Why a Geostationary Satellite Must Orbit Above the Equator

A satellite can complete one orbit every 23 hours, 56 minutes, and 4 seconds, travel in the same direction as Earth’s rotation, and still fail to remain above the same point on Earth. The missing condition is the orientation of its orbital plane.

A geostationary satellite must orbit in Earth’s equatorial plane because a freely orbiting satellite follows a path in a plane that passes through Earth’s centre, while a point fixed at any non-equatorial latitude travels in a circle in a plane that does not pass through Earth’s centre. The equator is the only latitude for which these two geometries can match.

This is a geometrical restriction, not simply a convention about where satellites are usually placed. Let’s work out why.

01Start with a point on Earth’s surface

Imagine placing a small marker at 30° south latitude on a globe. Now rotate the globe around its north-south axis.

What path does the marker follow?

It travels in a circle around Earth’s rotation axis. However, it does not travel in the equatorial plane. Its circular path lies in a parallel plane below the equator.

Try the same experiment with a marker placed directly on the equator. This time, its circular path lies in the equatorial plane itself, passing through Earth’s centre.

That difference is the key to the whole problem.

Describing the geometry precisely

Treat Earth as a perfect sphere with radius \(R_E\). Let \(\phi\) represent the latitude of a point on its surface.

As Earth rotates, the point travels around the rotation axis. Its distance from that axis is

\[ \rho=R_E\cos\phi \]

Its perpendicular distance from the equatorial plane is

\[ z=R_E\sin\phi \]

Here:

• \(R_E\) is Earth’s radius, in metres.
• \(\phi\) is the latitude, measured from the equator.
• \(\rho\) is the radius of the circular path around Earth’s rotation axis.
• \(z\) is the signed distance above or below the equatorial plane.

Notice what happens at the equator.

When \(\phi=0^\circ\),

\[ \rho=R_E,\qquad z=0 \]

The point travels in a circle centred on Earth’s rotation axis, in a plane passing through Earth’s centre.

Now consider a point at \(30^\circ\) south.

Its path is still circular, but its entire plane sits below the equatorial plane. It does not pass through Earth’s centre.

This gives us an important prediction:

A satellite that stays directly above a point at 30° south must also travel around Earth’s rotation axis in a circular path in a plane below the equator.

But can a freely orbiting satellite actually follow that path?

02Why a free satellite must follow a plane through Earth’s centre

A satellite in an ordinary unpowered orbit is primarily acted on by Earth’s gravitational force.

In the ideal HSC model, Earth is treated as a spherical body, so its gravitational force always points directly towards Earth’s centre.

This has an important consequence.

Imagine the satellite’s position, its velocity, and Earth’s centre at one instant. These three define a plane.

Because gravity points towards Earth’s centre, the gravitational force lies within this plane. The satellite therefore has no acceleration pushing it out of that plane.

It stays in the same orbital plane throughout its motion.

This is why an unpowered satellite cannot simply choose any circular path around Earth’s rotation axis. Its orbital plane must pass through Earth’s centre.

For a circular orbit around Earth, gravity supplies the centripetal force:

\[ \frac{GM_E m}{r^2}=\frac{mv^2}{r} \]

where:

• \(G\) is the universal gravitational constant,
• \(M_E\) is Earth’s mass,
• \(m\) is the satellite’s mass,
• \(r\) is the distance from Earth’s centre, and
• \(v\) is the satellite’s orbital speed.

The gravitational force points towards Earth’s centre, so the centre of the circular orbital path is also Earth’s centre.

An inclined circular orbit is possible. Its plane simply has to pass through Earth’s centre at an angle to the equatorial plane.

What is not possible is a freely orbiting satellite following a circular path in a plane offset from Earth’s centre.

Putting the two paths together

We now have two different geometrical requirements.

For a satellite to remain directly above a fixed point, its motion must match the motion of that point as Earth rotates.

Only at the equator can the required circular path also be a possible unpowered orbital path.

Therefore, a geostationary satellite must orbit in the equatorial plane.

03Why an inclined orbit cannot be geostationary

Let’s test a common prediction.

Suppose a satellite has all the following properties:

• A circular orbit.
• A prograde direction of motion, meaning west to east.
• An orbital period equal to one sidereal day.
• An orbital inclination of \(10^\circ\).

A sidereal day is the time Earth takes to rotate through \(360^\circ\) relative to distant stars:

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

You might predict that the satellite remains above the same point because it completes one revolution in exactly the same time as Earth.

That prediction is incomplete.

The satellite and Earth do complete one revolution in the same time. However, the satellite’s orbital plane is tilted relative to Earth’s equatorial plane.

As the satellite travels around its orbit, it moves north and south of the equator.

For a circular orbit with inclination \(i\), the satellite’s maximum latitude relative to the equator is equal to its inclination:

\[ \phi_{\max}=i \]

Therefore, our satellite reaches:

• \(10^\circ\) north at its maximum northern position.
• \(10^\circ\) south at its maximum southern position.

It cannot remain above one fixed equatorial location.

The important distinction: geosynchronous versus geostationary

A satellite with an orbital period equal to one sidereal day is called geosynchronous.

A geostationary satellite must satisfy additional conditions so that its position relative to Earth’s surface does not change.

For the question we are investigating, the critical difference is the orbital inclination.

The inclined satellite is geosynchronous, but not geostationary.

Its ground track repeats after one sidereal day because Earth and the satellite return to the same relative configuration. However, the satellite does not remain at the same location throughout that day.

This is a useful distinction when answering questions that give you an orbital period but do not specify the orbital plane.

For a fuller comparison of the two types of orbit, see Geosynchronous vs Geostationary Orbits: How to Tell Them Apart.

04Testing the idea with a physical model

The geometry can be investigated using a rotating globe and a circular hoop.

Consider the apparatus:

• A globe is mounted on a vertical spindle representing Earth’s rotation axis.
• A transparent circular hoop passes through the centre of the globe.
• A small bead is attached to the hoop to represent a satellite.
• A thin radial pointer extends from the globe’s centre towards the bead, indicating the satellite’s sub-satellite point on Earth’s surface.

The hoop represents the orbital plane. The bead moves around it, while the globe rotates beneath it.

What should the model show?

First, place the bead on an equatorial hoop.

Move the bead around the hoop once while rotating the globe once in the same direction.

Because the two rotations match, the bead remains aligned with the same equatorial longitude.

Now tilt the hoop by \(10^\circ\), keeping it centred on the globe.

Repeat the motion with the same period and direction.

What changes?

The bead now moves above and below the equatorial plane. Its projected position on the globe travels north and south, even though the bead and globe complete their respective rotations in the same time.

The model demonstrates that matching periods does not remove the effects of orbital inclination.

What does the model actually establish?

There is an important limitation.

The hoop forces the bead to follow a particular path. It does not demonstrate that gravity would naturally produce that path.

The central-force argument from the previous section supplies that missing physics: an unpowered satellite’s orbital plane must pass through Earth’s centre.

The model demonstrates the geometrical consequence of tilting that plane. The gravitational argument explains why the plane must pass through the centre in the first place.

05How could an observer test whether a satellite is geostationary?

The same distinction can be tested using a ground station.

Imagine a receiving station with a satellite dish pointed towards a satellite. The dish receives a radio signal and sends it through a receiver, where the signal strength can be monitored.

If the satellite is geostationary, its direction relative to the station should remain constant in the ideal model.

A fixed dish should therefore continue pointing towards it without requiring changes to its orientation.

The first measurement and its limitation

A fixed dish is a reasonable starting point.

If the signal weakens or disappears, however, that observation does not prove that the satellite has moved. Changes in transmission strength or signal propagation could produce a similar result.

The original measurement cannot reliably separate changes in signal strength from changes in satellite direction.

Improving the measurement

A better arrangement uses a motorised tracking dish with angle sensors.

The dish follows the strongest received signal, while the sensors record its azimuth and elevation.

These angles describe the direction in which the dish must point to receive the signal.

The predictions are now clearer:

• For an ideal geostationary satellite, the required pointing angles remain constant.
• For an inclined geosynchronous satellite, the pointing angles change throughout the day.
• If the inclined satellite has the correct sidereal period, its apparent position returns to its starting position after one sidereal day.

This is a proposed observational test, rather than a claim about a particular historical experiment.

It also illustrates why a satellite can be useful for communications without being geostationary: a tracking antenna can follow a satellite whose apparent position changes.

06Worked examples

07A useful connection: orbital period does not determine inclination

There is another subtle point worth understanding before attempting exam questions.

For a circular orbit around Earth, the relationship between orbital period and orbital radius is

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

where:

• \(T\) is the orbital period, in seconds,
• \(r\) is the orbital radius measured from Earth’s centre, in metres,
• \(G\) is the universal gravitational constant, and
• \(M_E\) is Earth’s mass.

Notice what is missing from this equation.

There is no inclination term.

For circular orbits around the same central mass, the orbital period determines the orbital radius, but it does not determine the orientation of the orbital plane.

This means two satellites can have:

• The same orbital radius.
• The same orbital speed.
• The same orbital period.
• Different orbital inclinations.

One could be geostationary, while the other is merely geosynchronous.

This is why calculating the correct geostationary orbital radius does not, by itself, establish that a satellite is geostationary.

If you need to practise the period-radius calculation, see How to Calculate Geostationary Altitude from Orbital Period.

08Questions and solutions

Question 1

A satellite travels in a circular orbit in Earth’s equatorial plane. It moves west to east and completes one orbit in \(86\,164\text{ s}\).

Assuming an ideal spherical Earth, explain why the satellite can remain above one fixed equatorial longitude. State the latitude of its sub-satellite point.

Solution 1

The satellite can remain above one fixed equatorial longitude, and its sub-satellite point has a latitude of \(0^\circ\).

There are two important reasons.

First, its orbital plane is the equatorial plane, which passes through Earth’s centre. This is a physically possible orbital plane because Earth’s gravitational force acts towards its centre.

Second, the satellite completes one revolution in the same time that Earth takes to rotate through \(360^\circ\) relative to distant stars.

Both move in the same direction and have the same angular velocity.

Consequently, the satellite maintains the same angular position relative to Earth’s surface.

It is not stationary in space. It is moving rapidly around Earth, but its position relative to the ground remains constant.

Question 2

A satellite travels in a circular prograde orbit with:

• Orbital inclination \(i=12^\circ\).
• Orbital period \(T=86\,164\text{ s}\).

At \(t=0\), the satellite crosses the equator travelling northwards. Its sub-satellite point is at longitude \(40^\circ\) east.

Determine the latitude and longitude of its sub-satellite point after a time of \(T/4\).

Explain why the satellite is not geostationary, even if its longitude has not changed.

Assume an ideal spherical Earth and uniform rotation.

Solution 2

After \(T/4\), the sub-satellite point is at \(12^\circ\) north latitude and \(40^\circ\) east longitude. The satellite is not geostationary because its latitude has changed.

Step 1: Determine the satellite’s position in its orbital plane.

During one complete orbit, the satellite travels through \(360^\circ\).

After one-quarter of an orbit, it travels through

\[ \theta=\frac{360^\circ}{4}=90^\circ \]

Starting at the ascending node, it reaches its maximum northern latitude.

For a circular orbit, this maximum latitude equals the orbital inclination:

\[ \phi_{\max}=i=12^\circ \]

Therefore, the sub-satellite point is at \(12^\circ\) north.

Step 2: Determine how far Earth rotates.

Earth completes one rotation in the same period as the satellite.

During \(T/4\), Earth also rotates through

\[ \theta_E=90^\circ \]

Step 3: Compare their angular positions.

In inertial space, the satellite’s position advances by \(90^\circ\) in longitude from its initial ascending-node position.

Earth also rotates eastwards through \(90^\circ\).

These changes cancel when the satellite’s position is measured relative to Earth’s rotating surface.

Its longitude therefore remains \(40^\circ\) east at this particular instant.

Step 4: Evaluate whether the satellite is geostationary.

Although the longitude is unchanged, the satellite has moved from the equator to \(12^\circ\) north.

A geostationary satellite must maintain both its latitude and longitude relative to Earth.

The inclined satellite fails the latitude condition, so it is geosynchronous rather than geostationary.

This example exposes an important trap: a satellite can momentarily retain the same longitude without remaining above the same point on Earth.

Question 3

A satellite travels in a circular prograde orbit with an inclination of \(20^\circ\) and an orbital period of \(86\,164\text{ s}\).

Its orbital plane is oriented so that its sub-satellite point passes directly over a ground station at \(20^\circ\) south at \(t=0\). At this instant, the satellite is at its maximum southern latitude.

(a) Determine the latitude of the sub-satellite point after \(T/4\).

(b) Explain why the satellite is not geostationary, even though it passes directly over the ground station.

(c) After what time is its complete ground track guaranteed to repeat?

Solution 3

(a) The sub-satellite point is at \(0^\circ\) latitude after \(T/4\).

(b) The satellite is not geostationary because it moves away from the ground station as it continues around its inclined orbit.

(c) Its complete ground track repeats after \(86\,164\text{ s}\), or one sidereal day.

Step 1: Determine the satellite’s angular displacement.

After one-quarter of its orbital period, it has travelled through

\[ \theta=\frac{360^\circ}{4}=90^\circ \]

The satellite starts at its maximum southern latitude.

As it moves prograde, it travels towards the ascending equatorial crossing.

Step 2: Determine its latitude.

At its maximum southern position, its latitude is

\[ \phi=-20^\circ \]

After one-quarter of an orbit, it reaches the equatorial plane.

Therefore,

\[ \phi=0^\circ \]

The satellite is no longer directly above the ground station at \(20^\circ\) south.

Step 3: Explain why it is not geostationary.

A geostationary satellite must maintain the same position relative to the ground throughout its orbit.

This satellite only passes over the station at a particular instant. It subsequently moves northwards, crosses the equator, and eventually travels into the northern hemisphere.

Passing over a location is not the same as remaining above it.

Step 4: Determine when the ground track repeats.

The satellite completes one orbit in \(86\,164\text{ s}\).

During this time, Earth also completes one rotation relative to distant stars.

Both therefore return to the same relative configuration.

The entire ground track repeats after

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

The satellite returns to its original ground-track position after one sidereal day.

This is an example of a repeating geosynchronous ground track, not a stationary satellite.

Question 4

Two satellites orbit Earth in circular prograde orbits.

Both have an orbital period of \(86\,164\text{ s}\), and both have the same orbital radius.

Satellite A has an inclination of \(0^\circ\), while satellite B has an inclination of \(15^\circ\).

An engineer argues:

“Satellite B should also be geostationary. It completes one revolution in the same time as Earth, so its ground track returns to exactly the same position every day.”

Evaluate this statement.

Calculate the maximum latitude reached by satellite B’s sub-satellite point and the total north-to-south distance covered along Earth’s surface.

Use \(R_E=6.37\times10^6\text{ m}\). Assume a spherical Earth.

Solution 4

The engineer is incorrect. Satellite B is geosynchronous, but it is not geostationary. Its sub-satellite point reaches \(15^\circ\) north and \(15^\circ\) south, giving a total north-to-south surface range of approximately \(3335\text{ km}\).

Step 1: Separate the two orbital conditions.

Both satellites have the same orbital period and move in the same direction as Earth’s rotation.

Satellite A has an equatorial orbit, so its sub-satellite point remains at latitude \(0^\circ\).

Satellite B has an inclined orbit, so its sub-satellite point moves north and south.

The engineer has confused a repeating ground track with a stationary ground position.

Step 2: Calculate the maximum latitude.

For a circular orbit, the maximum latitude equals the orbital inclination.

Therefore,

\[ \phi_{\max}=15^\circ \]

Satellite B reaches both

\[ 15^\circ\text{ N} \]

and

\[ 15^\circ\text{ S} \]

Step 3: Calculate the total angular range.

The satellite travels from \(15^\circ\) south to \(15^\circ\) north.

The total angular range is

\[ \Delta\phi=30^\circ \]

Convert this to radians:

\[ \Delta\phi=30^\circ\times\frac{\pi}{180^\circ} \]

\[ \Delta\phi=0.5236\text{ rad} \]

Step 4: Calculate the north-to-south surface distance.

Using the arc-length equation,

\[ s=R_E\Delta\phi \]

Substitute:

\[ \begin{aligned} s&=(6.37\times10^6)(0.5236)\\ &=3.335\times10^6\text{ m} \end{aligned} \]

Therefore,

\[ \boxed{s\approx3335\text{ km}} \]

Step 5: Explain why the engineer’s argument fails.

Satellite B returns to the same Earth-relative position after one sidereal day.

However, throughout that day, it travels through different latitudes and longitudes.

The ground track repeats, but the satellite is not stationary along that track.

Satellite A can be geostationary because its orbital plane matches the equatorial plane. Satellite B cannot, despite having the same orbital radius, speed, and period.

Question 5

An engineer proposes an unusual alternative to a conventional geostationary satellite.

The engineer wants a platform to remain directly above a fixed location at \(30^\circ\) north latitude. The platform is placed at a distance \(r_g\) from Earth’s centre, where \(r_g\) is the radius of the conventional geostationary orbit.

Unlike an ordinary satellite, the platform has a continuously operating thruster.

Assume:

• Earth is spherical.
• Earth’s gravitational field is central.
• The platform remains at a constant distance \(r_g\) from Earth’s centre.
• The platform rotates with Earth at angular velocity \(\Omega\).
• Gravitational acceleration at \(r_g\) has magnitude \(g=0.224\text{ m s}^{-2}\).
• The values satisfy \(\Omega^2r_g=0.224\text{ m s}^{-2}\).
• No other forces act on the platform apart from gravity and the thruster.

(a) Is it physically possible for the platform to remain fixed relative to the ground at \(30^\circ\) north if the thruster operates continuously?

(b) Calculate the required thrust acceleration per unit mass and state its direction.

(c) Would this platform be following an ordinary unpowered geostationary orbit? Explain the distinction.

Solution 5

(a) Yes. Continuous thrust can keep the platform fixed above \(30^\circ\) north.

(b) The required thrust acceleration is \(0.112\text{ m s}^{-2}\), directed northwards, parallel to Earth’s rotation axis.

(c) No. It would be following a powered, Earth-fixed circular path, not an ordinary unpowered geostationary orbit.

This question tests an important assumption hidden in the earlier explanations: the equatorial restriction applies to a freely orbiting satellite acted on by central gravitational force alone.

Step 1: Determine the acceleration required to remain fixed relative to Earth.

The platform rotates with Earth, so in inertial space it travels in a circle around Earth’s rotation axis.

Its distance from the rotation axis is

\[ \rho=r_g\cos30^\circ \]

Its required centripetal acceleration is therefore

\[ a_c=\Omega^2\rho \]

Substituting:

\[ a_c=\Omega^2r_g\cos30^\circ \]

Given that \(\Omega^2r_g=0.224\text{ m s}^{-2}\),

\[ \begin{aligned} a_c&=0.224\cos30^\circ\\ &=0.194\text{ m s}^{-2} \end{aligned} \]

This acceleration must point horizontally towards Earth’s rotation axis.

Step 2: Resolve the gravitational acceleration.

Gravity acts directly towards Earth’s centre.

At \(30^\circ\) north, its direction is inclined relative to the equatorial plane.

Resolve gravitational acceleration into two perpendicular components:

• A component parallel to the equatorial plane, directed towards Earth’s rotation axis.
• A component parallel to Earth’s rotation axis, directed towards the equatorial plane.

The first component is

\[ g_{\perp}=g\cos30^\circ \]

\[ g_{\perp}=0.224\cos30^\circ \]

\[ g_{\perp}=0.194\text{ m s}^{-2} \]

The second component is

\[ g_{\parallel}=g\sin30^\circ \]

\[ g_{\parallel}=0.224\sin30^\circ \]

\[ g_{\parallel}=0.112\text{ m s}^{-2} \]

Step 3: Compare the required acceleration with gravity.

The platform requires a centripetal acceleration of

\[ 0.194\text{ m s}^{-2} \]

towards Earth’s rotation axis.

Gravity already provides exactly this horizontal component:

\[ g_{\perp}=a_c \]

However, gravity also pulls the platform towards the equatorial plane with an acceleration of

\[ 0.112\text{ m s}^{-2} \]

The platform must have zero acceleration parallel to Earth’s rotation axis if it is to maintain its fixed latitude.

Gravity alone would cause it to accelerate towards the equatorial plane.

Step 4: Determine the required thrust.

The thruster must cancel the remaining gravitational component.

Therefore,

\[ a_{\text{thrust}}=g_{\parallel} \]

\[ \boxed{a_{\text{thrust}}=0.112\text{ m s}^{-2}} \]

Its direction must be parallel to Earth’s rotation axis, away from the equatorial plane.

Because the platform is north of the equator, the thrust must act northwards.

Step 5: Decide whether this is a geostationary orbit.

The platform can remain fixed relative to Earth’s surface because the thruster supplies the acceleration that gravity alone cannot provide.

However, it is not following an ordinary unpowered geostationary orbit.

Without continuous thrust, gravity would pull it into a natural orbital plane passing through Earth’s centre. It would then move away from the fixed position above \(30^\circ\) north.

The original equator restriction has not been disproved. We have changed the physical system by introducing an additional force.

This is the deeper principle: the equatorial requirement follows from the geometry of an unpowered orbit under central gravity, not from an absolute prohibition on keeping an object above another latitude.

09Where to go next

When you see a question about geostationary satellites, separate two decisions:

1. Does the satellite have the correct orbital period and direction?
2. Does its orbital plane allow it to remain above one fixed point on Earth’s surface?

A correct period alone is not enough. The orbit must also be circular and equatorial.

Once you understand why the equatorial plane is essential, the next useful step is to calculate the orbital radius required for that period. From there, you can investigate how changing a satellite’s orbit affects its speed and mechanical energy.