How to Compare Orbital Periods Without Knowing Central Mass

Two satellites orbit the same planet in circular paths. Satellite B is 1.5 times farther from the planet’s centre than satellite A. You need to compare their orbital periods, but the question doesn’t give you the planet’s mass.

Are you missing information?

No. Because both satellites orbit the same central body, its mass is identical in both cases. When you compare their orbital periods using a ratio, the central mass cancels.

The result is:

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

This is the key to comparing orbital periods without knowing the central mass.

For the example above, satellite B takes approximately 1.84 times as long to complete an orbit. Notice that you didn’t need to know the planet’s mass, calculate its gravitational field, or find either satellite’s actual speed.

There is one important condition: both objects must be in circular orbits around the same central body. Let’s see why that condition matters and how to use the relationship in HSC-style problems.

01Why the central mass cancels

Start with the physical situation.

Imagine two satellites orbiting Earth. Satellite A follows a smaller circular path, while satellite B follows a larger one.

Before looking at the mathematics, make a prediction: if B’s orbital radius doubles, will its orbital period also double?

That sounds reasonable. A circle with twice the radius has twice the circumference, so perhaps the satellite needs twice as long to travel around it.

But there is another change.

A satellite in a larger circular orbit travels more slowly than one in a smaller circular orbit around the same planet. It has a longer path to travel and a lower orbital speed.

These two effects combine to make the period increase by more than the radius does.

The starting equation

For a small satellite in a circular orbit around a central body of mass \(M\), the orbital period is:

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

Where:

• \(T\) is the orbital period, measured in seconds (s).
• \(r\) is the orbital radius, measured from the central body’s centre, in metres (m).
• \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\).
• \(M\) is the mass of the central body, measured in kilograms (kg).

This equation describes a circular orbit. It assumes the orbiting satellite has a negligible effect on the motion of the central body.

You can explore the force balance and full derivation in Orbital Period: Radius and Central Mass for HSC Physics.

Comparing two satellites

Suppose satellite A has orbital radius \(r_A\) and period \(T_A\), while satellite B has orbital radius \(r_B\) and period \(T_B\).

Both orbit the same central body, so both have the same central mass \(M\).

Write the period equation for each satellite:

\[ T_A=2\pi\sqrt{\frac{r_A^3}{GM}} \]

\[ T_B=2\pi\sqrt{\frac{r_B^3}{GM}} \]

Now divide the second equation by the first:

\[ \frac{T_B}{T_A} = \frac{2\pi\sqrt{\frac{r_B^3}{GM}}} {2\pi\sqrt{\frac{r_A^3}{GM}}} \]

The common factors \(2\pi\), \(G\), and \(M\) cancel.

This leaves:

\[ \frac{T_B}{T_A} = \sqrt{\frac{r_B^3}{r_A^3}} \]

Therefore:

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

This is the circular-orbit comparison relationship you need.

The important point is that you haven’t assumed the central mass is zero or irrelevant. Its effect is present in both orbital periods, so it disappears when you divide one period by the other.

What does the power of 3/2 mean?

For circular orbits around the same central body:

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

The orbital period is proportional to the orbital radius raised to the power of \(3/2\).

Here are some useful comparisons:

Notice something interesting: doubling the radius doesn’t double the period. It increases the period by approximately 2.83 times.

The reason is that the satellite also travels more slowly in the larger orbit.

If you want to understand that speed change in more detail, see Orbital Velocity: Derivation and Calculations for HSC Physics.

Reversing the relationship

Sometimes an exam question gives you the orbital periods and asks you to compare the orbital radii instead.

You can rearrange the same equation:

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

Raise both sides to the power of \(2/3\):

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

For example, if satellite B takes eight times as long to complete an orbit as satellite A:

\[ \frac{r_B}{r_A}=8^{2/3}=4 \]

Satellite B must have four times the orbital radius.

This reverse relationship is particularly useful when a question provides observations of orbital periods but asks you to determine an unknown orbital distance.

02The observational idea behind the relationship

The ability to compare orbital periods without knowing the central mass has an interesting history.

It comes from a relationship that astronomers identified by studying planetary motion long before Newton developed his theory of universal gravitation.

Kepler’s problem

In the early 1600s, astronomers were trying to explain how planets moved through the sky.

The German astronomer Johannes Kepler worked with observations collected by Tycho Brahe, whose measurements of planetary positions were remarkably accurate for observations made without telescopes.

Brahe’s problem was observational: how could astronomers turn repeated measurements of planetary positions into a reliable model of their orbits?

The measurement process was different from what you might expect.

An observer aligned a sighting instrument with a planet and read its angular position against the instrument’s graduated scale. By repeating these observations over time, astronomers could build a record of how the planet appeared to move relative to the stars.

The instrument measured direction, not the planet’s actual distance from the Sun or its mass.

Kepler was particularly interested in Mars. The observed positions did not agree accurately enough with the circular-orbit models he was trying to use.

The prediction was clear: if the circular model correctly described Mars’s motion, its calculated positions should match the observations.

They didn’t.

Kepler eventually revised the orbital model, recognising that planetary orbits are elliptical rather than perfectly circular. His first two laws were published in 1609.

In 1619, he published his third law, identifying a mathematical relationship between the time a planet takes to orbit the Sun and the size of its orbit.

The important historical distinction is that the measuring instruments were not the failed part. The original circular model was incomplete, and the observations forced a better mathematical description.

Kepler’s third law was an empirical relationship. Newton later explained its physical basis through universal gravitation.

What did Kepler’s third law reveal?

For planets orbiting the Sun, Kepler found that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

The semi-major axis, represented by \(a\), is half the longest diameter of an elliptical orbit.

For a perfectly circular orbit, the semi-major axis is equal to the orbital radius:

\[ a=r \]

For the circular-orbit comparisons in this article, we can therefore write:

\[ T^2\propto r^3 \]

This is the underlying relationship that allows us to compare orbital periods without knowing the Sun’s mass or another central mass.

Checking the pattern with planetary data

The following table uses rounded modern orbital values for three planets orbiting the Sun. These are illustrative reference values, not Tycho Brahe’s original measurements.

AU means astronomical unit, approximately the mean distance between Earth and the Sun.

We can test the relationship using Mars:

\[ \frac{T^2}{a^3} = \frac{(1.881)^2}{(1.524)^3} \]

\[ \frac{T^2}{a^3}\approx1.00 \]

Using the rounded values, the same result is obtained for Earth and approximately for Jupiter.

The important observation is that the ratio is approximately constant for objects orbiting the same central body.

There is an important qualification here. Real planetary orbits are elliptical, so Kepler’s original law uses the semi-major axis rather than a constant orbital radius. The circular-orbit relationship used in HSC-style calculations is the simpler case where \(a=r\).

The historical result gives us the bigger idea: orbital periods and orbital sizes are linked in a predictable way, even when the central mass has not been measured.

03When can you use the mass-free comparison?

Before substituting numbers, check that the question actually satisfies the conditions.

The central mass only cancels when both objects orbit the same central body.

The contrasting case: different central masses

Consider two satellites at exactly the same orbital radius.

Satellite A orbits planet X, while satellite B orbits planet Y.

Suppose planet Y has four times the mass of planet X.

Would their periods be equal because their orbital radii are equal?

No.

At a fixed orbital radius, the circular-orbit relationship gives:

\[ T\propto\frac{1}{\sqrt{M}} \]

Therefore:

\[ \frac{T_B}{T_A} = \sqrt{\frac{M_X}{M_Y}} \]

Substituting the mass ratio:

\[ \frac{T_B}{T_A} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

Satellite B takes half as long to complete an orbit.

The more massive planet produces a stronger gravitational attraction at the same radius, so the satellite must travel faster to maintain its circular orbit.

The central mass has not cancelled because the two satellites are orbiting different planets.

This is the key distinction:

• Same central body: you can compare periods using orbital radii alone.
• Different central bodies: you need information about their masses as well.

A quick decision process for exam questions

Before reaching for your calculator, follow these steps:

1. Identify the central body. Are both objects orbiting the same planet or star?
2. Check the orbit shape. Is the question describing circular orbits?
3. Identify the orbital radii. Are distances measured from the centre, rather than the surface?
4. Choose the correct ratio. Use the \(3/2\) power for period comparisons or the \(2/3\) power for radius comparisons.
5. Check the result. A larger circular orbital radius around the same central body must produce a longer period.

One final detail: orbital radius and altitude are not interchangeable. This becomes particularly important when a question describes satellites as being a certain height above a planet’s surface.

04Applying the relationship

05Questions and solutions

The following questions move from straightforward period comparisons to more involved problems involving different central masses, unknown orbital distances, and the limits of what orbital ratios can actually tell you.

Question 1

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

Satellite A has an orbital radius of \(9.0\times10^6\text{ m}\) and a period of \(3.0\text{ h}\).

Satellite B has an orbital radius of \(1.8\times10^7\text{ m}\).

Calculate the orbital period of satellite B.

Solution 1

Satellite B takes approximately \(8.5\text{ h}\) to complete one orbit.

Both satellites orbit the same planet, so the central mass cancels. We can use the period comparison relationship directly.

Step 1: Write the comparison equation.

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

Step 2: Substitute the orbital radii.

\[ \frac{T_B}{3.0} = \left( \frac{1.8\times10^7} {9.0\times10^6} \right)^{3/2} \]

\[ \frac{T_B}{3.0}=2^{3/2} \]

\[ \frac{T_B}{3.0}=2.828 \]

Step 3: Calculate the period.

\[ T_B=3.0(2.828) \]

\[ \boxed{T_B=8.5\text{ h}} \]

Satellite B takes approximately 8.5 hours per orbit.

A common mistake is to double the period because the radius has doubled. That would only be correct if the orbital speed remained constant. In a circular orbit around the same planet, the larger orbit also has a lower orbital speed.

Question 2

Planet P has a mass \(M\), while planet Q has a mass \(9M\).

A small satellite A, with mass \(m\), orbits planet P in a circular orbit of radius \(r\). Its orbital period is \(12\text{ h}\).

A second satellite B, with mass \(100m\), orbits planet Q in a circular orbit of the same radius \(r\).

Calculate satellite B’s orbital period.

Explain why the different satellite masses do not affect your answer.

Solution 2

Satellite B takes \(4.0\text{ h}\) to complete one orbit.

Although both satellites have the same orbital radius, they orbit different central bodies. The central masses therefore do not cancel.

Step 1: Identify the relevant relationship.

At a fixed orbital radius, the period is inversely proportional to the square root of the central mass:

\[ T\propto\frac{1}{\sqrt{M}} \]

Therefore:

\[ \frac{T_B}{T_A} = \sqrt{\frac{M_P}{M_Q}} \]

Step 2: Substitute the central masses.

\[ \frac{T_B}{12} = \sqrt{\frac{M}{9M}} \]

The common mass factor cancels:

\[ \frac{T_B}{12} = \frac{1}{3} \]

Step 3: Calculate the period.

\[ T_B=\frac{12}{3} \]

\[ \boxed{T_B=4.0\text{ h}} \]

Satellite B completes an orbit in four hours.

Why doesn’t its mass matter?

The satellite’s mass appears in both the gravitational force and the centripetal force required for circular motion. It cancels when those forces are equated.

The central mass is different, however, so it does not cancel between the two systems.

This is an important distinction: changing the mass of the orbiting satellite does not change its circular orbital period in this model, but changing the mass of the central body does.

Question 3

A spherical planet has a radius of \(6.0\times10^6\text{ m}\).

Satellite A orbits the planet in a circular path at an altitude of \(2.0\times10^6\text{ m}\) above its surface.

Satellite B also orbits the same planet in a circular path. Its orbital period is 3.375 times the period of satellite A.

Calculate the altitude of satellite B above the planet’s surface.

Solution 3

Satellite B is at an altitude of:

\[ \boxed{1.2\times10^7\text{ m}} \]

The main challenge is working backwards from the period ratio to the orbital-radius ratio, then converting the result into an altitude.

Step 1: Calculate satellite A’s orbital radius.

The orbital radius is measured from the planet’s centre.

\[ r_A=R+h_A \]

\[ r_A=6.0\times10^6+2.0\times10^6 \]

\[ r_A=8.0\times10^6\text{ m} \]

Step 2: Rearrange the period comparison equation.

We are given the period ratio:

\[ \frac{T_B}{T_A}=3.375 \]

Use:

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

Substitute:

\[ \frac{r_B}{r_A}=3.375^{2/3} \]

Since \(3.375=27/8\):

\[ \frac{r_B}{r_A} = \left(\frac{27}{8}\right)^{2/3} \]

\[ \frac{r_B}{r_A}=\frac{9}{4}=2.25 \]

Step 3: Calculate satellite B’s orbital radius.

\[ r_B=2.25r_A \]

\[ r_B=2.25(8.0\times10^6) \]

\[ r_B=1.8\times10^7\text{ m} \]

Step 4: Convert orbital radius into altitude.

Rearrange:

\[ h_B=r_B-R \]

Substitute:

\[ h_B=1.8\times10^7-6.0\times10^6 \]

\[ \boxed{h_B=1.2\times10^7\text{ m}} \]

The altitude of satellite B is \(12\,000\text{ km}\).

The tempting mistake is to calculate a radius ratio and treat it as an altitude ratio. They are different measurements. The planet’s radius must be subtracted to obtain the altitude.

Question 4

Two different planets, P and Q, each have a small probe in a circular orbit.

The following measurements are recorded:

Answer the following:

(a) Explain why the mass-free comparison for two satellites orbiting the same central body cannot be applied directly.

(b) Determine the ratio of the central masses \(M_Q/M_P\).

(c) A second probe, C, moves in a circular orbit around planet Q at a radius of \(4.00\times10^7\text{ m}\). Calculate its orbital period.

Solution 4

(a) The mass-free comparison cannot be applied directly because the probes orbit different central bodies, which have different masses.

(b) Planet Q has twice the mass of planet P:

\[ \boxed{\frac{M_Q}{M_P}=2.00} \]

(c) Probe C has an orbital period of:

\[ \boxed{T_C=22.6\text{ h}} \]

Let’s work through the reasoning.

Step 1: Identify why the central mass does not cancel.

For probe A:

\[ T_A=2\pi\sqrt{\frac{r_A^3}{GM_P}} \]

For probe B:

\[ T_B=2\pi\sqrt{\frac{r_B^3}{GM_Q}} \]

Because the probes orbit different planets, we cannot cancel \(M_P\) against \(M_Q\).

Dividing the equations gives:

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

Step 2: Substitute the measurements.

\[ \frac{8.00}{4.00} = \left( \frac{2.00\times10^7} {1.00\times10^7} \right)^{3/2} \sqrt{\frac{M_P}{M_Q}} \]

Simplifying:

\[ 2.00=2^{3/2}\sqrt{\frac{M_P}{M_Q}} \]

\[ 2.00=2.828\sqrt{\frac{M_P}{M_Q}} \]

Step 3: Rearrange to find the mass ratio.

\[ \sqrt{\frac{M_P}{M_Q}} = \frac{2.00}{2.828} \]

\[ \sqrt{\frac{M_P}{M_Q}} =0.7071 \]

Squaring both sides:

\[ \frac{M_P}{M_Q}=0.500 \]

Therefore:

\[ \boxed{M_Q=2.00M_P} \]

Planet Q has twice the mass of planet P.

Notice why simply comparing the orbital radii would have given the wrong conclusion. If both planets had equal masses, probe B would have a period approximately 2.83 times that of probe A, not twice as long.

The fact that its measured period is only twice as long tells us that planet Q is more massive.

Step 4: Calculate the period of probe C.

Probes B and C orbit the same planet, Q. We can now use the mass-free comparison.

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

Substitute:

\[ \frac{T_C}{8.00} = \left( \frac{4.00\times10^7} {2.00\times10^7} \right)^{3/2} \]

\[ \frac{T_C}{8.00}=2^{3/2} \]

\[ T_C=8.00(2.828) \]

\[ \boxed{T_C=22.6\text{ h}} \]

Probe C takes approximately 22.6 hours per orbit.

This question illustrates an important strategy: when the central mass is unknown but common to two objects, use a ratio. When the central masses differ, additional information is needed.

Question 5

Two small probes, A and B, orbit the same spherical planet in circular paths.

Their measured orbital periods are:

\[ T_A=2.00\text{ h} \]

\[ T_B=5.657\text{ h} \]

At one instant, the probes are positioned on opposite sides of the planet, along a straight line passing through its centre.

Their centre-to-centre separation is measured to be:

\[ d=9.00\times10^6\text{ m} \]

Assume the probes have negligible mass compared with the planet and that the measured periods are accurate.

(a) Determine the ratio of their orbital radii.

(b) Calculate the individual orbital radii of A and B.

(c) Calculate the mass of the planet.

(d) Explain why the period ratio alone would not have been sufficient to determine the planet’s mass.

Use:

\[ G=6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2} \]

Solution 5

The orbital radii are:

\[ \boxed{r_A=3.00\times10^6\text{ m}} \]

\[ \boxed{r_B=6.00\times10^6\text{ m}} \]

The planet’s mass is:

\[ \boxed{M=3.08\times10^{23}\text{ kg}} \]

The key is to combine the period ratio with the geometric information. The period ratio gives the relative orbital radii, while the measured separation supplies the missing absolute distance.

Step 1: Calculate the ratio of the orbital periods.

\[ \frac{T_B}{T_A} = \frac{5.657}{2.00} \]

\[ \frac{T_B}{T_A}=2.8285 \]

Step 2: Determine the ratio of the orbital radii.

Both probes orbit the same planet in circular paths, so:

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

Substitute:

\[ \frac{r_B}{r_A} = (2.8285)^{2/3} \]

\[ \boxed{\frac{r_B}{r_A}\approx2.00} \]

Therefore:

\[ r_B=2r_A \]

At this stage, we know the relative sizes of the orbits, but not their actual radii.

Step 3: Use the measured separation to find the individual radii.

This is where the geometry matters.

The probes are on opposite sides of the planet, along the same straight line through its centre.

Therefore, the distance between them is the sum of their orbital radii:

\[ d=r_A+r_B \]

We know:

\[ r_B=2r_A \]

Substitute:

\[ 9.00\times10^6=r_A+2r_A \]

\[ 9.00\times10^6=3r_A \]

Rearrange:

\[ r_A=\frac{9.00\times10^6}{3} \]

\[ \boxed{r_A=3.00\times10^6\text{ m}} \]

Now calculate the radius of B:

\[ r_B=2(3.00\times10^6) \]

\[ \boxed{r_B=6.00\times10^6\text{ m}} \]

Step 4: Calculate the planet’s mass.

We now have an absolute orbital radius and an absolute orbital period.

Return to the circular-orbit equation:

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

Rearrange to make \(M\) the subject:

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

Step 5: Convert satellite A’s period into seconds.

\[ T_A=2.00\times3600 \]

\[ T_A=7200\text{ s} \]

Step 6: Substitute the known values.

Using satellite A:

\[ M= \frac{4\pi^2(3.00\times10^6)^3} {(6.67\times10^{-11})(7200)^2} \]

Simplifying:

\[ M= \frac{1.066\times10^{21}} {3.458\times10^{-3}} \]

\[ \boxed{M=3.08\times10^{23}\text{ kg}} \]

Step 7: Explain why the period ratio alone was insufficient.

The period ratio gave us:

\[ \frac{r_B}{r_A}=2 \]

This tells us that B’s orbital radius is twice A’s, but it doesn’t tell us the actual size of either orbit.

For example, orbital radii of \(3\text{ million m}\) and \(6\text{ million m}\) have the same ratio as orbital radii of \(30\text{ million m}\) and \(60\text{ million m}\).

The period ratio alone cannot distinguish between these absolute scales.

The measured separation provided the missing length measurement. Once we knew an actual orbital radius and an actual orbital period, we could calculate the planet’s mass.

The tempting mistake is to use the measured separation as one satellite’s orbital radius. It isn’t. Because the probes are on opposite sides of the planet, their separation is the sum of the two radii.

This is the deeper lesson: cancelling the central mass makes relative comparisons possible, but it also removes the information needed to determine that mass. You need an additional absolute measurement to recover it.

06Where to go next

You now have two distinct tools for orbital-period questions.

When two objects orbit the same central body in circular paths, compare their periods directly using:

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

When a question asks for an actual orbital period or central mass, you need additional information about the central body’s gravitational influence.

The next useful step is connecting orbital period to orbital velocity. Once you understand how the period changes with radius, you can predict how quickly a satellite must travel to maintain its circular orbit.