How to Find a Satellite’s Orbital Radius from Its Period

A satellite completes one orbit around Earth every 2.00 hours. How far from Earth’s centre is it?

If you know Earth’s mass and can assume the satellite follows a circular orbit, you can calculate its orbital radius using:

\[ \boxed{r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3}} \]

The important detail is that you need the planet’s mass, not the satellite’s mass. You also need to use the orbital period in seconds.

The result is the distance from the planet’s centre to the satellite. It is not the satellite’s altitude above the surface.

Before calculating, make a prediction. If a satellite takes twice as long to complete an orbit around the same planet, does its orbital radius double?

It doesn’t. Its orbital radius increases by a factor of \(2^{2/3}\), or approximately \(1.59\). The reason is that a satellite farther from a planet travels around a larger circle, but also moves more slowly because gravity is weaker.

01The equation for finding orbital radius

For a satellite in a circular orbit, the orbital period is related to the orbital radius and the central body’s mass by:

\[ T=2\pi\sqrt{\frac{r^3}{GM}} \]

This relationship comes from equating gravitational force with the centripetal force required for circular motion.

If you need a refresher on where the equation comes from, see Orbital Period: Radius and Central Mass for HSC Physics.

For this question, we want to rearrange the equation to make \(r\) the subject.

Step 1: Square both sides.

\[ T^2=\frac{4\pi^2r^3}{GM} \]

Step 2: Multiply both sides by \(GM\).

\[ GMT^2=4\pi^2r^3 \]

Step 3: Divide by \(4\pi^2\).

\[ r^3=\frac{GMT^2}{4\pi^2} \]

Step 4: Take the cube root.

\[ \boxed{r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3}} \]

That’s the equation you need.

Understanding the symbols

Use:

\[ G=6.67\times10^{-11} \ \mathrm{m^3\,kg^{-1}\,s^{-2}} \]

Notice that the satellite’s mass does not appear in the final equation. It cancels when gravitational force is equated with centripetal force.

This means that, under the usual HSC approximation, two satellites orbiting the same planet at the same radius have the same orbital period, regardless of their individual masses.

02A reliable method for finding orbital radius

When you see a question asking you to find a satellite’s orbital radius from its period, work through these steps.

1. Identify the central mass.

Find the mass of the planet or other central body being orbited.

Do not accidentally use the satellite’s mass.

2. Convert the orbital period into seconds.

The equation uses SI units.

Some useful conversions are:

3. Substitute into the rearranged equation.

Use:

\[ r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \]

Keep the units consistent.

4. Check whether the question wants orbital radius or altitude.

This is an important distinction.

Orbital radius is measured from the centre of the planet. Altitude is measured from its surface.

If the question gives the planet’s physical radius \(R\), then:

\[ \boxed{h=r-R} \]

where \(h\) is the satellite’s altitude.

If the question gives altitude instead of orbital radius, you must add the planet’s radius:

\[ \boxed{r=R+h} \]

The orbital equation always uses \(r\), not \(h\).

A useful check: Does the result make physical sense?

For satellites orbiting the same central body, rearranging the equation gives:

\[ r\propto T^{2/3} \]

This means that increasing the orbital period increases the orbital radius, but not in direct proportion.

For example:

You can also compare satellites orbiting different central bodies.

Starting with the orbital equation, the general comparison relationship is:

\[ \boxed{ \frac{r_B}{r_A} = \left[ \frac{M_B}{M_A} \left(\frac{T_B}{T_A}\right)^2 \right]^{1/3} } \]

This is useful when a question gives you ratios rather than actual masses.

For the same central mass, the mass ratio becomes 1, leaving:

\[ \frac{r_B}{r_A} = \left(\frac{T_B}{T_A}\right)^{2/3} \]

For the same orbital period:

\[ \frac{r_B}{r_A} = \left(\frac{M_B}{M_A}\right)^{1/3} \]

These shortcuts will save you time in more complicated HSC questions.

03Worked examples

The following examples use illustrative orbital data. Earth’s mass and radius are approximate standard values.

04Common mistakes to avoid

Before moving to the questions, make sure you can recognise these three mistakes.

Mistake 1: Using altitude instead of orbital radius

If a satellite is \(500\text{ km}\) above a planet with a radius of \(6000\text{ km}\), its orbital radius is:

\[ r=6000+500=6500\text{ km} \]

You must use \(6500\text{ km}\), not \(500\text{ km}\), in the orbital equation.

Mistake 2: Forgetting to convert the period into seconds

If a question gives \(T=5.00\text{ h}\), use:

\[ T=5.00\times3600=18000\text{ s} \]

Do not substitute 5.00 directly.

Because the period is squared in the equation, this mistake produces a very large error.

Mistake 3: Assuming the period alone determines the radius

It doesn’t.

The central mass also matters:

\[ r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \]

Two satellites can have identical periods but different orbital radii if they orbit central bodies of different masses.

You need to know the central mass, or have enough additional information to determine it.

05Questions and solutions

The following questions progress from straightforward calculations to more demanding applications. The final question explores what happens when the assumption of a circular orbit is removed.

Question 1

A satellite moves in a circular orbit around Earth with an orbital period of \(3.00\text{ h}\).

Calculate its orbital radius from Earth’s centre.

Use:

\[ M_E=5.97\times10^{24}\text{ kg} \]

\[ G=6.67\times10^{-11} \ \mathrm{m^3\,kg^{-1}\,s^{-2}} \]

Solution 1

The satellite’s orbital radius is:

\[ \boxed{r=1.06\times10^7\text{ m}} \]

To find this, first convert the period into seconds.

\[ \begin{aligned} T&=3.00\times3600\\ &=10800\text{ s} \end{aligned} \]

Use the rearranged orbital equation:

\[ r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \]

Substitute:

\[ \begin{aligned} r^3&=\frac{ (6.67\times10^{-11}) (5.97\times10^{24}) (10800)^2 }{4\pi^2}\\ &=1.18\times10^{21}\text{ m}^3 \end{aligned} \]

Taking the cube root:

\[ r=(1.18\times10^{21})^{1/3} \]

\[ \boxed{r=1.06\times10^7\text{ m}} \]

This is the distance from Earth’s centre, not its surface.

The most important steps here are converting the period into seconds and remembering to take the cube root at the end.

Question 2

A satellite moves in a circular orbit around a fictional planet, Planet K.

The following information is available:

• Mass of Planet K: \(8.00\times10^{24}\text{ kg}\).
• Radius of Planet K: \(6.00\times10^6\text{ m}\).
• Orbital period of the satellite: \(9.00\text{ h}\).

Calculate:

a) The satellite’s orbital radius.

b) Its altitude above Planet K’s surface.

Use \(G=6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\).

Solution 2

The orbital radius is \(2.42\times10^7\text{ m}\), and the altitude is \(1.82\times10^7\text{ m}\).

Part a: Find the orbital radius.

First, convert the period into seconds:

\[ \begin{aligned} T&=9.00\times3600\\ &=32400\text{ s} \end{aligned} \]

Now use:

\[ r=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \]

Substitute the given values:

\[ \begin{aligned} r^3&=\frac{ (6.67\times10^{-11}) (8.00\times10^{24}) (32400)^2 }{4\pi^2}\\ &=1.42\times10^{22}\text{ m}^3 \end{aligned} \]

Take the cube root:

\[ \begin{aligned} r&=(1.42\times10^{22})^{1/3}\\ &=2.42\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{r=2.42\times10^7\text{ m}} \]

Part b: Find the altitude.

The altitude is the distance from the planet’s surface, so subtract the planet’s physical radius.

\[ h=r-R_K \]

\[ \begin{aligned} h&=2.42\times10^7-6.00\times10^6\\ &=1.82\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{h=1.82\times10^7\text{ m}} \]

A common mistake is to report the orbital radius as the altitude. The two quantities have different reference points, so you must subtract the planet’s radius when the question asks for altitude.

Question 3

Two satellites, A and B, orbit the same planet in separate circular orbits.

Satellite A has an orbital radius of \(1.20\times10^7\text{ m}\) and an orbital period of \(4.00\text{ h}\).

Satellite B has an orbital period of \(10.0\text{ h}\).

Calculate the orbital radius of Satellite B.

A student argues that Satellite B must have an orbital radius of \(3.00\times10^7\text{ m}\) because its period is 2.50 times longer.

Explain why the student’s prediction is incorrect.

Solution 3

Satellite B has an orbital radius of:

\[ \boxed{r_B=2.21\times10^7\text{ m}} \]

The student’s prediction is incorrect because orbital radius is not directly proportional to orbital period.

Since both satellites orbit the same planet, the central mass is unchanged.

Therefore:

\[ \frac{r_B}{r_A} = \left(\frac{T_B}{T_A}\right)^{2/3} \]

Substitute:

\[ \begin{aligned} \frac{r_B}{r_A} &=\left(\frac{10.0}{4.00}\right)^{2/3}\\ &=(2.50)^{2/3}\\ &=1.84 \end{aligned} \]

Now calculate the orbital radius:

\[ \begin{aligned} r_B&=1.84(1.20\times10^7)\\ &=2.21\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{r_B=2.21\times10^7\text{ m}} \]

The tempting approach is to assume that a period 2.50 times longer requires a radius 2.50 times larger.

That would only be true if both satellites maintained the same orbital speed.

They don’t. Satellite B is farther from the planet, where gravity is weaker and the circular orbital speed is lower. This changes the relationship between orbital distance and time.

The correct relationship is:

\[ T\propto r^{3/2} \]

not \(T\propto r\).

Question 4

Two satellites orbit separate planets, A and B, in circular orbits.

Both satellites have the same orbital period of \(6.00\text{ h}\).

The planets have the following properties:

Satellite A has an altitude of \(9.00\times10^6\text{ m}\) above Planet A’s surface.

Calculate the altitude of Satellite B above Planet B’s surface.

Solution 4

Satellite B’s altitude is:

\[ \boxed{h_B=2.80\times10^7\text{ m}} \]

The key is to recognise that the satellites have the same orbital period but orbit planets with different masses.

Step 1: Find Satellite A’s orbital radius.

The orbital radius is the planet’s physical radius plus the satellite’s altitude.

\[ r_A=R_A+h_A \]

\[ \begin{aligned} r_A&=3.00\times10^6+9.00\times10^6\\ &=1.20\times10^7\text{ m} \end{aligned} \]

Step 2: Compare the orbital radii.

The orbital periods are equal, so:

\[ \frac{T_B}{T_A}=1 \]

Use the general comparison equation:

\[ \frac{r_B}{r_A} = \left[ \frac{M_B}{M_A} \left(\frac{T_B}{T_A}\right)^2 \right]^{1/3} \]

Substitute:

\[ \begin{aligned} \frac{r_B}{r_A} &=\left[27(1)^2\right]^{1/3}\\ &=3 \end{aligned} \]

Therefore:

\[ r_B=3r_A \]

\[ \begin{aligned} r_B&=3(1.20\times10^7)\\ &=3.60\times10^7\text{ m} \end{aligned} \]

Step 3: Convert the orbital radius into altitude.

For Planet B:

\[ h_B=r_B-R_B \]

\[ \begin{aligned} h_B&=3.60\times10^7-8.00\times10^6\\ &=2.80\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{h_B=2.80\times10^7\text{ m}} \]

The tempting mistake is to compare the satellites’ altitudes directly.

However, the orbital equation uses distances from the planets’ centres. Planet B’s satellite must orbit at three times Satellite A’s orbital radius to maintain the same period around a planet with 27 times the mass.

Only after finding that orbital radius can you subtract Planet B’s physical radius to obtain its altitude.

Question 5

Extension: Testing the circular-orbit assumption.

A probe orbits a planet with a mass of:

\[ M=6.00\times10^{24}\text{ kg} \]

The time between successive closest approaches to the planet is measured to be \(10.0\text{ h}\).

Further tracking reveals that the probe’s distance from the planet’s centre at its closest approach is:

\[ r_p=1.77\times10^7\text{ m} \]

An analyst uses the circular-orbit equation and reports an orbital radius of approximately \(2.36\times10^7\text{ m}\).

However, the probe is not necessarily following a circular orbit.

Assume it follows an ideal elliptical orbit, with the planet at one focus. For an elliptical orbit, use Kepler’s third law with the semi-major axis \(a\):

\[ T^2=\frac{4\pi^2a^3}{GM} \]

Answer the following:

a) Calculate the semi-major axis of the probe’s orbit.

b) Determine the eccentricity of the orbit.

c) Calculate the maximum distance between the probe and the planet’s centre.

d) Explain why the analyst cannot describe the calculated semi-major axis as the probe’s actual orbital radius at every point in its journey.

Solution 5

The analyst’s numerical result is correct as the semi-major axis, but it is not the probe’s constant distance from the planet.

The results are:

\[ \boxed{a=2.36\times10^7\text{ m}} \]

\[ \boxed{e=0.250} \]

\[ \boxed{r_a=2.95\times10^7\text{ m}} \]

The important distinction is that an elliptical orbit has a changing distance from the planet’s centre.

Part a: Calculate the semi-major axis.

First, convert the orbital period into seconds:

\[ \begin{aligned} T&=10.0\times3600\\ &=36000\text{ s} \end{aligned} \]

For an elliptical orbit, the semi-major axis \(a\) is used instead of a constant orbital radius.

Rearrange Kepler’s third law:

\[ a=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \]

Substitute:

\[ \begin{aligned} a^3&=\frac{ (6.67\times10^{-11}) (6.00\times10^{24}) (36000)^2 }{4\pi^2}\\ &=1.31\times10^{22}\text{ m}^3 \end{aligned} \]

Taking the cube root:

\[ \boxed{a=2.36\times10^7\text{ m}} \]

The semi-major axis is the distance halfway between the closest and furthest points of the elliptical orbit, measured along its longest diameter.

Part b: Determine the eccentricity.

The eccentricity \(e\) describes how stretched an elliptical orbit is.

For an elliptical orbit:

\[ a=\frac{r_p+r_a}{2} \]

where:

• \(r_p\) is the distance at closest approach, called periapsis.
• \(r_a\) is the distance at furthest approach, called apoapsis.
• \(a\) is the semi-major axis.

The eccentricity can be calculated using:

\[ e=\frac{r_a-r_p}{r_a+r_p} \]

We first need the apoapsis distance, which we will calculate in part c.

Part c: Calculate the maximum distance.

Rearrange:

\[ a=\frac{r_p+r_a}{2} \]

to make \(r_a\) the subject:

\[ r_a=2a-r_p \]

Substitute:

\[ \begin{aligned} r_a&=2(2.36\times10^7)-1.77\times10^7\\ &=4.72\times10^7-1.77\times10^7\\ &=2.95\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{r_a=2.95\times10^7\text{ m}} \]

Now calculate the eccentricity:

\[ e=\frac{r_a-r_p}{r_a+r_p} \]

\[ \begin{aligned} e&=\frac{2.95-1.77}{2.95+1.77}\\ &=\frac{1.18}{4.72}\\ &=0.250 \end{aligned} \]

Thus:

\[ \boxed{e=0.250} \]

Part d: Explain why the analyst’s interpretation is incomplete.

The value \(2.36\times10^7\text{ m}\) is the semi-major axis, not the probe’s instantaneous distance from the planet.

In a circular orbit, the distance from the planet’s centre remains constant. This allows us to use a single orbital radius \(r\).

In an elliptical orbit, however, the distance changes continuously as the probe travels around the planet.

For this probe, the distance varies between:

\[ 1.77\times10^7\text{ m} \]

at closest approach, and:

\[ 2.95\times10^7\text{ m} \]

at furthest approach.

The tempting assumption is that one measured period must correspond to one fixed orbital radius. The missing constraint is the shape of the orbit.

For a circular orbit, the semi-major axis and orbital radius are equal. For an elliptical orbit, they are not.

The period determines the semi-major axis, but additional information is needed to determine the probe’s instantaneous distance.

This is why you should always check whether a question specifies a circular orbit before applying the simple orbital-radius equation.

06Where to go next

Once you can calculate orbital radius from period and central mass, the next useful step is finding the satellite’s orbital speed.

For a circular orbit, the satellite travels one circumference in one period:

\[ v=\frac{2\pi r}{T} \]

You can use the radius you have just calculated to determine its speed.

For a closer look at the relationship between speed, radius, and period, see Speed, Radius and Period in Uniform Circular Motion.

The key decision is knowing which relationship to use: orbital period and central mass determine the circular orbital radius, while orbital speed connects that radius to the distance travelled during each revolution.