How to Calculate Geostationary Altitude from Orbital Period

Suppose a satellite takes exactly 24.0 hours to complete one circular orbit around Earth. You calculate its altitude and get approximately 35,900 km. But a reference gives the geostationary altitude as 35,786 km.

Which answer is correct?

Both can be correct. The difference comes from a small but important detail: an exact 24-hour day is not the same as the time Earth takes to complete one rotation relative to distant stars.

If the question specifies a period of exactly 24.0 hours, the calculated altitude is approximately 35,863 km. For a genuinely geostationary orbit, the correct period is approximately 23 hours, 56 minutes, and 4 seconds, giving an altitude of approximately 35,786 km.

Let’s work through the calculation, understand why these answers differ, and establish which period to use in an HSC Physics question.

01First, understand what altitude you are calculating

There are two distances that students commonly confuse:

• Orbital radius, \(r\): the distance from Earth’s centre to the satellite.
• Altitude, \(h\): the distance from Earth’s surface to the satellite.

These are not interchangeable.

Imagine drawing a straight line from Earth’s centre to a satellite. The total length is the orbital radius. Part of that distance is already occupied by Earth itself.

Therefore:

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

where:

• \(h\) is the satellite’s altitude above Earth’s surface, in metres.
• \(r\) is its orbital radius, measured from Earth’s centre, in metres.
• \(R_E\) is Earth’s equatorial radius, in metres.

For our calculations, we will use:

\[ R_E=6.3781\times10^6\text{ m} \]

Why the equatorial radius? A geostationary satellite orbits directly above Earth’s equator, so the equatorial radius is the appropriate reference.

For example, if your calculation gives an orbital radius of \(4.2241\times10^7\text{ m}\), you must still subtract Earth’s radius:

\[ h=4.2241\times10^7-6.3781\times10^6 \]

Simply reporting \(4.2241\times10^7\text{ m}\) as the altitude would be incorrect.

02Why can we calculate orbital radius from period?

The connection between orbital period and orbital radius comes from Newton’s law of universal gravitation.

There is an interesting historical development behind this relationship.

Kepler identified a mathematical pattern in planetary motion using astronomical observations. His third law showed that the square of a planet’s orbital period is proportional to the cube of its orbital size.

Newton later explained why this relationship occurs. His theory of gravitation showed that the gravitational attraction between two bodies can provide the inward acceleration needed for circular motion.

We apply that same reasoning to an artificial satellite orbiting Earth.

Gravity provides the centripetal force

Consider a satellite of mass \(m\) travelling in a circular orbit around Earth.

Earth attracts the satellite towards its centre. This gravitational force provides the centripetal force required to keep the satellite moving in a circle.

The gravitational force is:

\[ F_g=\frac{GM_Em}{r^2} \]

The centripetal force required for circular motion is:

\[ F_c=\frac{mv^2}{r} \]

Equating these gives:

\[ \frac{GM_Em}{r^2}=\frac{mv^2}{r} \]

Here:

• \(G\) is the universal gravitational constant.
• \(M_E\) is Earth’s mass.
• \(m\) is the satellite’s mass.
• \(r\) is the orbital radius.
• \(v\) is the satellite’s orbital speed.

Notice that the satellite’s mass appears on both sides. It cancels.

This is why the orbital period at a particular radius does not depend on whether the satellite is a small scientific instrument or a much heavier communications satellite.

Introduce the orbital period

The satellite travels one complete circumference during one orbit.

The circumference of a circle is \(2\pi r\).

If the satellite completes one orbit in time \(T\), its orbital speed is:

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

Substituting this into the gravitational force equation:

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

Rearranging:

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

Taking the cube root gives the equation we need:

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

This calculates orbital radius, not altitude.

To obtain altitude, subtract Earth’s radius:

\[ \boxed{ h=\left(\frac{GM_ET^2}{4\pi^2}\right)^{1/3}-R_E } \]

This is the circular-orbit relationship for a satellite orbiting a central mass \(M_E\).

If you want to explore the general period-to-radius derivation further, see How to Find a Satellite’s Orbital Radius from Its Period.

03The important distinction: 24 hours versus a sidereal day

Before substituting numbers, we need to settle which period belongs in the equation.

Earth has two relevant definitions of a day.

A sidereal day is approximately:

\[ T=23\text{ h }56\text{ min }4\text{ s} \]

Why are they different?

Earth rotates on its axis while also travelling around the Sun.

After Earth has rotated through 360° relative to distant stars, it has moved slightly along its orbit around the Sun. It must rotate a little further before the Sun returns to the same position in the sky.

That extra rotation makes a solar day slightly longer.

Which period does a geostationary satellite need?

A geostationary satellite must rotate at the same angular speed as Earth, measured relative to distant stars.

Therefore, its orbital period must match one sidereal day:

\[ T=86\,164\text{ s} \]

An orbit with an exact 24-hour period is very close, but it is not precisely geostationary.

The European Space Agency also gives the standard geostationary values as approximately 23 hours, 56 minutes, 4 seconds, and 35,786 km above Earth’s surface (ESA: Types of orbits).

Keep this distinction in mind:

• If a question explicitly gives \(T=24.0\text{ h}\), use \(86\,400\text{ s}\).
• If a question asks for the true geostationary altitude, use \(86\,164\text{ s}\), unless it specifies a different approximation.

You should never silently replace a period supplied in a question with another value.

04Worked method: calculating the altitude

For the following examples, use these approximate Earth values.

Using the gravitational parameter, \(\mu=GM_E\), makes the calculations shorter.

05A reliable decision process for HSC questions

When a question asks you to calculate a satellite’s altitude from its orbital period, follow this sequence.

There are also three mistakes worth watching for.

Mistake 1: Using 24 hours when the question requires a true geostationary orbit.

If the question specifically asks for the true geostationary altitude, use 86,164 seconds.

Mistake 2: Reporting orbital radius as altitude.

Always subtract Earth’s radius:

\[ h=r-R_E \]

Mistake 3: Including the satellite’s mass in the final equation.

The satellite’s mass cancels when gravitational force is equated with centripetal force. It is not needed to calculate the orbital radius.

One final limitation: this equation assumes a circular orbit around a specified central mass. For an elliptical orbit, the corresponding period relationship determines the semi-major axis, not a single constant orbital radius or altitude.

06Questions and solutions

The following questions progress from a straightforward orbital calculation to more demanding applications of the period-radius relationship.

Question 1

A satellite travels in a circular, equatorial, prograde orbit around Earth. Its orbital period is 20.0 hours.

Calculate:

a) Its orbital radius from Earth’s centre.

b) Its altitude above Earth’s surface.

c) Whether the satellite is geostationary.

Use:

\[ GM_E=3.9860\times10^{14}\text{ m}^3\text{ s}^{-2} \]

\[ R_E=6.3781\times10^6\text{ m} \]

Solution 1

The satellite’s altitude is approximately \(31\,028\text{ km}\), and it is not geostationary.

Part a: Orbital radius

First, convert the period into seconds:

\[ T=20.0\times3600=72\,000\text{ s} \]

Use:

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

Substitute:

\[ r= \left[ \frac{(3.9860\times10^{14})(72\,000)^2} {4\pi^2} \right]^{1/3} \]

Therefore:

\[ r=3.7407\times10^7\text{ m} \]

Part b: Altitude

Subtract Earth’s radius:

\[ h=r-R_E \]

\[ \begin{aligned} h&=3.7407\times10^7-6.3781\times10^6\\ &=3.1028\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{h\approx31\,028\text{ km}} \]

Part c: Is it geostationary?

No.

Its orbital period is 20.0 hours, whereas a true geostationary satellite requires a period of approximately 23 hours, 56 minutes, and 4 seconds.

Even though the satellite has the correct orbital direction and inclination, its angular speed does not match Earth’s rotation.

Question 2

A satellite orbits Earth in a circular path with a period exactly half the length of a sidereal day.

The true geostationary orbital radius is approximately \(42\,164\text{ km}\).

Calculate:

a) The ratio of the satellite’s orbital radius to the geostationary orbital radius.

b) The satellite’s orbital radius.

c) Its altitude above Earth’s surface.

Explain why halving the orbital period does not halve the altitude.

Use:

\[ R_E=6\,378\text{ km} \]

Solution 2

The satellite’s altitude is approximately \(20\,184\text{ km}\). This is not half the geostationary altitude.

Part a: Calculate the orbital-radius ratio

For satellites orbiting the same central mass:

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

Therefore:

\[ \frac{r_2}{r_1} = \left(\frac{T_2}{T_1}\right)^{2/3} \]

The second satellite has half the geostationary period:

\[ \frac{r_2}{r_1} = \left(\frac{1}{2}\right)^{2/3} \]

\[ \boxed{\frac{r_2}{r_1}=0.630} \]

Part b: Calculate the orbital radius

\[ r_2=0.630\times42\,164 \]

\[ \boxed{r_2\approx26\,562\text{ km}} \]

Part c: Calculate the altitude

Subtract Earth’s radius:

\[ h=r-R_E \]

\[ h=26\,562-6\,378 \]

\[ \boxed{h\approx20\,184\text{ km}} \]

Why isn’t the altitude also 63.0% of the geostationary altitude?

Because the orbital-radius relationship measures distance from Earth’s centre. Altitude measures distance from Earth’s surface.

You must subtract the same Earth radius from both orbital radii.

The geostationary altitude is approximately \(35\,786\text{ km}\), so:

\[ \frac{20\,184}{35\,786}\approx0.564 \]

The new altitude is approximately 56.4% of the geostationary altitude, not 63.0%.

The important distinction is that orbital radius scales with \(T^{2/3}\), but altitude does not follow that same proportional relationship.

Question 3

Two satellites are placed in the same circular orbit around Earth, with an orbital period of exactly 24.0 hours.

Satellite A has a mass of \(1.20\times10^3\text{ kg}\).

Satellite B has a mass of \(3.60\times10^3\text{ kg}\).

Calculate the gravitational force acting on each satellite.

Then explain why both satellites can maintain the same orbital period despite having different masses.

Use:

\[ GM_E=3.9860\times10^{14}\text{ m}^3\text{ s}^{-2} \]

Solution 3

The gravitational forces are approximately 268 N on Satellite A and 804 N on Satellite B. Both satellites have the same orbital period of 24.0 hours.

Step 1: Calculate the orbital radius.

From the 24-hour calculation:

\[ r=4.2241\times10^7\text{ m} \]

Step 2: Calculate the gravitational field strength at that radius.

The gravitational field strength is:

\[ g=\frac{GM_E}{r^2} \]

Substitute:

\[ g= \frac{3.9860\times10^{14}} {(4.2241\times10^7)^2} \]

\[ g=0.2234\text{ m s}^{-2} \]

Step 3: Calculate the gravitational force on Satellite A.

Using:

\[ F_g=mg \]

\[ F_A=(1.20\times10^3)(0.2234) \]

\[ \boxed{F_A\approx268\text{ N}} \]

Step 4: Calculate the gravitational force on Satellite B.

\[ F_B=(3.60\times10^3)(0.2234) \]

\[ \boxed{F_B\approx804\text{ N}} \]

Satellite B experiences three times the gravitational force because it has three times the mass.

But does that mean it must orbit faster?

No.

The required centripetal force is also proportional to the satellite’s mass:

\[ F_c=\frac{mv^2}{r} \]

A heavier satellite experiences a greater gravitational force, but it also requires a proportionally greater centripetal force to follow the same circular path.

When the two forces are equated, the satellite’s mass cancels.

Consequently, both satellites have the same orbital speed and period at the same orbital radius, regardless of their individual masses.

Question 4

An artificial satellite travels in a circular, equatorial, prograde orbit around Earth.

Its orbital period is 23 hours and 50 minutes.

Calculate:

a) Its orbital radius.

b) Its altitude above Earth’s surface.

c) Whether it will remain above the same longitude. If not, state the direction in which it drifts relative to the ground.

Use:

\[ GM_E=3.9860\times10^{14}\text{ m}^3\text{ s}^{-2} \]

\[ R_E=6.3781\times10^6\text{ m} \]

Solution 4

The satellite’s altitude is approximately \(35\,667\text{ km}\). It will drift eastwards relative to Earth’s surface.

Part a: Orbital radius

Convert the period into seconds:

\[ \begin{aligned} T&=(23\times3600)+(50\times60)\\ &=85\,800\text{ s} \end{aligned} \]

Use:

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

Substitute:

\[ r= \left[ \frac{(3.9860\times10^{14})(85\,800)^2} {4\pi^2} \right]^{1/3} \]

Therefore:

\[ \boxed{r=4.2045\times10^7\text{ m}} \]

Part b: Altitude

\[ h=r-R_E \]

\[ \begin{aligned} h&=4.2045\times10^7-6.3781\times10^6\\ &=3.5667\times10^7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{h\approx35\,667\text{ km}} \]

This is approximately 119 km below the true geostationary altitude.

Part c: Direction of drift

The satellite’s period is shorter than Earth’s sidereal rotation period.

This means the satellite completes a full orbit more quickly than Earth completes one rotation relative to distant stars.

Its angular speed is therefore greater than Earth’s rotational angular speed.

Because both move west to east, the satellite gradually moves ahead of the ground beneath it.

It will drift eastwards relative to Earth’s surface rather than remaining above one longitude.

This demonstrates that being at approximately the right altitude is not enough. The orbital period must also match Earth’s rotation.

Question 5

A hypothetical spherical planet, Planet B, has the following properties:

• Its mass is eight times Earth’s mass.
• Its equatorial radius is three times Earth’s equatorial radius.
• Its sidereal rotation period is exactly half Earth’s sidereal rotation period.

A satellite is placed in a circular, equatorial, prograde orbit around Planet B, with an orbital period matching the planet’s rotation.

Calculate the satellite’s altitude above Planet B’s surface.

Then compare this altitude with Earth’s true geostationary altitude.

Explain why a planet with a larger mass and a larger orbital radius does not necessarily produce a greater satellite altitude.

Assume Newtonian gravity and ignore gravitational perturbations.

Solution 5

Planet B’s synchronous satellite altitude is approximately \(33\,989\text{ km}\), which is about \(1\,797\text{ km}\) lower than Earth’s true geostationary altitude.

The key is to calculate orbital radius first, then account for Planet B’s much larger surface radius.

Step 1: Establish the reference values for Earth.

For Earth:

\[ r_E=42\,164\text{ km} \]

\[ R_E=6\,378.1\text{ km} \]

Therefore:

\[ h_E=42\,164-6\,378.1 \]

\[ h_E=35\,786\text{ km} \]

Step 2: Compare the orbital radii.

The period-radius relationship is:

\[ r\propto(MT^2)^{1/3} \]

Therefore:

\[ \frac{r_B}{r_E} = \left[ \frac{M_B}{M_E} \left(\frac{T_B}{T_E}\right)^2 \right]^{1/3} \]

Substitute the information about Planet B:

\[ \frac{r_B}{r_E} = \left[8\left(\frac12\right)^2\right]^{1/3} \]

Simplify:

\[ \frac{r_B}{r_E} = \sqrt[3]{2} \]

\[ \frac{r_B}{r_E}=1.260 \]

Step 3: Calculate Planet B’s orbital radius.

\[ r_B=1.260\times42\,164 \]

\[ \boxed{r_B\approx53\,123\text{ km}} \]

Notice that Planet B’s orbital radius is larger than Earth’s geostationary orbital radius.

Step 4: Calculate Planet B’s equatorial radius.

We are told that:

\[ R_B=3R_E \]

Therefore:

\[ R_B=3\times6\,378.1 \]

\[ R_B=19\,134.3\text{ km} \]

Step 5: Calculate the satellite’s altitude.

\[ h_B=r_B-R_B \]

\[ h_B=53\,123.5-19\,134.3 \]

\[ \boxed{h_B\approx33\,989\text{ km}} \]

The altitude is positive, so the calculated orbit lies outside Planet B’s surface.

Step 6: Compare with Earth’s geostationary altitude.

\[ \Delta h=35\,786-33\,989 \]

\[ \boxed{\Delta h\approx1\,797\text{ km}} \]

Planet B’s satellite is approximately 1,797 km closer to its planet’s surface than a geostationary satellite is to Earth’s surface.

Why is this surprising?

There are two competing effects.

First, Planet B has eight times Earth’s mass. However, its rotation period is half as long. These changes combine to increase the required orbital radius by a factor of only \(\sqrt[3]{2}\), rather than doubling it.

Second, Planet B’s surface radius is three times Earth’s. That means a much larger portion of its orbital radius is already occupied by the planet itself.

The tempting conclusion is that a larger orbital radius must mean a greater altitude.

But altitude is measured from the surface, not the centre. The larger planetary radius changes the answer.

This is why you must keep orbital radius and altitude separate throughout your calculations.

The next step is to determine whether a satellite at the calculated altitude is genuinely geostationary. Its period must match Earth’s sidereal rotation, but its orbit must also be circular, equatorial, and prograde.

For a closer look at those conditions and why each one matters, read Geostationary Orbits for HSC Physics: Conditions and Uses.