Why a Polar Satellite Cannot Be Geostationary

At \(t=0\), a satellite is directly above a marked point A on Earth’s equator. Half a sidereal day later, it can be directly above A again, even though it has travelled over the North Pole in between.

Does returning to the same point make it geostationary?

No. Returning to a position is not the same as remaining there.

A polar satellite cannot be geostationary because its orbital plane contains Earth’s rotation axis, causing its latitude to change as it travels. A geostationary satellite must remain in Earth’s equatorial plane, above one fixed longitude.

You could give a polar satellite exactly the same orbital period as Earth’s rotation. It would then be geosynchronous, but it still would not be geostationary. The difference comes down to the orientation of its orbit, not simply how long it takes to complete a revolution.

01The experiment: Does the satellite actually stay in one position?

Suppose someone claims that a satellite is geostationary. Rather than accepting its stated orbital period, we want to test whether it actually remains in the same position relative to Earth.

A simple observation experiment can distinguish the two types of motion.

Setting up the observation

Imagine a telescope mounted on a fixed platform on Earth’s surface.

A camera is attached to the telescope, and a clock records the time of every exposure. The telescope is calibrated to measure the satellite’s position using its azimuth (horizontal direction) and elevation (angle above the horizon).

The light path is straightforward:

1. Sunlight reflects off the satellite.
2. The reflected light enters the telescope.
3. The telescope’s lens focuses the light onto the camera sensor.
4. The camera records the satellite’s apparent position and the clock records the time.

The telescope remains fixed relative to the ground. It does not automatically follow the satellite.

What should we observe?

Consider two possible predictions.

Prediction 1: The satellite is geostationary.

Its position relative to the observing station should remain constant.

Its azimuth and elevation do not change. The telescope can continue pointing in the same direction and the satellite remains in view, provided it is above the observer’s horizon.

Prediction 2: The satellite is in a polar orbit.

Its position relative to the observing station changes.

As it travels along its polar orbit, its position relative to Earth’s surface moves through different latitudes. An observer sees its apparent direction change.

This observation would rule out geostationary motion.

Why two photographs might give the wrong conclusion

There is a problem with a simple experimental design.

Suppose we take one photograph at the beginning of an orbital period and another exactly one period later.

The satellite might appear in precisely the same position in both photographs.

That does not prove it was stationary between the observations.

A geosynchronous polar satellite can return to its original position relative to Earth after one sidereal day. It has completed an orbit, and Earth has completed one rotation.

The experiment has measured the starting and finishing positions, but missed everything in between.

The improved design: Record the satellite’s position at regular intervals throughout the orbit. If necessary, combine observations from several ground stations to reconstruct its full path.

This reveals the difference between a satellite that remains fixed and one that follows a repeating path.

02The geometry: Why polar and geostationary orbits are incompatible

To understand the result, we need to picture two different orbital planes.

An orbital plane is the flat plane containing a satellite’s circular path and Earth’s centre.

Earth has two important reference features:

• Its equatorial plane, which passes through the equator.
• Its rotation axis, which runs through the North and South Poles.

A geostationary satellite travels in the equatorial plane.

A polar satellite travels in an orbital plane that contains Earth’s rotation axis.

These planes have fundamentally different orientations.

What does orbital inclination mean?

Orbital inclination, represented by \(i\), measures the angle between a satellite’s orbital plane and Earth’s equatorial plane.

Two important cases are:

• Equatorial orbit: \(i=0^\circ\).
• Polar orbit: \(i=90^\circ\).

A near-polar orbit has an inclination close to \(90^\circ\).

For an exactly polar orbit, the satellite’s path passes over both poles. Its position relative to Earth’s surface changes from the equator towards the North Pole, then back towards the equator, and eventually towards the South Pole.

A geostationary satellite cannot do this.

It must remain above the equator at all times.

Why Earth’s rotation cannot fix the problem

Here is the important physical detail.

Earth rotates around its North-South axis. This rotation changes the longitude of locations on Earth’s surface, but it does not change their latitude.

For example, imagine standing at a point on the equator.

As Earth rotates, you move eastwards around Earth’s axis. Your latitude remains zero.

A polar satellite behaves differently. Its orbital motion takes it north and south relative to the equator.

Earth’s rotation can change the satellite’s longitude relative to the surface, but it cannot cancel its changing latitude.

This is why adjusting the satellite’s speed or orbital period cannot make a polar orbit geostationary.

A more precise explanation using angular velocity

Angular velocity describes how quickly an object rotates through an angle.

Its magnitude is:

\[ \omega=\frac{2\pi}{T} \]

where:

• \(\omega\) is angular speed in radians per second.
• \(T\) is the time for one complete revolution in seconds.

Earth’s rotation and a geosynchronous satellite can have the same angular speed.

However, angular velocity is a vector. Its direction matters as well as its magnitude.

The direction of an orbit’s angular momentum vector is perpendicular to its orbital plane.

For a geostationary satellite, this direction is parallel to Earth’s rotation axis.

For an exactly polar satellite, it is perpendicular to Earth’s rotation axis.

Therefore, even if the two objects have identical angular speeds, they are rotating around different axes.

Matching the number of seconds per revolution does not make their motions equivalent.

03Why having the correct orbital period is not enough

Let’s connect this geometry to the circular-orbit equations used in HSC Physics.

For a satellite in a circular orbit around Earth, gravitational force provides the centripetal force.

Therefore:

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

where:

• \(G\) is the universal gravitational constant, in \(\mathrm{N\,m^2\,kg^{-2}}\).
• \(M_E\) is Earth’s mass, in kilograms.
• \(m\) is the satellite’s mass, in kilograms.
• \(r\) is the orbital radius measured from Earth’s centre, in metres.
• \(v\) is orbital speed, in \(\mathrm{m\,s^{-1}}\).

For a circular orbit, the satellite travels a circumference of \(2\pi r\) in one period \(T\).

Thus:

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

Substituting this into the gravitational force equation gives:

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

Rearranging:

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

This relationship applies to circular orbits around Earth.

Notice something important: the equation contains no orbital inclination.

The period determines the radius of a circular orbit around Earth, but it does not determine the orientation of the orbital plane.

That means a circular polar orbit and a circular equatorial orbit can have exactly the same orbital period and radius.

For a geostationary satellite, the required period is one sidereal day:

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

This is approximately 23 hours, 56 minutes, and 4 seconds.

Using this period gives an orbital radius of approximately \(4.216\times10^7\text{ m}\) from Earth’s centre.

A polar satellite can theoretically occupy a circular orbit at this same radius. It simply will not be geostationary.

Comparing three satellite orbits

The middle example is particularly important.

The satellite has the correct period, but its orbital plane is wrong.

For a complete explanation of the other geostationary requirements, see Geostationary Orbits for HSC Physics: Conditions and Uses.

You can also read Geosynchronous vs Geostationary Orbits: How to Tell Them Apart for a broader comparison.

06Why polar orbits are useful anyway

It would be easy to assume that geostationary orbits are always better because they remain above one location.

But the two types of orbits serve different purposes.

A geostationary satellite continuously observes the same large region of Earth. This is useful for communications and monitoring weather systems.

A polar satellite travels across different latitudes. Meanwhile, Earth rotates beneath its orbital plane, allowing successive passes to cover different longitudes.

This makes polar and near-polar orbits useful for Earth observation, mapping, and environmental monitoring.

The important distinction is what the satellite needs to observe.

• If it needs to remain above one fixed region, a geostationary orbit is useful.
• If it needs to observe different regions across a wide range of latitudes, a polar orbit is useful.

The inability to remain stationary is not a disadvantage for every satellite. Sometimes it is exactly what the mission requires.

07Questions and solutions

Question 1

A satellite travels in a circular polar orbit around Earth. Its orbital period is exactly one sidereal day.

A student claims that the satellite must be geostationary because its orbital period matches Earth’s rotation period.

Explain why the student’s conclusion is incorrect.

Solution 1

The satellite is geosynchronous, but it is not geostationary.

Although its orbital period matches Earth’s sidereal rotation period, its orbital plane is polar rather than equatorial.

Its latitude therefore changes as it travels around Earth, passing over both poles in an ideal exactly polar orbit.

A geostationary satellite must remain above one fixed equatorial longitude.

Matching the orbital period is necessary, but it does not compensate for an incorrect orbital inclination.

Question 2

A satellite follows a circular polar orbit with an orbital period of \(100\text{ min}\).

Assume:

• Earth’s sidereal rotation period is \(86\,164\text{ s}\).
• The satellite’s orbital plane remains fixed relative to distant stars.
• Earth’s rotation is uniform.

Calculate the change in the Earth-fixed longitude of the satellite’s ascending equator crossing between two consecutive orbits.

Take eastward longitude as positive.

Solution 2

The next ascending equator crossing occurs approximately \(25.1^\circ\) west of the previous crossing.

The orbital plane remains fixed relative to distant stars, but Earth rotates underneath it.

Step 1: Convert the orbital period into seconds.

\[ T=100\times60 \]

\[ T=6000\text{ s} \]

Step 2: Calculate Earth’s angular rotation during one satellite orbit.

Earth completes \(360^\circ\) in \(86\,164\text{ s}\).

Therefore:

\[ \Delta\theta_E= 360^\circ\left(\frac{6000}{86\,164}\right) \]

\[ \Delta\theta_E\approx25.1^\circ \]

Step 3: Determine the direction of the ground-track shift.

The ascending node remains in the same inertial direction because we are assuming the orbital plane does not precess.

However, Earth rotates eastward by approximately \(25.1^\circ\) during the satellite’s orbit.

Therefore, the ascending node appears to move westward relative to Earth’s surface.

Using east-positive longitude:

\[ \Delta\lambda=-25.1^\circ \]

The negative sign indicates a westward shift.

Step 4: Interpret the result.

Each successive ascending equator crossing occurs approximately \(25.1^\circ\) west of the previous one.

This explains how a polar satellite can gradually cover different longitudes.

The important distinction is that the orbital plane remains fixed in space while Earth’s surface rotates beneath it.

Question 3

A satellite follows a circular orbit around Earth with:

• Orbital inclination \(i=60^\circ\).
• Orbital period \(T=86\,164\text{ s}\).

Assume the satellite moves in the prograde direction and Earth is spherical.

Determine:

1. The maximum northern and southern latitudes reached by the satellite’s sub-satellite point.
2. Whether the satellite is geosynchronous.
3. Whether the satellite is geostationary.

Solution 3

The satellite reaches a maximum latitude of \(60^\circ\) north and \(60^\circ\) south.

It is geosynchronous, but it is not geostationary.

Step 1: Determine the maximum latitude.

For a circular orbit inclined between \(0^\circ\) and \(90^\circ\), the maximum latitude reached by the satellite’s sub-satellite point equals its orbital inclination.

Therefore:

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

\[ \boxed{\phi_{\max}=60^\circ} \]

The satellite moves between approximately \(60^\circ\) north and \(60^\circ\) south.

It does not pass directly over either pole.

Step 2: Determine whether it is geosynchronous.

A geosynchronous satellite has an orbital period equal to Earth’s sidereal rotation period.

Here:

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

Therefore, the satellite is geosynchronous.

Step 3: Determine whether it is geostationary.

A geostationary satellite must remain above one fixed equatorial longitude.

This satellite reaches latitudes of \(60^\circ\) north and south, so its position relative to Earth’s surface changes.

Therefore, it is not geostationary.

The tempting mistake is to use the correct period as the only classification test.

The period matches Earth’s rotation, but the inclination prevents the satellite from remaining above one fixed location.

Question 4

A satellite initially travels in a circular equatorial orbit around Earth with an orbital period of \(86\,164\text{ s}\).

At an equatorial crossing, an ideal instantaneous impulse rotates its velocity direction by \(90^\circ\), changing it from due east to due north.

The magnitude of its velocity remains unchanged.

Assume Earth is spherical and ignore orbital perturbations.

Determine:

1. The new orbital plane.
2. Whether the new orbit remains circular.
3. Whether its orbital period changes.
4. Whether it remains geostationary.
5. Whether its total orbital mechanical energy changes.

Solution 4

The satellite enters a circular polar orbit with the same orbital radius and period as before. It is no longer geostationary, and its total orbital mechanical energy remains unchanged.

Step 1: Determine the new orbital plane.

At the equator, the radial direction from Earth’s centre to the satellite lies in the equatorial plane.

The new velocity is directed due north, perpendicular to this radial direction.

The plane containing the radial direction and the northward velocity also contains Earth’s rotation axis.

Therefore, the new orbital plane is polar.

Step 2: Determine whether the orbit remains circular.

For a circular orbit, the satellite’s velocity must be perpendicular to its radial position vector and have the required circular orbital speed.

The impulse changes the direction of velocity but not its magnitude.

The satellite is still moving perpendicular to the radial direction, with exactly the same circular orbital speed as before.

Therefore, the new orbit remains circular.

Step 3: Determine whether the orbital period changes.

The orbital radius and central mass have not changed.

For a circular orbit around Earth:

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

Since \(r\) and \(M_E\) remain unchanged:

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

The orbital period does not change.

Step 4: Determine whether the satellite is geostationary.

The new orbit is polar.

The satellite now moves north and south relative to Earth’s surface rather than remaining above the equator.

Therefore, it is no longer geostationary.

It is still geosynchronous because its orbital period remains equal to one sidereal day.

Step 5: Determine the change in total mechanical energy.

Total mechanical energy is the sum of kinetic and gravitational potential energy:

\[ E=K+U \]

Kinetic energy depends on speed:

\[ K=\frac12mv^2 \]

The speed has not changed, so kinetic energy is unchanged.

Gravitational potential energy depends on the distance from Earth’s centre in the spherical central-mass model.

The orbital radius has not changed, so gravitational potential energy is also unchanged.

Therefore:

\[ \boxed{\Delta E=0\text{ J}} \]

The impulse changes the direction of the satellite’s angular momentum, not its magnitude or total orbital mechanical energy.

This is an important distinction: changing an orbital plane does not necessarily mean changing the satellite’s orbital energy.

Question 5

An ideal satellite follows a circular polar orbit around Earth with an orbital period of exactly \(86\,164\text{ s}\).

At \(t=0\), the satellite is directly above a marked point A on Earth’s equator and is moving northwards.

Assume:

• The satellite’s orbital plane remains fixed relative to distant stars.
• Earth rotates uniformly through \(360^\circ\) during one orbital period.
• The satellite completes a circular orbit at constant angular speed.
• Earth is spherical, and orbital perturbations are ignored.

An observer records the satellite’s position at three times:

• \(t=0\).
• \(t=T/2\).
• \(t=T\).

Surprisingly, the satellite is directly above point A at all three observations.

A student concludes that the satellite is geostationary because it repeatedly appears above the same ground location.

Is this conclusion correct?

Determine the satellite’s position relative to Earth’s surface at:

1. \(t=T/4\).
2. \(t=T/2\).
3. \(t=3T/4\).
4. \(t=T\).

Then explain why returning to point A does not make the satellite geostationary.

Solution 5

The student is incorrect. The satellite is geosynchronous, but it is not geostationary.

It passes over the North Pole, returns above A moving south, passes over the South Pole, and eventually returns to A moving north.

The key is that the satellite and Earth can return to the same relative position at particular times without remaining aligned throughout the motion.

Step 1: Determine the satellite’s position at \(t=T/4\).

The satellite completes one full revolution in time \(T\).

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

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

\[ \theta=90^\circ \]

It started at the equator moving north.

Therefore, after one-quarter of its orbit, it is directly above the North Pole.

\[ \boxed{t=T/4:\text{ North Pole}} \]

Step 2: Determine the position at \(t=T/2\).

After half an orbit, the satellite has travelled through:

\[ \theta=\frac{360^\circ}{2} \]

\[ \theta=180^\circ \]

It is now on the opposite side of its circular orbital path, crossing the equator while moving south.

But something else has happened.

During half a sidereal day, Earth has also rotated through \(180^\circ\).

The marked ground point A has therefore moved to the opposite side of Earth relative to its original inertial position.

The satellite has also moved to the opposite side of its orbital circle.

These two changes place the satellite directly above A again.

\[ \boxed{t=T/2:\text{ Point A, moving south}} \]

This is the unusual part of the problem. The satellite returns above the same ground point after half an orbit, despite not being stationary.

Step 3: Determine the position at \(t=3T/4\).

After three-quarters of its orbit, the satellite has travelled through:

\[ \theta=270^\circ \]

It has passed the equator heading south and is now directly above the South Pole.

\[ \boxed{t=3T/4:\text{ South Pole}} \]

Step 4: Determine the position at \(t=T\).

After one complete orbit, the satellite has travelled through:

\[ \theta=360^\circ \]

Earth has also completed one full sidereal rotation.

Both return to their original relative orientation.

Therefore:

\[ \boxed{t=T:\text{ Point A, moving north}} \]

The sequence is:

Step 5: Explain why the satellite is not geostationary.

A geostationary satellite must remain above the same point on Earth’s surface continuously.

This satellite only returns above A at particular times.

Between those observations, it travels over both poles and changes its latitude dramatically.

The student has confused a repeating position with a stationary position.

That is the crucial distinction.

A polar satellite can have the correct orbital period, return to the same ground point, and repeat its ground track regularly. None of these properties makes it geostationary.

The next time a satellite question gives you an orbital period, check the orbital plane before classifying it. Once you know the inclination, you can use the circular-orbit equations to calculate its radius, speed, and other orbital properties without confusing a repeating orbit with a stationary one.