversion is as follows: 241×601×601=86400.hoursdayminuteshoursecondsminutesecondsday

Applying this to our value of 𝑇, 𝑇=1.77×86400=152928.dayssecondsdayseconds

We will substitute this 𝑇 value into the orbital period equation to solve for 𝑣. But first, because we are given a value of 𝑟 expressed in kilometres, we need to convert 𝑟 into metres. Recall that 1=1000kmm, so 𝑟=422000000m, but this value is better expressed in scientific notation as 𝑟=4.22×10m. Let us now substitute these values into the orbital period equation to solve for 𝑣: 𝑣=2𝜋𝑟𝑇=2𝜋4.22×10152928=17338/.msms

Now that we have values for 𝑣, 𝑟, and 𝐺, we can substitute them into the rearranged orbital speed equation to calculate 𝑀, the mass of Jupiter: 𝑀=𝑟𝑣𝐺=4.22×10(17338/)6.67×10/=1.90×10.mmsmkgskg

Thus, we have found the mass of Jupiter to be 1.90×10 kg.

Let us finish by summarizing a few important concepts.

Key Points

  • For an object to move in a circular orbit around a large object of mass 𝑀 at a radial distance 𝑟, it must have an orbital speed 𝑣 given by 𝑣=𝐺𝑀𝑟.
  • Orbital speed does not depend on the mass of the object in orbit.
  • We can use 𝑣=𝐺𝑀𝑟 to solve for 𝑀 or 𝑟𝑀=𝑟𝑣𝐺, and 𝑟=𝐺𝑀𝑣.

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