Why Adding Energy Can Reduce a Satellite’s Circular Speed

A satellite can gain orbital mechanical energy and still finish moving more slowly. The key is that the extra energy does not all become kinetic energy. Some of it increases the satellite’s gravitational potential energy, making it less negative.

For a circular orbit, the larger the orbital radius, the lower the speed required to maintain that orbit. So when a satellite moves into a higher circular orbit, its final speed decreases even though its total mechanical energy increases.

Consider a satellite moving from a circular orbit of radius \(7.00\times10^6\,\mathrm{m}\) to one of radius \(1.40\times10^7\,\mathrm{m}\), measured from Earth’s centre. Its circular speed falls from approximately \(7.55\,\mathrm{km\,s^{-1}}\) to \(5.34\,\mathrm{km\,s^{-1}}\).

Yet its total orbital energy per kilogram increases from \(-28.5\,\mathrm{MJ\,kg^{-1}}\) to \(-14.25\,\mathrm{MJ\,kg^{-1}}\).

How can both statements be true?

The answer is that kinetic energy and total mechanical energy are different quantities. Understanding the difference also explains why a rocket can accelerate a satellite twice during an orbital transfer, yet leave it travelling more slowly than when it started.

01A model investigation: does adding energy always increase speed?

Imagine testing this situation using a computer model of a satellite orbiting Earth.

The model places Earth at the centre of the coordinate system. A satellite begins on a circular path at a radius of \(7.00\times10^6\,\mathrm{m}\). Its velocity is tangential to the orbit.

A simulated rocket engine then applies a short thrust in the direction the satellite is travelling. The satellite moves onto an elliptical transfer trajectory, eventually reaching a distance of \(1.40\times10^7\,\mathrm{m}\) from Earth’s centre.

At this furthest point, a second short thrust adjusts its speed so that it enters a circular orbit.

The model tracks the satellite’s position, speed, kinetic energy, gravitational potential energy, and total mechanical energy.

This is an idealised numerical investigation, not a record of a particular satellite mission.

What should the first test show?

Suppose you only measure the satellite’s speed immediately before and after the first engine burn.

You should observe an increase in speed. The engine does positive work on the satellite, increasing its kinetic energy.

A student might conclude that the satellite must therefore finish its journey moving faster.

But this first test has a limitation: it only measures the immediate effect of the engine. It doesn’t measure what happens while the satellite travels towards its higher orbit.

To answer the original question, the investigation needs to track the entire transfer and compare the initial and final circular orbits.

What does the complete investigation show?

The satellite passes through several distinct stages.

The important observation is that the satellite slows down during the outward coast, even though its total mechanical energy stays constant during that part of the journey.

The second engine burn then increases its speed again. However, the final circular speed is still lower than the original circular speed.

This is the apparent contradiction we need to explain.

01Start with gravitational potential energy

The first thing to understand is why gravitational potential energy is negative.

We choose gravitational potential energy to be zero when two objects are infinitely far apart. This is our reference point.

For a satellite of mass \(m\), orbiting a planet of mass \(M\), gravitational potential energy is:

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

where:

• \(U\) is gravitational potential energy in joules (\(\mathrm{J}\)).
• \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\).
• \(M\) is the planet’s mass in kilograms.
• \(m\) is the satellite’s mass in kilograms.
• \(r\) is the distance between the centres of the two objects in metres.

The negative sign is essential.

A satellite in a gravitationally bound orbit has negative gravitational potential energy. As it moves further away from Earth, its potential energy increases because it becomes less negative.

For example:

\[ -60\,\mathrm{MJ}\longrightarrow -30\,\mathrm{MJ} \]

This is an increase of \(30\,\mathrm{MJ}\), even though both values are negative.

Think of a cyclist travelling uphill. As the cyclist climbs, some kinetic energy is converted into gravitational potential energy. The cyclist slows down even though the total mechanical energy remains constant.

The analogy helps explain the energy transfer during the satellite’s outward coast. However, a satellite isn’t travelling along a physical hill. Its changing speed is also governed by the condition required to maintain an orbit.

What does negative total energy mean?

A satellite’s total mechanical energy is the sum of its kinetic and gravitational potential energies:

\[ E=K+U \]

where kinetic energy is:

\[ K=\frac12 mv^2 \]

Kinetic energy is positive, while gravitational potential energy is negative.

For an ideal satellite under the influence of gravity alone:

• \(E<0\): the satellite is gravitationally bound.
• \(E=0\): the satellite is at the parabolic escape threshold.
• \(E>0\): the satellite has enough energy to escape with some speed remaining at infinity.

A negative total energy does not mean the satellite has negative kinetic energy. It means its negative gravitational potential energy has a greater magnitude than its positive kinetic energy.

A satellite can therefore gain total mechanical energy while remaining gravitationally bound.

02Why higher circular orbits have lower speeds

Now we need to connect orbital radius to orbital speed.

Imagine a satellite travelling in a circular orbit around Earth. Gravity provides the centripetal force that continually changes the direction of its velocity.

The gravitational force is:

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

The centripetal force required for circular motion is:

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

For a circular orbit, these forces are equal:

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

Cancel the satellite’s mass and rearrange:

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

Therefore, the circular orbital speed is:

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

Here, \(v_c\) is the circular orbital speed in metres per second.

Notice that the satellite’s mass cancels out. Two satellites at the same orbital radius have the same circular speed, regardless of their masses.

More importantly, the equation shows that:

\[ v_c\propto\frac{1}{\sqrt r} \]

As orbital radius increases, circular orbital speed decreases.

Why?

At a greater distance from Earth, the gravitational force is weaker. A satellite requires a lower speed to maintain a circular orbit at that larger radius.

What does this mean for kinetic energy?

Kinetic energy is:

\[ K=\frac12mv^2 \]

Substitute the circular orbital speed equation:

\[ K=\frac12m\left(\frac{GM}{r}\right) \]

Therefore:

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

As orbital radius increases, the satellite’s kinetic energy decreases.

But remember that its gravitational potential energy increases because it becomes less negative:

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

Now combine the two energy expressions.

\[ E=K+U \]

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

Therefore, for a circular orbit:

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

This equation is the key to understanding the original question.

A satellite in a larger circular orbit has:

• Lower orbital speed.
• Lower kinetic energy.
• Higher gravitational potential energy.
• Higher total mechanical energy.

The last two quantities increase even though the first two decrease.

For circular orbits, there is also a useful relationship:

\[ U=-2K \]

and therefore:

\[ E=-K \]

The increase in gravitational potential energy is twice the decrease in kinetic energy when comparing two circular orbital states.

That leaves a net increase in total mechanical energy.

03Why the satellite speeds up during the burns

There is one more important detail.

If adding energy can reduce a satellite’s final circular speed, why do rocket engines speed it up?

The answer is that the satellite’s speed changes throughout the transfer. We need to distinguish the immediate effect of an engine burn from the final state of the satellite.

Consider a transfer from a lower circular orbit to a higher circular orbit.

First burn: increase the speed

The satellite begins in a circular orbit.

Its engine fires in the direction of motion. This is called a prograde burn.

Because the burn happens over a short time, we approximate the satellite’s position as unchanged during the burn.

Therefore, its gravitational potential energy stays approximately constant.

The engine increases its kinetic energy and total mechanical energy.

However, the satellite is now moving too quickly to remain in its original circular orbit. It enters an elliptical transfer orbit.

Outward coast: speed decreases

The satellite continues travelling outwards with its engine switched off.

Gravity is now the only force doing work in our idealised model. Because gravity is conservative, total mechanical energy remains constant.

As the satellite moves further from Earth:

• Gravitational potential energy increases.
• Kinetic energy decreases.
• Total mechanical energy remains constant.
• Speed decreases.

The satellite is effectively converting kinetic energy into gravitational potential energy as it climbs away from Earth.

Second burn: circularise the orbit

Eventually, the satellite reaches the furthest point of its transfer trajectory.

It is travelling more slowly than the circular speed required at that radius.

A second prograde burn increases its speed until it matches the circular orbital speed at the new radius.

This increases its kinetic energy and total mechanical energy again.

However, the final circular speed can still be lower than the original circular speed.

This resolves the apparent contradiction.

Both engine burns increase the satellite’s speed at the instant they occur. The satellite then slows during its outward coast. Its second burn increases its speed again, but only to the lower circular speed required at its new, larger orbital radius.

Adding energy does not mean the satellite must have a greater speed at every later point.

The result depends on both its position and its motion.

An important distinction: circular orbit versus transfer orbit

You must also be careful about which equation you use.

For a circular orbit:

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

But this equation is not valid at an arbitrary point along an elliptical transfer orbit.

For a satellite at any position, its general mechanical energy is:

\[ \boxed{E=\frac12mv^2-\frac{GMm}{r}} \]

The general equation accounts for the satellite’s actual speed and position.

During an engine-off transfer, this total remains constant, even though the kinetic and potential energies change.

This distinction is particularly useful when a question gives you the satellite’s speed and distance from Earth but does not state that its orbit is circular.

04The contrasting case: removing energy can increase final speed

Now reverse the situation.

Suppose a satellite moves from a higher circular orbit into a lower circular orbit.

Its final orbital radius is smaller.

Using:

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

a smaller radius means a greater circular orbital speed.

At the same time, its total mechanical energy becomes more negative:

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

Therefore, the satellite can lose total mechanical energy while finishing with a higher speed.

For example, if its orbital radius is halved:

\[ \frac{v_f}{v_i}=\sqrt{\frac{r_i}{r_f}} \]

\[ =\sqrt{\frac{r_i}{0.5r_i}} \]

\[ =\sqrt{2} \]

Its final circular speed is approximately \(1.414\) times its initial speed.

Yet its total mechanical energy becomes twice as negative.

This is the reverse of the higher-orbit situation.

The general lesson is that orbital speed and total mechanical energy do not increase and decrease together in a simple way.

05A useful decision process for HSC questions

When you see a satellite energy question, first identify what the question is actually describing.

Two common mistakes are worth avoiding.

Mistake 1: Assuming a positive energy change means a positive final total energy.

A satellite can gain energy while remaining gravitationally bound. For example, changing its total energy from \(-30\,\mathrm{MJ}\) to \(-20\,\mathrm{MJ}\) is a positive energy change, even though the final energy remains negative.

Mistake 2: Using the circular-orbit energy equation during an elliptical transfer.

The equation \(E=-GMm/(2r)\) relies on the satellite being in a circular orbit. During an elliptical transfer, use the general mechanical-energy equation instead.

For a more detailed treatment of energy changes between orbital states, see Energy Changes When Satellites Change Orbit: HSC Physics.

09Questions and solutions

Question 1

A satellite moves from a circular orbit of radius \(8.00\times10^6\,\mathrm{m}\) to another circular orbit of radius \(1.20\times10^7\,\mathrm{m}\) around Earth.

Use:

\[ GM=3.99\times10^{14}\,\mathrm{m^3\,s^{-2}} \]

Calculate:

a) The initial circular orbital speed.

b) The final circular orbital speed.

c) The percentage decrease in speed.

Solution 1

The satellite’s final circular speed decreases by approximately \(18.4\%\), from \(7.06\,\mathrm{km\,s^{-1}}\) to \(5.77\,\mathrm{km\,s^{-1}}\).

For a circular orbit:

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

Part a: Initial speed

\[ \begin{aligned} v_i&=\sqrt{\frac{3.99\times10^{14}}{8.00\times10^6}}\\ &=7.06\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part b: Final speed

\[ \begin{aligned} v_f&=\sqrt{\frac{3.99\times10^{14}}{1.20\times10^7}}\\ &=5.77\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part c: Percentage decrease

Use:

\[ \%\text{ decrease}=\frac{v_i-v_f}{v_i}\times100 \]

Substitute:

\[ \begin{aligned} \%\text{ decrease} &=\frac{7062-5766}{7062}\times100\\ &=18.4\% \end{aligned} \]

The final speed is lower because the satellite is now in a larger circular orbit.

Notice that the orbital speed depends on the square root of the inverse radius, not directly on the satellite’s total mechanical energy.

Question 2

A \(750\,\mathrm{kg}\) satellite initially travels in a circular orbit of radius \(8.00\times10^6\,\mathrm{m}\) around Earth.

A brief prograde engine burn increases its speed by \(6.0\%\). Assume the satellite’s position does not change appreciably during the burn.

Use:

\[ GM=3.99\times10^{14}\,\mathrm{m^3\,s^{-2}} \]

Calculate:

a) The satellite’s speed immediately before and after the burn.

b) The change in its total orbital mechanical energy during the burn.

c) Whether the satellite remains gravitationally bound immediately after the burn.

Solution 2

The burn increases the satellite’s speed from approximately \(7.06\,\mathrm{km\,s^{-1}}\) to \(7.49\,\mathrm{km\,s^{-1}}\), adding approximately \(2.31\times10^9\,\mathrm{J}\) of orbital mechanical energy. The satellite remains gravitationally bound.

Part a: Calculate the speeds

The initial circular speed is:

\[ \begin{aligned} v_i&=\sqrt{\frac{3.99\times10^{14}}{8.00\times10^6}}\\ &=7.06\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

The burn increases the speed by \(6.0\%\):

\[ v_f=1.06v_i \]

\[ v_f=1.06(7.062\times10^3) \]

\[ v_f=7.49\times10^3\,\mathrm{m\,s^{-1}} \]

Part b: Calculate the change in mechanical energy

During the short burn, the satellite’s position is approximately constant.

Therefore, its gravitational potential energy does not change appreciably:

\[ \Delta U\approx0 \]

This means:

\[ \Delta E\approx\Delta K \]

Use:

\[ \Delta E=\frac12m(v_f^2-v_i^2) \]

Substitute:

\[ \begin{aligned} \Delta E &=\frac12(750) \left[(1.06v_i)^2-v_i^2\right]\\ &=\frac12(750)(7.062\times10^3)^2\\ &\quad\times[(1.06)^2-1]\\ &=2.31\times10^9\,\mathrm{J} \end{aligned} \]

The positive result means orbital mechanical energy has been added.

Part c: Determine whether the satellite remains bound

Calculate its total mechanical energy per kilogram immediately after the burn:

\[ \epsilon=\frac12v_f^2-\frac{GM}{r} \]

Substitute:

\[ \begin{aligned} \epsilon &=\frac12(7.486\times10^3)^2\\ &\quad-\frac{3.99\times10^{14}}{8.00\times10^6}\\ &=-2.19\times10^7\,\mathrm{J\,kg^{-1}} \end{aligned} \]

The result is negative.

Therefore, the satellite remains gravitationally bound.

The tempting mistake is to assume that adding energy must make the satellite escape. However, the satellite must gain enough energy to reach zero total mechanical energy before reaching the escape threshold.

Question 3

Using the satellite from Question 2, suppose it now switches off its engines and coasts outwards under gravity alone.

When it reaches a distance of \(1.00\times10^7\,\mathrm{m}\) from Earth’s centre:

a) Calculate its speed.

b) Calculate the circular orbital speed at that radius.

c) Explain why the two speeds are different.

Assume the satellite is still moving outwards and experiences no other forces.

Solution 3

The satellite’s speed at the larger radius is approximately \(6.01\,\mathrm{km\,s^{-1}}\), while the circular speed there is approximately \(6.32\,\mathrm{km\,s^{-1}}\). The satellite is not in a circular orbit at that point.

Part a: Calculate the speed during the outward coast

The satellite experiences no external work during the coast, so its total mechanical energy remains constant.

From Question 2, its specific mechanical energy is:

\[ \epsilon=-2.1855\times10^7\,\mathrm{J\,kg^{-1}} \]

Use the general energy equation per kilogram:

\[ \epsilon=\frac12v^2-\frac{GM}{r} \]

Rearrange:

\[ v=\sqrt{2\left(\epsilon+\frac{GM}{r}\right)} \]

Substitute the new radius:

\[ \begin{aligned} v&=\sqrt{2\left[ -2.1855\times10^7+ \frac{3.99\times10^{14}}{1.00\times10^7} \right]}\\ &=\sqrt{3.609\times10^7}\\ &=6.01\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part b: Calculate the circular speed

Use:

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

\[ \begin{aligned} v_c&=\sqrt{\frac{3.99\times10^{14}}{1.00\times10^7}}\\ &=6.32\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part c: Explain the difference

The satellite is moving outwards along an elliptical trajectory, not a circular one.

It therefore has a radial component of velocity and a tangential component that differs from the circular orbital speed.

The circular speed equation assumes that the satellite moves tangentially at exactly the speed needed to maintain a circular orbit.

The general energy equation, however, accounts for the satellite’s actual motion.

The important point is that knowing the satellite’s radius does not automatically mean it is in a circular orbit.

Question 4

A satellite is initially in a circular orbit around Earth at a radius of \(8.00\times10^6\,\mathrm{m}\).

An idealised tangential engine burn increases its speed to exactly \(\sqrt{2}\) times its original circular orbital speed.

a) Show that the satellite’s total mechanical energy becomes zero per kilogram.

b) Explain what this means physically.

c) If its final speed were instead \(1.01\) times the escape speed, calculate its specific mechanical energy and determine whether it is bound.

Use:

\[ GM=3.99\times10^{14}\,\mathrm{m^3\,s^{-2}} \]

Solution 4

At exactly \(\sqrt{2}\) times the circular speed, the satellite has zero total mechanical energy and is at the escape threshold. At \(1.01\) times escape speed, its specific mechanical energy is positive, so it is unbound.

Part a: Show that total energy is zero

The circular orbital speed is:

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

The escape-speed condition in this question gives:

\[ v=\sqrt2v_c \]

Squaring:

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

The total mechanical energy per kilogram is:

\[ \epsilon=\frac12v^2-\frac{GM}{r} \]

Substitute:

\[ \begin{aligned} \epsilon &=\frac12\left(\frac{2GM}{r}\right) -\frac{GM}{r}\\ &=0\,\mathrm{J\,kg^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{\epsilon=0} \]

Part b: Interpret the result

Zero total mechanical energy represents the ideal escape threshold.

The satellite has enough kinetic energy to reach infinite separation from Earth, with its speed approaching zero at infinity.

It is not gravitationally bound.

Importantly, zero total mechanical energy does not mean zero kinetic energy at the satellite’s current position.

Part c: Calculate the energy at \(1.01\) times escape speed

The escape speed is:

\[ v_{\mathrm{esc}}=\sqrt2v_c \]

The new speed is:

\[ v=1.01v_{\mathrm{esc}} \]

Therefore:

\[ v^2=(1.01)^2\frac{2GM}{r} \]

Substitute into the specific mechanical energy equation:

\[ \begin{aligned} \epsilon &=\frac12(1.01)^2\frac{2GM}{r} -\frac{GM}{r}\\ &=[(1.01)^2-1]\frac{GM}{r}\\ &=0.0201\left(\frac{3.99\times10^{14}} {8.00\times10^6}\right)\\ &=1.00\times10^6\,\mathrm{J\,kg^{-1}} \end{aligned} \]

The specific mechanical energy is positive.

Therefore, the satellite is unbound and will escape with some kinetic energy remaining at infinity.

This question highlights why the sign of total mechanical energy is useful: it tells you whether a satellite is gravitationally bound, even when its speed and kinetic energy are changing.

Question 5

Two satellites, A and B, initially travel in the same circular orbit around Earth.

Their initial orbital radius is:

\[ r_i=8.00\times10^6\,\mathrm{m} \]

Satellite A has a mass of \(500\,\mathrm{kg}\), while satellite B has a mass of \(1000\,\mathrm{kg}\).

Both satellites receive the same net increase in orbital mechanical energy:

\[ \Delta E=+2.00\times10^9\,\mathrm{J} \]

After their orbital manoeuvres, both are in circular orbits.

Assume the stated energy changes refer to the satellite-Earth orbital mechanical energy, and ignore atmospheric drag and other external influences.

A student argues:

“The circular orbital speed equation does not contain satellite mass. Both satellites start at the same radius and receive the same energy, so they must finish with the same circular speed.”

Another student argues:

“The lighter satellite receives more energy per kilogram, so it should finish in a higher orbit with a lower circular speed.”

Determine which student is correct.

Calculate the final orbital radius and circular speed of each satellite. Explain why the mass cancellation in the circular speed equation does not produce the result predicted by the first student.

Solution 5

The second student is correct. Satellite A finishes in a higher circular orbit and therefore has a lower final circular speed than satellite B.

The key is that both satellites receive the same total energy in joules, not the same energy per kilogram.

Step 1: Calculate their initial specific mechanical energy.

Specific mechanical energy is total mechanical energy divided by satellite mass:

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

For a circular orbit:

\[ \epsilon_i=-\frac{GM}{2r_i} \]

Substitute:

\[ \begin{aligned} \epsilon_i &=-\frac{3.99\times10^{14}} {2(8.00\times10^6)}\\ &=-2.49375\times10^7\,\mathrm{J\,kg^{-1}} \end{aligned} \]

Both satellites begin with the same specific mechanical energy because they have the same initial orbital radius.

Step 2: Calculate satellite A’s final specific energy.

Satellite A receives \(2.00\times10^9\,\mathrm{J}\).

Its energy increase per kilogram is:

\[ \Delta\epsilon_A=\frac{\Delta E}{m_A} \]

\[ \begin{aligned} \Delta\epsilon_A &=\frac{2.00\times10^9}{500}\\ &=4.00\times10^6\,\mathrm{J\,kg^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} \epsilon_{Af} &=\epsilon_i+\Delta\epsilon_A\\ &=-24.9375+4.00\\ &=-20.9375\,\mathrm{MJ\,kg^{-1}} \end{aligned} \]

Step 3: Calculate satellite A’s final radius.

Since the final orbit is circular:

\[ \epsilon_{Af}=-\frac{GM}{2r_{Af}} \]

Rearrange:

\[ r_{Af}=-\frac{GM}{2\epsilon_{Af}} \]

Substitute:

\[ \begin{aligned} r_{Af} &=\frac{3.99\times10^{14}} {2(20.9375\times10^6)}\\ &=9.53\times10^6\,\mathrm{m} \end{aligned} \]

Satellite A finishes in a larger orbit than its original orbit.

Step 4: Calculate satellite A’s final speed.

Use:

\[ v_{Af}=\sqrt{\frac{GM}{r_{Af}}} \]

\[ \begin{aligned} v_{Af} &=\sqrt{\frac{3.99\times10^{14}}{9.53\times10^6}}\\ &=6.47\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Step 5: Repeat the calculation for satellite B.

Satellite B has twice the mass, so the same energy increase gives it a smaller increase in energy per kilogram:

\[ \begin{aligned} \Delta\epsilon_B &=\frac{2.00\times10^9}{1000}\\ &=2.00\times10^6\,\mathrm{J\,kg^{-1}} \end{aligned} \]

Its final specific mechanical energy is:

\[ \begin{aligned} \epsilon_{Bf} &=-24.9375+2.00\\ &=-22.9375\,\mathrm{MJ\,kg^{-1}} \end{aligned} \]

Calculate its final radius:

\[ \begin{aligned} r_{Bf} &=\frac{3.99\times10^{14}} {2(22.9375\times10^6)}\\ &=8.70\times10^6\,\mathrm{m} \end{aligned} \]

Now calculate its final circular speed:

\[ \begin{aligned} v_{Bf} &=\sqrt{\frac{3.99\times10^{14}}{8.70\times10^6}}\\ &=6.77\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Step 6: Compare the final states.

Satellite A receives twice as much energy per kilogram as satellite B.

It therefore moves into a higher circular orbit and finishes with a lower circular speed.

The tempting argument was that satellite mass cancels from:

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

That cancellation is correct. At the same orbital radius, both satellites have the same circular speed.

However, they do not finish at the same orbital radius because they receive different energy increases per kilogram.

If both satellites had received the same energy per kilogram instead, they would have finished at the same circular radius and speed.

This is the important distinction between total energy and specific energy.

10What to understand before your next orbital-energy question

The original puzzle comes down to keeping three quantities separate:

1. Kinetic energy tells you about the satellite’s speed.
2. Gravitational potential energy tells you about its position relative to Earth, using zero potential energy at infinity.
3. Total mechanical energy combines both and determines whether the satellite is gravitationally bound in the ideal two-body model.

When a satellite moves into a higher circular orbit, its kinetic energy decreases, but its gravitational potential energy increases by a greater amount. Its total mechanical energy therefore increases while its final circular speed decreases.

The next useful step is learning to distinguish a bound elliptical orbit from an escape trajectory using total mechanical energy and escape speed. This is particularly important when a question gives you the satellite’s actual speed and position rather than simply describing a change between two circular orbits.

For that calculation, see How to Calculate Escape Speed from a Planet’s Mass and Radius.