Why Satellite Mass Doesn’t Affect Circular Orbital Speed

Put two satellites into the same circular orbit around Earth. One has a mass of \(100\text{ kg}\), and the other has a mass of \(1000\text{ kg}\). Earth’s gravity pulls ten times harder on the heavier satellite. Shouldn’t that make it orbit faster?

No. For an ideal circular orbit, satellite mass does not affect orbital speed. Both satellites must travel at the same speed if they orbit the same central mass at the same orbital radius.

The heavier satellite experiences a greater gravitational force, but it also has greater inertia, meaning it takes more force to change its velocity. These two effects increase in exactly the same proportion.

01A lunar experiment explains the first half of the puzzle

Imagine dropping a hammer and a feather from the same height.

On Earth, the hammer reaches the ground first. It’s tempting to conclude that heavier objects naturally accelerate faster under gravity.

But there’s a problem with that experiment: air resistance.

The feather has relatively little mass compared with the area it presents to the air. Air resistance has a much greater effect on its motion than on the hammer’s motion.

How could we remove that complication?

Apollo 15: Removing air resistance

In 1971, during the Apollo 15 mission, astronaut David Scott demonstrated this principle on the Moon.

He held a geological hammer and a feather at approximately the same height, around \(1.6\text{ m}\) above the lunar surface.

He released them simultaneously.

What should happen if heavier objects accelerate faster under gravity?

The hammer should arrive first.

Instead, the hammer and feather struck the lunar surface at essentially the same time, within the accuracy of the demonstration. NASA’s Apollo 15 Preliminary Science Report records this observation.

The Moon has almost no atmosphere, so air resistance was negligible. Removing that complication allowed the difference between gravitational force and gravitational acceleration to become clear.

The hammer experiences a greater gravitational force because it has more mass. However, its greater inertia means that the larger force produces the same gravitational acceleration as the feather experiences.

We can even estimate the falling time.

Assume both objects start from rest at height \(h=1.6\text{ m}\), experience constant lunar gravitational acceleration \(g_{\mathrm{Moon}}=1.62\text{ m s}^{-2}\), and have negligible air resistance.

Using \(h=\frac12gt^2\), where \(t\) is falling time in seconds,

\[ \begin{aligned} t&=\sqrt{\frac{2h}{g_{\mathrm{Moon}}}}\\ &=\sqrt{\frac{3.2}{1.62}}\\ &\approx1.4\text{ s}. \end{aligned} \]

Both objects have the same predicted falling time. Their masses never enter the calculation.

This is an approximate theoretical estimate, not a measured Apollo 15 falling time. The demonstration was also not a precision test of gravitational theory.

Most importantly, dropping objects vertically does not directly measure satellite orbital speed. It demonstrates the mass-independent gravitational acceleration that helps explain orbital motion.

02From falling objects to orbiting satellites

Long before Apollo 15, Isaac Newton had considered a related problem.

Imagine a cannon firing a projectile horizontally from an extremely high mountain. Ignore air resistance, and assume Earth is spherical.

With a relatively low launch speed, the projectile curves downwards and hits the ground.

Increase its horizontal speed, and it travels farther before landing.

At a particular speed, its downward motion matches the curvature of Earth’s surface. It continues falling towards Earth but never reaches the ground.

Newton’s cannon was a thought experiment, not apparatus that he actually built.

Now imagine firing two projectiles with the same initial position and velocity, but very different masses. Would the heavier projectile follow a different path?

Under the ideal assumptions, no. Both experience the same gravitational acceleration, so their trajectories coincide.

A satellite is essentially a projectile that keeps falling around its central body.

There’s an important limit to this simple picture. Falling at the same rate does not mean that any sideways speed produces a circular orbit. At a particular orbital radius, only one speed produces the required circular motion.

That is where centripetal acceleration comes in.

03Why greater gravitational force doesn’t mean greater orbital speed

A satellite travelling around Earth continually changes the direction of its velocity.

Even when its speed is constant, its velocity is changing because velocity includes direction.

Changing velocity requires acceleration. For circular motion, this acceleration points towards the centre of the circle.

It is called centripetal acceleration.

Gravity provides this inward acceleration for an ideal satellite orbit.

The two equations that explain the cancellation

Newton’s law of universal gravitation gives the magnitude of the gravitational force between a spherical planet and an external satellite:

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

Here:

• \(F_g\) is the gravitational force in newtons (N).
• \(G=6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\) is the universal gravitational constant.
• \(M\) is the mass of the central body in kilograms (kg).
• \(m\) is the satellite’s mass in kilograms (kg).
• \(r\) is the distance from the planet’s centre to the satellite in metres (m).

Notice that gravitational force is proportional to satellite mass.

Doubling \(m\), while keeping \(M\) and \(r\) constant, doubles \(F_g\).

But what force is needed to keep the satellite moving in a circle?

For circular motion,

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

Here, \(F_c\) is the required net inward or centripetal force in newtons, and \(v\) is the satellite’s orbital speed in metres per second (\(\text{m s}^{-1}\)).

The same satellite mass \(m\) appears in this equation.

A satellite with twice the mass needs twice the inward force to maintain the same circular speed at the same radius.

This is similar to pushing a loaded supermarket trolley. Doubling the trolley’s mass means you need twice the net force to produce the same acceleration.

The difference is that gravity automatically provides a larger force to a more massive satellite. A person pushing a trolley doesn’t automatically double their effort, and trolley friction has no direct equivalent in the ideal orbital model.

Setting gravity equal to the required centripetal force

For an ideal circular gravitational orbit, gravity is the only force and provides the required centripetal force.

Therefore,

\[ F_g=F_c \]

Substituting the two expressions,

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

The satellite mass \(m\) appears on both sides.

Because \(m\) is non-zero, we can divide both sides by it:

\[ \frac{GM}{r^2}=\frac{v^2}{r} \]

Multiplying by \(r\) gives

\[ v^2=\frac{GM}{r} \]

Taking the positive square root, because speed is a non-negative magnitude,

\[ v=\sqrt{\frac{GM}{r}} \]

This is the circular orbital-speed equation.

The satellite’s mass has disappeared. The speed depends on the central mass \(M\) and orbital radius \(r\), not the satellite’s own mass.

All forces and accelerations in this derivation are positive magnitudes. Their physical direction is radially inward, towards the planet’s centre. The satellite’s instantaneous velocity is perpendicular to that direction.

The equation assumes a spherical or spherically symmetric central body, a circular orbit, negligible air resistance and propulsion, and a satellite whose mass is much smaller than the central body’s mass. We’ll return to that last assumption.

For a broader treatment of the equation and its applications, see Orbital Velocity: Derivation and Calculations for HSC Physics.

04What actually changes when satellite mass increases?

The important distinction is between force, acceleration, and speed.

Suppose two satellites have the same circular orbital radius and orbit the same planet.

If one satellite has ten times the mass of the other:

• Its gravitational force is ten times greater.
• Its required centripetal force is ten times greater.
• Its gravitational acceleration is unchanged.
• Its circular orbital speed is unchanged.

We can see why by dividing gravitational force by mass.

From Newton’s second law,

\[ a=\frac{F_g}{m} \]

Substitute the gravitational force equation:

\[ \begin{aligned} a&=\frac{GMm}{mr^2}\\ &=\frac{GM}{r^2}. \end{aligned} \]

Here, \(a\) is the gravitational acceleration in \(\text{m s}^{-2}\).

The mass cancels again.

A heavier satellite is not pulled more strongly per kilogram. It experiences the same gravitational acceleration as a lighter satellite at the same position.

05How to recognise when mass independence applies

The quickest approach in an HSC question is to identify exactly what has changed.

These are comparisons between ideal circular orbits. They do not describe what necessarily happens immediately after a spacecraft fires its engines or collides with another object.

Three situations where the simple conclusion needs care

1. Air resistance acts on the spacecraft.

In a sufficiently low orbit, Earth’s thin upper atmosphere produces drag.

Two satellites at the same altitude may experience different drag accelerations because drag depends on properties such as cross-sectional area, shape, speed, atmospheric density, and mass.

If equal drag forces act on a light and heavy satellite, the light satellite experiences the greater deceleration.

This does not mean the lighter satellite requires a different circular orbital speed at the same radius under gravity alone.

It means the real motion is no longer an ideal gravity-only circular orbit.

2. A spacecraft changes mass during a manoeuvre or collision.

Suppose a spacecraft receives a payload.

If the payload docks while matching the spacecraft’s position and velocity, the combined spacecraft can continue at the same orbital speed.

But if the payload arrives with a different velocity, the docking process can change the spacecraft’s momentum and speed.

The resulting change is caused by the interaction during docking, not by the satellite mass suddenly appearing in the circular orbital-speed equation.

If the new velocity does not match the circular speed required at that radius, the trajectory will no longer be circular.

3. The orbiting object’s mass is not negligible compared with the central mass.

The usual equation assumes that the central body remains effectively fixed.

This is an excellent approximation for an artificial satellite orbiting Earth.

However, when two gravitating bodies have comparable masses, both move around their common centre of mass.

The more complete two-body model depends on both masses, and we must distinguish the speed of one body around the common centre of mass from the speed of the bodies relative to each other.

We’ll use that distinction in the final challenge.

06Questions and solutions

Question 1

Two satellites travel in circular orbits of radius \(8.00\times10^6\text{ m}\) around a spherical planet of mass \(3.00\times10^{24}\text{ kg}\).

Satellite A has a mass of \(40.0\text{ kg}\), and satellite B has a mass of \(120\text{ kg}\).

Assume ideal circular motion, with \(G=6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\).

Calculate:

a) The orbital speed of each satellite.

b) The gravitational force acting on each satellite.

Solution 1

Both satellites travel at approximately \(5.00\times10^3\text{ m s}^{-1}\), but satellite B experiences three times the gravitational force acting on A.

For part (a), circular orbital speed is

\[ v=\sqrt{\frac{GM}{r}}. \]

Substitute the planet’s mass and orbital radius:

\[ \begin{aligned} GM&=(6.67\times10^{-11})\\ &\quad\times(3.00\times10^{24})\\ &=2.001\times10^{14} \text{ m}^3\text{s}^{-2}. \end{aligned} \]

Therefore,

\[ \begin{aligned} v&=\sqrt{\frac{2.001\times10^{14}} {8.00\times10^6}}\\ &=5.00\times10^3\text{ m s}^{-1}. \end{aligned} \]

This applies to both satellites because satellite mass is absent from the equation.

For part (b), first calculate the gravitational acceleration:

\[ \begin{aligned} a&=\frac{GM}{r^2}\\ &=\frac{2.001\times10^{14}} {(8.00\times10^6)^2}\\ &=3.13\text{ m s}^{-2}. \end{aligned} \]

For satellite A,

\[ \begin{aligned} F_A&=m_Aa\\ &=(40.0)(3.12656)\\ &=125\text{ N}. \end{aligned} \]

For satellite B,

\[ \begin{aligned} F_B&=m_Ba\\ &=(120)(3.12656)\\ &=375\text{ N}. \end{aligned} \]

Satellite B has three times the mass and therefore experiences three times the gravitational force.

The tempting mistake is to assume that the greater gravitational force produces a greater orbital speed. It doesn’t, because B also has three times the inertia. The same gravitational acceleration acts on both.

Question 2

Two satellites are travelling in the same ideal circular orbit around a spherical planet.

The planet’s mass is \(6.00\times10^{24}\text{ kg}\), and the orbital radius is \(1.00\times10^7\text{ m}\).

Satellite A has a mass of \(80.0\text{ kg}\), while satellite B has a mass of \(800\text{ kg}\).

Each satellite contains an ideal accelerometer fixed inside its cabin. Both accelerometers read approximately zero, despite the satellites continuously changing direction.

a) Calculate the gravitational acceleration experienced by each satellite.

b) Calculate the gravitational force acting on each.

c) Explain how both accelerometers can read zero while the satellites experience non-zero gravitational forces and continuously accelerate towards the planet.

Assume no drag, thrust, or significant tidal effects across either spacecraft.

Solution 2

Both satellites experience an inward gravitational acceleration of approximately \(4.00\text{ m s}^{-2}\), and the heavier satellite experiences ten times the gravitational force. Their onboard accelerometers can nevertheless read zero because they measure acceleration caused by non-gravitational forces, not the gravitational acceleration of a freely falling spacecraft.

For part (a),

\[ a=\frac{GM}{r^2}. \]

Substituting,

\[ \begin{aligned} a&=\frac{(6.67\times10^{-11}) (6.00\times10^{24})} {(1.00\times10^7)^2}\\ &=4.002\text{ m s}^{-2}. \end{aligned} \]

Both satellites accelerate towards the planet at approximately \(4.00\text{ m s}^{-2}\).

For part (b), use \(F_g=ma\).

Satellite A:

\[ \begin{aligned} F_A&=(80.0)(4.002)\\ &=3.20\times10^2\text{ N}. \end{aligned} \]

Satellite B:

\[ \begin{aligned} F_B&=(800)(4.002)\\ &=3.20\times10^3\text{ N}. \end{aligned} \]

The gravitational force on B is ten times greater.

For part (c), the distinction is between gravitational acceleration and the acceleration measured by an accelerometer in free fall.

An accelerometer fixed inside a satellite measures the acceleration associated with forces such as a push from the floor, engine thrust, or aerodynamic drag.

In ideal free fall, the instrument and its surrounding spacecraft accelerate together under gravity.

There is no supporting surface force required to keep the instrument moving with the spacecraft. Consequently, an ideal accelerometer reads zero.

But an observer tracking the satellite relative to the planet sees its velocity continually changing direction. Its gravitational acceleration is non-zero.

The misconception is that a zero accelerometer reading means gravity has disappeared.

Gravity is present and provides the inward acceleration required for the orbit. The spacecraft is weightless in the sense of experiencing no support force, not in the sense of being unaffected by gravity.

Question 3

Two satellites, A and B, initially travel in the same circular orbit around a spherical planet.

Their masses are:

• Satellite A: \(100\text{ kg}\)
• Satellite B: \(400\text{ kg}\)

They have identical external shapes and cross-sectional areas.

Both encounter a thin atmospheric region that initially exerts a drag force of \(2.0\text{ N}\), opposite their direction of motion.

For the next \(10.0\text{ s}\), assume the drag force remains approximately constant, and differences in position and gravitational acceleration are negligible when estimating their relative changes in tangential velocity.

a) Calculate the drag-induced tangential acceleration of each satellite, taking the direction of motion as positive.

b) Estimate which satellite has the greater speed after the \(10.0\text{ s}\) interval, and by how much.

Later, after separate orbital manoeuvres, both spacecraft are again in ideal circular orbits without atmospheric drag. A is at radius \(0.81r_0\), while B is at radius \(r_0\), where \(r_0\) is their original orbital radius.

c) Which satellite now has the greater circular orbital speed? Calculate the ratio \(v_A/v_B\).

d) Explain why the speed ranking can reverse without contradicting the mass independence of ideal circular orbital speed.

Solution 3

Satellite B is approximately \(0.15\text{ m s}^{-1}\) faster after the brief drag interval. However, in the later circular orbits, A travels about \(11.1\%\) faster than B because A has the smaller orbital radius.

For part (a), Newton’s second law gives tangential acceleration due to drag:

\[ a_t=-\frac{F_d}{m}, \]

where \(F_d\) is the positive magnitude of drag force and the negative sign indicates that drag acts against the motion.

For satellite A,

\[ \begin{aligned} a_{t,A}&=-\frac{2.0}{100}\\ &=-0.020\text{ m s}^{-2}. \end{aligned} \]

For satellite B,

\[ \begin{aligned} a_{t,B}&=-\frac{2.0}{400}\\ &=-0.0050\text{ m s}^{-2}. \end{aligned} \]

The lighter satellite experiences four times the drag-induced deceleration.

For part (b), using the short-interval approximation \(\Delta v_t=a_t\Delta t\),

\[ \begin{aligned} \Delta v_A&=(-0.020)(10.0)\\ &=-0.20\text{ m s}^{-1},\\ \Delta v_B&=(-0.0050)(10.0)\\ &=-0.050\text{ m s}^{-1}. \end{aligned} \]

Since both started at the same orbital speed, B is approximately

\[ 0.20-0.050=0.15\text{ m s}^{-1} \]

faster after this brief interval.

This is an estimate of the difference caused by drag, not an exact description of their changing orbital trajectories.

For part (c), the spacecraft are now in different circular orbits.

Circular orbital speed is

\[ v=\sqrt{\frac{GM}{r}}. \]

For the same central planet,

\[ \begin{aligned} \frac{v_A}{v_B} &=\sqrt{\frac{r_B}{r_A}}\\ &=\sqrt{\frac{r_0}{0.81r_0}}\\ &=\frac{1}{0.9}\\ &\approx1.11. \end{aligned} \]

A is now approximately \(11.1\%\) faster than B.

For part (d), there are two different physical situations.

During the initial drag interval, both satellites are being slowed by an external force. Their motion is not ideal circular motion, and their different masses produce different drag accelerations.

In the later comparison, both have returned to ideal circular motion, but their orbital radii are different.

A is faster because its circular orbital radius is smaller, not because it has less mass.

The tempting route is to treat a slowing force as proof that every later orbital speed must be smaller. That ignores the change of orbit.

Drag initially reduces velocity and orbital energy. A spacecraft subsequently moving in a lower circular orbit has a greater circular orbital speed, even though reaching that orbit required a loss of orbital energy. The final circularisation manoeuvres specified in the question make the two endpoint orbits well-defined.

Mass independence still applies at each fixed circular radius.

Question 4

A \(600\text{ kg}\) satellite travels in an ideal circular orbit around a spherical planet.

The planet has mass \(5.40\times10^{24}\text{ kg}\), and the satellite’s orbital radius is \(1.00\times10^7\text{ m}\).

A \(200\text{ kg}\) cargo module is being held at rest relative to the planet’s centre by its own engines, at a position directly in the satellite’s path.

The satellite catches the module, and they lock together in a perfectly inelastic docking event.

Assume:

• The docking occurs in a negligibly short time.
• External gravitational impulse during docking is negligible.
• The cargo module’s engines switch off.
• Tangential momentum is conserved during the collision.
• The subsequent motion is gravity-only, and the trajectory does not intersect the planet’s surface.

a) Calculate the circular orbital speed immediately before docking.

b) Calculate the speed of the combined \(800\text{ kg}\) spacecraft immediately after docking.

c) Determine whether the combined spacecraft remains in the original circular orbit.

d) Explain whether the docking point becomes the closest or furthest point from the planet on its new trajectory.

Solution 4

The satellite initially travels at approximately \(6.00\times10^3\text{ m s}^{-1}\). Docking reduces its speed to approximately \(4.50\times10^3\text{ m s}^{-1}\), so the combined spacecraft leaves the original circular orbit and begins an elliptical trajectory. The docking point is the furthest point from the planet on that new orbit.

For part (a),

\[ v_c=\sqrt{\frac{GM}{r}}. \]

First calculate \(GM\):

\[ \begin{aligned} GM&=(6.67\times10^{-11})\\ &\quad\times(5.40\times10^{24})\\ &=3.6018\times10^{14} \text{ m}^3\text{s}^{-2}. \end{aligned} \]

Therefore,

\[ \begin{aligned} v_c&=\sqrt{\frac{3.6018\times10^{14}} {1.00\times10^7}}\\ &=6.00\times10^3\text{ m s}^{-1}. \end{aligned} \]

For part (b), the docking event changes the spacecraft’s momentum.

Let \(v_f\) be the combined spacecraft’s speed immediately after capture.

Tangential momentum conservation gives

\[ m_sv_c+m_pv_p=(m_s+m_p)v_f, \]

where \(m_s=600\text{ kg}\) is the original spacecraft mass, \(m_p=200\text{ kg}\) is the payload mass, and the payload’s initial tangential velocity is \(v_p=0\).

Thus,

\[ \begin{aligned} v_f&=\frac{600(6.00\times10^3)} {600+200}\\ &=4.50\times10^3\text{ m s}^{-1}. \end{aligned} \]

The combined spacecraft has a smaller speed than the original satellite.

For part (c), the important comparison is not between the old and new masses. It is between the actual post-docking speed and the speed required for a circular orbit at the same radius.

Immediately after docking, the orbital radius is still \(1.00\times10^7\text{ m}\).

The circular orbital speed required at that radius is still approximately \(6.00\times10^3\text{ m s}^{-1}\).

The combined spacecraft is travelling too slowly to maintain that circle.

We can check this using acceleration.

Gravity provides

\[ \begin{aligned} a_g&=\frac{GM}{r^2}\\ &=\frac{3.6018\times10^{14}} {(1.00\times10^7)^2}\\ &=3.60\text{ m s}^{-2}. \end{aligned} \]

At the actual post-docking speed, a circle of the original radius would require

\[ \begin{aligned} a_c&=\frac{v_f^2}{r}\\ &=\frac{(4.50\times10^3)^2} {1.00\times10^7}\\ &=2.03\text{ m s}^{-2}. \end{aligned} \]

The gravitational acceleration is greater than the acceleration required to follow the original circle at this reduced speed.

The spacecraft therefore begins moving inward rather than following the original circular path.

For part (d), immediately after docking, the velocity is purely tangential. Its radial velocity is zero.

Because it is travelling below the required circular speed, its distance from the planet begins decreasing.

That makes the docking position the furthest point, or apocentre, of the new elliptical orbit.

The tempting argument is that doubling or otherwise increasing satellite mass cannot change orbital speed because satellite mass cancels from the orbital equation.

But that equation describes the speed required for an existing ideal circular orbit. It does not say that a collision must preserve the spacecraft’s velocity.

The cargo module arrived with a different velocity. Capturing it changed the system’s momentum distribution, and the resulting speed no longer satisfied the original circular-orbit condition.

If the cargo had matched the satellite’s velocity before docking, the conclusion would have been different.

Question 5

Extension challenge: When the central body cannot be treated as fixed

A student argues:

“Satellite mass never affects circular orbital speed because it cancels from Newton’s equations.”

Another student replies:

“That cannot be exactly true when the orbiting object is massive enough to move the planet itself.”

Consider an isolated system containing a spherical planet of mass \(M\) and a moon of mass \(m=M/4\).

The two bodies move in circular paths around their common centre of mass, maintaining a constant centre-to-centre separation \(r\).

Both bodies have the same angular speed, \(\omega\), measured in radians per second.

Let:

• \(d_M\) be the planet’s distance from the common centre of mass.
• \(d_m\) be the moon’s distance from the common centre of mass.

You are given

\[ Md_M=md_m \]

and

\[ d_M+d_m=r. \]

For comparison, imagine a satellite of negligible mass moving around a fixed planet of mass \(M\) at circular orbital radius \(r\).

Its speed would be

\[ v_0=\sqrt{\frac{GM}{r}}. \]

Determine:

a) The distances \(d_M\) and \(d_m\) in terms of \(r\).

b) The speed of the moon relative to the planet, expressed as a multiple of \(v_0\).

c) The moon’s speed around the common centre of mass, in the inertial frame where that centre is stationary, expressed as a multiple of \(v_0\).

d) Whether the moon’s relative speed can be greater than \(v_0\) while its speed around the common centre of mass is less than \(v_0\).

e) Exactly which assumption in the original mass-cancellation argument fails, and why the two different speed comparisons give different results.

Solution 5

Both students have identified part of the truth, but the statement that satellite mass never matters is too broad.

The moon’s speed relative to the planet is approximately \(1.118v_0\), while its speed around the common centre of mass is approximately \(0.894v_0\).

Both results are correct because they describe different motions and use different reference points.

Part (a): Locate the common centre of mass.

The two bodies are separated by distance \(r\), and

\[ Md_M=md_m. \]

Since \(m=M/4\),

\[ Md_M=\frac{M}{4}d_m. \]

Cancelling \(M\),

\[ d_m=4d_M. \]

Using the total separation,

\[ \begin{aligned} d_M+d_m&=r\\ d_M+4d_M&=r\\ 5d_M&=r. \end{aligned} \]

Therefore,

\[ d_M=\frac r5 \]

and

\[ d_m=\frac{4r}{5}. \]

The common centre of mass lies closer to the more massive planet.

Part (b): Find the moon’s speed relative to the planet.

The gravitational force magnitude between the bodies is

\[ F_g=\frac{GMm}{r^2}. \]

That force provides the centripetal force for the moon as it moves around the common centre of mass.

Importantly, the radius of the moon’s circular path is \(d_m\), not \(r\).

Therefore,

\[ \frac{GMm}{r^2}=m\omega^2d_m. \]

Cancel \(m\):

\[ \frac{GM}{r^2}=\omega^2d_m. \]

Substituting \(d_m=4r/5\),

\[ \begin{aligned} \frac{GM}{r^2} &=\omega^2\frac{4r}{5}\\ \omega^2&=\frac{5GM}{4r^3}. \end{aligned} \]

The two bodies rotate with the same angular speed \(\omega\) and remain separated by \(r\).

Their relative tangential speed is therefore

\[ v_{\mathrm{rel}}=\omega r. \]

Hence,

\[ \begin{aligned} v_{\mathrm{rel}} &=\sqrt{\frac{5GM}{4r^3}}\,r\\ &=\sqrt{\frac54}\sqrt{\frac{GM}{r}}\\ &=\sqrt{\frac54}\,v_0\\ &\approx1.118v_0. \end{aligned} \]

The moon’s speed relative to the planet is approximately \(11.8\%\) greater than the negligible-mass satellite’s circular speed around a fixed planet of mass \(M\) at separation \(r\).

Part (c): Find the moon’s speed around the common centre of mass.

The moon follows a circle of radius \(d_m=4r/5\).

Its speed in the centre-of-mass frame is

\[ v_m=\omega d_m. \]

Therefore,

\[ \begin{aligned} v_m&=\omega\frac{4r}{5}\\ &=\frac45v_{\mathrm{rel}}\\ &=\frac45\sqrt{\frac54}\,v_0\\ &\approx0.894v_0. \end{aligned} \]

The moon’s speed around the common centre of mass is approximately \(10.6\%\) smaller than \(v_0\).

Part (d): Explain why the two answers can point in opposite directions.

The relative speed describes how quickly the moon moves around the planet when their separation is considered.

Because the planet is also moving, their relative motion combines the motions of both bodies.

The moon’s speed in the centre-of-mass frame is different. It measures only the moon’s motion around the fixed common centre of mass.

In this frame, the moon moves on a circle of radius \(4r/5\), while the planet moves on a smaller circle of radius \(r/5\).

Both bodies have the same angular speed, but their linear speeds differ because their circular path radii differ.

The moon’s relative speed exceeds \(v_0\) because both masses contribute to the gravitational two-body motion.

Yet its individual speed around the common centre of mass is below \(v_0\), because its circular path has a smaller radius than the full separation \(r\).

The two statements are not contradictory.

Part (e): Identify the failed assumption.

The original circular-speed derivation used

\[ \frac{GMm}{r^2}=\frac{mv^2}{r}. \]

It treated the gravitational separation \(r\) and the satellite’s circular path radius as the same distance.

That is appropriate when the central body can be treated as fixed and the satellite mass is negligible.

But in this two-body problem, the planet also moves.

The gravitational separation is \(r\), while the moon’s radius around the common centre of mass is only \(4r/5\).

Using \(r\) on both sides of the original equation would therefore describe the wrong geometry.

The correct centripetal equation is

\[ \frac{GMm}{r^2}=m\omega^2d_m. \]

The mass \(m\) still cancels explicitly, but \(d_m\) itself depends on the mass ratio. Substituting the correct centre-of-mass distance introduces the combined mass dependence.

For two bodies in a circular gravitational orbit, the relative speed is

\[ v_{\mathrm{rel}} =\sqrt{\frac{G(M+m)}{r}}. \]

This is a two-body result, where \(r\) is the centre-to-centre separation and \(v_{\mathrm{rel}}\) is the relative speed, not necessarily the speed of either body in an inertial frame.

When \(m\) is negligible compared with \(M\),

\[ M+m\approx M, \]

and the familiar result is recovered:

\[ v\approx\sqrt{\frac{GM}{r}}. \]

This is why the ordinary HSC satellite model works so well for artificial satellites around planets.

The fair conclusion is not that satellite mass can never matter. It is that satellite mass does not affect ideal circular orbital speed when the central mass and orbital radius are fixed and the satellite is sufficiently light that the central body can be treated as stationary.

07The decision to make in an orbital-speed question

When a question changes a satellite’s mass, don’t immediately change its orbital speed.

First decide whether the satellite is still moving freely in an ideal circular orbit at the same radius.

If the central mass and orbital radius are unchanged, and the satellite remains a negligible-mass test object, its circular orbital speed is unchanged. A heavier satellite experiences a greater gravitational force, but exactly the same gravitational acceleration.

If a collision, engine burn, or atmospheric drag changes its velocity, the motion may stop being circular. If the orbital radius changes, the required circular speed changes. And if both gravitating bodies have substantial mass, their motion around the common centre of mass must be considered.

There’s one further consequence of mass cancellation. Two ideal satellites at the same circular radius also have the same orbital period, because each travels the same circumference at the same speed.

That connection between circular speed, orbital radius, and the time taken for one revolution is developed in Orbital Period: Radius and Central Mass for HSC Physics.