Why Total Energy Is Negative for a Bound Circular Orbit

A satellite can be travelling at several kilometres per second, so its kinetic energy is clearly positive. Yet when we calculate its total mechanical energy in a circular orbit, we get a negative number.

How can a moving object have negative total energy?

The answer is that gravitational potential energy is negative when we choose infinity as our zero-energy reference. For a circular orbit, its magnitude is exactly twice the satellite’s kinetic energy. When we add the two energies, the negative potential energy wins:

\[ E=K+U=-K<0 \]

The negative result tells us something important: the satellite is gravitationally bound. It does not have enough total mechanical energy to escape to infinity without receiving additional energy.

Let’s build that result from the physical situation rather than treating the minus sign as something to memorise.

01What does negative gravitational energy actually mean?

Start by separating two different types of mechanical energy.

A satellite has kinetic energy because it is moving:

\[ K=\frac{1}{2}mv^2 \]

Here:

• \(K\) is kinetic energy, measured in joules (J).
• \(m\) is the satellite’s mass, in kilograms (kg).
• \(v\) is its speed, in metres per second (\(\mathrm{m\,s^{-1}}\)).

Kinetic energy is never negative. The speed is squared, so \(K\geq 0\).

The satellite also has gravitational potential energy because it is interacting with the planet.

For two objects separated by a distance \(r\), gravitational potential energy is

\[ U=-\frac{GMm}{r} \]

where:

• \(U\) is gravitational potential energy, in joules.
• \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\) is the universal gravitational constant.
• \(M\) is the mass of the planet, in kilograms.
• \(m\) is the satellite’s mass, in kilograms.
• \(r\) is the centre-to-centre separation, in metres.

Why is there a minus sign?

Choosing zero at infinity

Gravitational potential energy needs a reference point. We choose the potential energy of the satellite-planet system to be zero when the objects are infinitely far apart:

\[ U(\infty)=0 \]

Now imagine bringing the satellite in from infinity towards Earth.

Gravity attracts the satellite, so gravity does positive work as the satellite moves closer. Its gravitational potential energy decreases.

At any finite separation, its potential energy is therefore below zero:

\[ U=-\frac{GMm}{r}<0 \]

The negative sign is a consequence of our chosen reference and the attractive nature of gravity.

It does not mean that the satellite has negative kinetic energy, or that energy has somehow disappeared.

In fact, the negative potential energy tells us how much energy must be supplied to separate the satellite and planet to infinity, if the satellite starts from rest at that position.

The total mechanical energy is the sum of kinetic and gravitational potential energy:

\[ E=K+U \]

Therefore, for an object at any position and speed,

\[ \boxed{ E=\frac{1}{2}mv^2-\frac{GMm}{r} } \]

This is the general energy equation for an isolated Newtonian gravitational system, treating the planet as much more massive than the satellite.

Keep this equation in mind. It will become particularly useful when we compare circular and non-circular motion.

02Why does a circular orbit produce negative total energy?

We now need one extra piece of information: the satellite cannot travel at just any speed and maintain a circular orbit of a fixed radius.

At a particular radius, there is exactly one required circular orbital speed.

Think about a satellite travelling around Earth. Its velocity points along the tangent to its circular path, while gravity pulls it towards Earth’s centre.

Gravity provides the centripetal force that continually changes the direction of the satellite’s velocity.

For a circular orbit:

\[ F_g=F_c \]

Substituting the gravitational force and centripetal force equations gives

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

Cancel the satellite mass and rearrange:

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

Therefore, the circular orbital speed is

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

Here, \(v_c\) represents the speed required to maintain a circular orbit at radius \(r\).

If you need to refresh why gravity supplies the centripetal force, see Gravity as Centripetal Force in Circular Orbits for HSC Physics.

The important substitution

We already know that kinetic energy is

\[ K=\frac{1}{2}mv^2 \]

But for a circular orbit, we can substitute \(v^2=GM/r\):

\[ K=\frac{1}{2}m\left(\frac{GM}{r}\right) \]

So,

\[ K=\frac{GMm}{2r} \]

Now compare this with gravitational potential energy:

\[ U=-\frac{GMm}{r} \]

Notice that the magnitude of the potential energy is twice the kinetic energy:

\[ U=-2K \]

This is the crucial relationship.

The satellite has positive kinetic energy, but its negative gravitational potential energy is twice as large in magnitude.

Adding them gives:

\[ \begin{aligned} E&=K+U\\ &=K-2K\\ &=-K \end{aligned} \]

Therefore,

\[ \boxed{E=-\frac{GMm}{2r}} \]

This is the total mechanical energy of a satellite in a circular gravitational orbit.

The minus sign is unavoidable because the circular-orbit condition fixes the kinetic energy at exactly half the magnitude of the gravitational potential energy.

Keep the three energy values separate

There is a useful extra relationship:

\[ E=-K \]

The total energy of a circular orbit is equal to the negative of its kinetic energy.

That does not mean kinetic energy is being removed. Both quantities remain constant in an ideal circular orbit.

Gravity continually changes the direction of the satellite’s velocity, but because the gravitational force is perpendicular to its instantaneous velocity, it does no work on the satellite.

03How negative energy tells us whether a satellite can escape

The negative total energy becomes much easier to interpret when we compare it with escape speed.

Imagine giving a satellite enough energy to travel infinitely far away from Earth.

At infinity, our chosen reference gives

\[ U=0 \]

If the satellite just manages to escape, its speed approaches zero at infinity. Therefore,

\[ K\rightarrow0 \]

The total mechanical energy at this boundary is

\[ E=0 \]

This gives us a useful classification for an isolated Newtonian gravitational system.

A negative total energy means the satellite and planet form a gravitationally bound system.

The satellite can continue moving, but it cannot reach infinity without receiving additional energy.

Compare circular speed with escape speed

At a particular radius, escape speed is found by setting the total energy equal to zero:

\[ 0=\frac{1}{2}mv_e^2-\frac{GMm}{r} \]

Rearranging:

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

Therefore,

\[ v_e=\sqrt{\frac{2GM}{r}} \]

Compare this with circular orbital speed:

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

We find:

\[ v_e=\sqrt{2}\,v_c \]

So escape speed is approximately 1.414 times the circular orbital speed at the same radius.

A satellite in a circular orbit is moving more slowly than the escape speed. It therefore has negative total energy.

A common misconception: you need to add the entire magnitude of potential energy to escape

A student might argue that because gravitational potential energy is

\[ U=-\frac{GMm}{r} \]

we must add \(GMm/r\) joules to escape.

That is not correct for a satellite already in a circular orbit.

The satellite already has positive kinetic energy:

\[ K=\frac{GMm}{2r} \]

Its total energy is

\[ E=-\frac{GMm}{2r} \]

To reach the escape boundary, its total energy must increase from this negative value to zero.

Therefore, the minimum energy increase is

\[ \Delta E=0-E \]

\[ \boxed{\Delta E=\frac{GMm}{2r}} \]

This is only half the magnitude of its gravitational potential energy.

The satellite does not need to replace all of its negative potential energy because it already has kinetic energy.

That is an important distinction between the energy required to escape from rest at a given radius and the energy required to escape from an existing circular orbit.

05How can we check the prediction using satellite measurements?

The negative total energy is a prediction of Newtonian gravitational mechanics. It is not a number that a sensor directly reads from a satellite.

Instead, we can reconstruct the satellite’s position and velocity, then calculate its mechanical energy.

Consider a simplified radio-tracking arrangement.

The measurement process works as follows:

1. A ground station transmits a radio signal towards the satellite.
2. The satellite’s transponder returns a signal to the ground station.
3. The round-trip travel time provides the distance between the station and satellite, using the speed of light.
4. The Doppler shift provides information about the satellite’s velocity component along the line of sight.
5. Repeated measurements as the satellite moves allow its position and full velocity to be reconstructed.

There is an important limitation here.

A single Doppler measurement only gives the velocity component along the line of sight. It does not establish the satellite’s entire velocity vector, so one measurement cannot prove that the orbit is circular.

The improved approach is to track the satellite repeatedly as its position relative to the ground station changes. This provides enough information to reconstruct its orbital motion and test whether its radius and speed are consistent with a circular orbit.

An illustrative set of tracking data

Suppose repeated tracking measurements produce the following orbital information.

These are illustrative values, not measurements from a particular satellite.

We can first check the predicted circular orbital speed:

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

Substituting gives approximately

\[ v_c=7.542\times10^3\ \mathrm{m\,s^{-1}} \]

This agrees with the reconstructed speed.

Now calculate the total energy per kilogram. This is called the specific mechanical energy:

\[ \epsilon=\frac{E}{m} \]

Using the general energy equation:

\[ \epsilon=\frac{1}{2}v^2-\frac{GM}{r} \]

Substituting:

\[ \begin{aligned} \epsilon &=\frac{1}{2}(7.542\times10^3)^2\\ &\quad-\frac{(6.67\times10^{-11})(5.97\times10^{24})} {7.00\times10^6} \end{aligned} \]

Therefore,

\[ \epsilon\approx-2.84\times10^7\ \mathrm{J\,kg^{-1}} \]

The calculated specific energy is negative.

The reconstructed speed also agrees with the circular-orbit prediction, supporting the circular-orbit model.

This is how the theoretical relationship connects to measurement: the tracking system provides position and velocity, and the Newtonian model tells us what those measurements imply about the satellite’s energy.

06Does negative total energy always mean circular motion?

No. This is an important distinction.

A negative total mechanical energy indicates that an object is gravitationally bound, assuming the isolated Newtonian model and zero potential energy at infinity.

It does not automatically mean the orbit is circular.

For an object at any position, the general energy equation remains

\[ E=\frac{1}{2}mv^2-\frac{GMm}{r} \]

The circular-orbit expression

\[ E=-\frac{GMm}{2r} \]

is obtained only when the circular-motion condition is satisfied.

An object can have negative total energy while travelling in an elliptical orbit. Its speed and distance from the planet change throughout the orbit, but its total mechanical energy remains constant if no non-conservative forces do work.

This distinction matters in exam questions.

If a question gives you an object moving at a particular radius and speed, do not immediately use the circular-orbit formula. First establish whether the motion is actually circular.

07Questions and solutions

The following questions move from straightforward energy calculations to situations where you need to decide which energy relationship is actually valid.

Question 1

A \(300\ \mathrm{kg}\) satellite travels in a circular orbit around a planet of mass \(4.00\times10^{24}\ \mathrm{kg}\). Its orbital radius is \(8.00\times10^6\ \mathrm{m}\).

Using \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\), calculate:

a) Its kinetic energy.

b) Its gravitational potential energy, taking zero potential energy at infinity.

c) Its total mechanical energy.

Solution 1

The total mechanical energy is \(-5.00\times10^9\ \mathrm{J}\). The gravitational potential energy has twice the magnitude of the kinetic energy, leaving a negative total.

Step 1: Calculate the kinetic energy.

For a circular orbit:

\[ K=\frac{GMm}{2r} \]

Substitute:

\[ K= \frac{(6.67\times10^{-11})(4.00\times10^{24})(300)} {2(8.00\times10^6)} \]

\[ \boxed{K=5.00\times10^9\ \mathrm{J}} \]

Step 2: Calculate gravitational potential energy.

\[ U=-\frac{GMm}{r} \]

\[ U= -\frac{(6.67\times10^{-11})(4.00\times10^{24})(300)} {8.00\times10^6} \]

\[ \boxed{U=-1.00\times10^{10}\ \mathrm{J}} \]

Step 3: Calculate total mechanical energy.

\[ E=K+U \]

\[ E=5.00\times10^9-1.00\times10^{10} \]

\[ \boxed{E=-5.00\times10^9\ \mathrm{J}} \]

The satellite is gravitationally bound because its total energy is negative.

Question 2

Two satellites, A and B, travel in the same circular orbit around a planet of mass \(4.80\times10^{24}\ \mathrm{kg}\).

Their masses are:

• Satellite A: \(200\ \mathrm{kg}\)
• Satellite B: \(600\ \mathrm{kg}\)

Both have an orbital radius of \(1.20\times10^7\ \mathrm{m}\).

A student argues that satellite B must travel faster because it experiences three times the gravitational force.

Evaluate this claim.

Calculate the orbital speed of each satellite and compare their kinetic energies, gravitational potential energies, and total mechanical energies. Explain how the different masses affect the energy required for each satellite to escape.

Solution 2

Both satellites have the same circular orbital speed, but satellite B has three times the magnitude of each energy term and requires three times as much additional energy to escape.

Step 1: Calculate the circular orbital speed.

The circular orbital speed is independent of the satellite’s mass:

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

Substitute:

\[ v_c= \sqrt{ \frac{(6.67\times10^{-11})(4.80\times10^{24})} {1.20\times10^7} } \]

\[ \boxed{v_c=5.17\times10^3\ \mathrm{m\,s^{-1}}} \]

Therefore,

\[ v_A=v_B=5.17\times10^3\ \mathrm{m\,s^{-1}} \]

Step 2: Calculate the energy of satellite A.

For satellite A, \(m_A=200\ \mathrm{kg}\).

\[ K_A=\frac{GMm_A}{2r} \]

\[ K_A= \frac{(6.67\times10^{-11})(4.80\times10^{24})(200)} {2(1.20\times10^7)} \]

\[ K_A=2.67\times10^9\ \mathrm{J} \]

Its gravitational potential energy is

\[ U_A=-\frac{GMm_A}{r} \]

\[ U_A=-5.34\times10^9\ \mathrm{J} \]

Therefore,

\[ E_A=K_A+U_A \]

\[ \boxed{E_A=-2.67\times10^9\ \mathrm{J}} \]

Step 3: Calculate the energy of satellite B.

Satellite B has three times the mass of satellite A.

Since all three energy expressions are directly proportional to satellite mass:

\[ K_B=3K_A \]

\[ K_B=8.00\times10^9\ \mathrm{J} \]

Similarly,

\[ U_B=3U_A \]

\[ U_B=-1.60\times10^{10}\ \mathrm{J} \]

And

\[ E_B=3E_A \]

\[ \boxed{E_B=-8.00\times10^9\ \mathrm{J}} \]

Step 4: Compare the energy required to escape.

The minimum energy increase required to escape is the magnitude of the original total mechanical energy.

For satellite A:

\[ \Delta E_A=2.67\times10^9\ \mathrm{J} \]

For satellite B:

\[ \Delta E_B=8.00\times10^9\ \mathrm{J} \]

Satellite B requires three times as much additional energy.

Step 5: Explain why the student’s prediction is incorrect.

The gravitational force is

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

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

However, its inertia is also three times greater.

Its gravitational acceleration is

\[ g=\frac{F_g}{m}=\frac{GM}{r^2} \]

The satellite mass cancels.

Therefore, both satellites require the same circular orbital speed, even though their gravitational forces and total energies are different.

Question 3

A \(500\ \mathrm{kg}\) probe is travelling tangentially around a planet of mass \(5.00\times10^{24}\ \mathrm{kg}\).

Its distance from the planet’s centre is \(9.00\times10^6\ \mathrm{m}\), and its speed is \(6.00\times10^3\ \mathrm{m\,s^{-1}}\).

Assume an ideal gravitational system with no additional forces.

a) Calculate the probe’s kinetic energy, gravitational potential energy, and total mechanical energy.

b) Determine whether the probe is gravitationally bound.

c) Is it in a circular orbit? Justify your answer using the appropriate orbital speeds.

d) Calculate the minimum additional energy needed for the probe to escape to infinity with zero final speed.

Use \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).

Solution 3

The probe has negative total mechanical energy of approximately \(-9.53\times10^9\ \mathrm{J}\), so it is bound, but it is not in a circular orbit.

Step 1: Calculate kinetic energy.

\[ K=\frac{1}{2}mv^2 \]

\[ K=\frac{1}{2}(500)(6.00\times10^3)^2 \]

\[ \boxed{K=9.00\times10^9\ \mathrm{J}} \]

Step 2: Calculate gravitational potential energy.

\[ U=-\frac{GMm}{r} \]

\[ U= -\frac{(6.67\times10^{-11})(5.00\times10^{24})(500)} {9.00\times10^6} \]

\[ \boxed{U=-1.85\times10^{10}\ \mathrm{J}} \]

Step 3: Calculate total mechanical energy.

\[ E=K+U \]

\[ E=9.00\times10^9-1.85\times10^{10} \]

\[ \boxed{E=-9.53\times10^9\ \mathrm{J}} \]

Since \(E<0\), the probe is gravitationally bound.

Step 4: Determine whether its orbit is circular.

Calculate the circular orbital speed:

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

\[ v_c= \sqrt{ \frac{(6.67\times10^{-11})(5.00\times10^{24})} {9.00\times10^6} } \]

\[ v_c=6.09\times10^3\ \mathrm{m\,s^{-1}} \]

The probe is travelling at

\[ v=6.00\times10^3\ \mathrm{m\,s^{-1}} \]

This is slightly below the required circular orbital speed.

Therefore, the orbit is not circular.

Because its velocity is tangential and below circular orbital speed, the probe is at the furthest point of its elliptical orbit, known as its apocentre.

Step 5: Calculate the minimum additional energy required to escape.

For escape with zero final speed at infinity:

\[ E_f=0 \]

Therefore,

\[ \Delta E=0-E_i \]

\[ \boxed{\Delta E=9.53\times10^9\ \mathrm{J}} \]

The negative initial energy tells us exactly how much additional mechanical energy is needed to reach the escape boundary.

Question 4

A satellite initially travels in a circular orbit of radius \(r\) around a planet of mass \(M\).

It has mass \(m\) and circular orbital speed \(v_c\).

A short retrograde engine burn reduces its speed to \(0.80v_c\). The burn occurs tangentially, opposite to the satellite’s original direction of travel.

Assume:

• The satellite’s position does not change during the short burn.
• Its velocity remains tangential immediately after the burn.
• There is no atmospheric drag or further engine thrust.
• The planet’s mass remains constant.

a) Express the satellite’s initial total mechanical energy in terms of \(GMm/r\).

b) Calculate its kinetic energy immediately after the burn in terms of \(GMm/r\).

c) Calculate its new total mechanical energy.

d) Determine the change in total mechanical energy.

e) Explain whether the satellite remains bound and describe what happens to its motion immediately after the burn.

Solution 4

The satellite remains gravitationally bound, and its new total mechanical energy is

\[ \boxed{E'=-0.68\frac{GMm}{r}} \]

The burn removes mechanical energy, making the total energy more negative.

Step 1: Find the initial total energy.

For a circular orbit:

\[ E_i=-\frac{GMm}{2r} \]

Therefore,

\[ E_i=-0.50\frac{GMm}{r} \]

Step 2: Calculate the new kinetic energy.

The new speed is

\[ v'=0.80v_c \]

Therefore,

\[ K'=\frac{1}{2}m(0.80v_c)^2 \]

\[ K'=0.64\left(\frac{1}{2}mv_c^2\right) \]

Since the original circular kinetic energy is

\[ K_i=\frac{GMm}{2r} \]

we obtain

\[ K'=0.64\left(\frac{GMm}{2r}\right) \]

Hence,

\[ \boxed{K'=0.32\frac{GMm}{r}} \]

Step 3: Calculate the new total mechanical energy.

The satellite has not changed position during the burn, so its gravitational potential energy is unchanged:

\[ U'=-\frac{GMm}{r} \]

Therefore,

\[ E'=K'+U' \]

\[ E'=0.32\frac{GMm}{r}-\frac{GMm}{r} \]

\[ \boxed{E'=-0.68\frac{GMm}{r}} \]

Step 4: Calculate the change in total mechanical energy.

\[ \Delta E=E'-E_i \]

Substitute:

\[ \Delta E= -0.68\frac{GMm}{r} -\left(-0.50\frac{GMm}{r}\right) \]

Therefore,

\[ \boxed{\Delta E=-0.18\frac{GMm}{r}} \]

The negative result means the engine burn has reduced the satellite’s total mechanical energy.

This makes sense because the satellite loses kinetic energy while its gravitational potential energy remains unchanged during the brief burn.

Step 5: Determine the satellite’s new motion.

The new total energy is still negative:

\[ E'<0 \]

Therefore, the satellite remains gravitationally bound.

It is also no longer in a circular orbit because its new speed is below the circular orbital speed at that radius.

Immediately after the burn, its radial velocity is zero, but gravity produces a greater inward acceleration than would be needed for circular motion at its reduced speed.

The satellite therefore begins moving towards the planet.

Its current position becomes the apocentre of a new elliptical orbit.

This question demonstrates why you must distinguish between an instantaneous change in kinetic energy and the subsequent motion of the entire gravitational system.

Question 5

Two identical satellites, A and B, are located at the same distance \(r\) from the centre of a planet of mass \(M\).

Both satellites have mass \(m\), and both have the same speed:

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

However, their velocity directions are different.

• Satellite A travels exactly tangentially to its circular path.
• Satellite B travels at an angle of \(30^\circ\) away from the local tangent, towards the outward radial direction.

Neither satellite receives any further thrust, and the only force acting on either satellite is the planet’s gravitational attraction.

A student makes the following argument:

Both satellites have the circular orbital speed at the same radius. Therefore, both must have the same circular orbit and the same total mechanical energy.

Evaluate the student’s argument.

In your response:

a) Determine the total mechanical energy of each satellite at the stated instant.

b) Explain whether each satellite is in a circular orbit.

c) Resolve the velocity of satellite B into tangential and radial components.

d) Explain why the two satellites can have identical total mechanical energy but follow different orbital paths.

Solution 5

The student’s argument is incorrect. Both satellites have the same negative total mechanical energy, but only satellite A follows a circular orbit. Satellite B has a radial component of velocity, so its distance from the planet changes immediately.

This is a subtle but important distinction: total mechanical energy depends on the speed and position, but not on the direction of the velocity.

Step 1: Calculate the total energy of satellite A.

The general energy equation is

\[ E=\frac{1}{2}mv^2-\frac{GMm}{r} \]

Substitute the given speed:

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

Therefore,

\[ E_A= \frac{1}{2}m\left(\frac{GM}{r}\right) -\frac{GMm}{r} \]

Simplifying:

\[ \boxed{E_A=-\frac{GMm}{2r}} \]

Satellite A has negative total mechanical energy.

Because its velocity is exactly tangential and its speed is the required circular orbital speed, it follows a circular orbit.

Step 2: Calculate the total energy of satellite B.

Satellite B has the same mass, position, and speed magnitude as satellite A.

Although its velocity points in a different direction, kinetic energy depends on the square of the speed magnitude:

\[ K=\frac{1}{2}mv^2 \]

Therefore,

\[ E_B= \frac{1}{2}m\left(\frac{GM}{r}\right) -\frac{GMm}{r} \]

Hence,

\[ \boxed{E_B=-\frac{GMm}{2r}} \]

Both satellites have identical total mechanical energy.

Step 3: Resolve satellite B’s velocity.

Satellite B’s velocity is directed \(30^\circ\) away from the tangent.

Its tangential component is

\[ v_t=v\cos30^\circ \]

Therefore,

\[ v_t=\frac{\sqrt{3}}{2}v \]

Its outward radial component is

\[ v_r=v\sin30^\circ \]

Therefore,

\[ v_r=\frac{1}{2}v \]

Substituting the given speed:

\[ v_t=\frac{\sqrt{3}}{2}\sqrt{\frac{GM}{r}} \]

\[ v_r=\frac{1}{2}\sqrt{\frac{GM}{r}} \]

The important result is that satellite B has a non-zero radial velocity.

Step 4: Determine why the orbital paths are different.

Satellite A has no radial velocity.

Its distance from the planet remains constant, and gravity provides exactly the centripetal acceleration required for circular motion.

Satellite B is different.

Its outward radial velocity means its distance from the planet is already increasing at the instant described.

It cannot be following a circular path.

Its angular momentum is also different.

Angular momentum for motion about the planet’s centre is

\[ L=mrv_t \]

For satellite A:

\[ L_A=mrv \]

For satellite B:

\[ L_B=mrv\cos30^\circ \]

Therefore,

\[ L_B=\frac{\sqrt{3}}{2}L_A \]

Satellite B has less angular momentum, even though both satellites have the same total mechanical energy.

Under the ideal Newtonian gravitational model, satellite B follows a bound elliptical orbit rather than a circular one.

Step 5: Identify the mistake in the original argument.

The tempting conclusion is that having the circular orbital speed automatically guarantees circular motion.

But that speed only produces a circular orbit when the entire velocity is tangential.

Satellite B has the same speed magnitude but a different velocity direction. Its radial component changes its distance from the planet, while its smaller tangential component changes its angular momentum.

Both satellites are gravitationally bound because

\[ E<0 \]

But only satellite A is in a circular orbit.

This is the deeper lesson: negative total energy tells you whether the system is gravitationally bound, not the exact shape of its orbit.

The key relationship to carry into HSC Physics is

\[ \boxed{E=-\frac{GMm}{2r}} \]

but remember where it comes from: the gravitational potential energy is negative, and the circular orbital condition fixes kinetic energy at half its magnitude.

When you encounter an unfamiliar orbital problem, start with the general equation

\[ E=\frac{1}{2}mv^2-\frac{GMm}{r} \]

Then establish whether the orbit is actually circular before simplifying it.

The next useful connection is how total mechanical energy changes when a satellite moves between two circular orbital radii. See Total Energy in Circular Orbits for HSC Physics for that comparison.