How a Sudden Speed Change Affects a Satellite’s Orbit

A satellite is travelling in a circular orbit when its engine gives it a sudden forward push. Does it immediately move into a higher circular orbit?

No. Its position hasn’t changed at the instant of the push, but its velocity has. If the push is tangential to its original path, the satellite begins following a new elliptical orbit that reaches farther from Earth. The point where the engine fired becomes the ellipse’s closest point to Earth.

A backward push has the opposite effect. The satellite enters an ellipse that dips closer to Earth, with the engine firing point now its farthest point.

That is the basic answer, but there are two important limits. A sufficiently large forward push can put the satellite on an escape trajectory instead of an ellipse. And if the engine changes the satellite’s direction as well as its speed, the result cannot be predicted from speed alone.

The useful question is: what velocity does the satellite have immediately after the engine switches off?

01Newton’s cannonball: why a faster satellite doesn’t simply fly away

Imagine firing a cannonball horizontally from an extremely high mountain, above Earth’s atmosphere.

Newton used this kind of thought experiment to investigate a strange possibility: could something keep falling towards Earth without ever reaching the ground?

Start with a relatively slow cannonball.

It moves forwards, but gravity continuously accelerates it towards Earth’s centre. Its curved path eventually intersects the ground.

Now increase its horizontal launch speed.

The cannonball travels farther before hitting Earth. Increase the speed again, and its path bends around more of the planet.

At one particular speed, the cannonball falls around Earth at exactly the rate Earth’s surface curves away beneath it. It completes a circular orbit.

What happens if we increase the speed beyond that value?

Gravity still acts, but the path no longer curves tightly enough to remain on the original circle. The cannonball travels farther from Earth before eventually turning back, provided its speed remains below escape speed.

This was a thought experiment, not a record of cannonballs being launched into orbit. Its purpose was to isolate the relationship between sideways speed and gravitational motion.

The model also ignores air resistance, terrain, and the complications of launching from a rotating planet.

But its central insight is correct: an orbiting satellite doesn’t stay up because gravity has disappeared. It stays in orbit because its sideways motion and gravitational acceleration produce a path around Earth.

Changing that sideways motion changes the path.

02How could we tell that the satellite’s orbit had changed?

Picture a satellite already travelling around Earth. It has a small rocket thruster, an onboard clock, and a radio transmitter.

A ground tracking station has an antenna that can send signals to the spacecraft and receive signals back.

We want to distinguish two events:

• The satellite’s velocity changing during a brief engine firing.
• Its position gradually departing from the original circular path afterwards.

The first happens during the burn. The second becomes apparent as the satellite continues moving under gravity.

The first measurement isn’t enough

Suppose we record only the satellite’s speed immediately before and after the engine fires.

We might discover that its speed increased by 200 m/s.

But that doesn’t tell us the complete shape of the new orbit.

Was the engine firing directly forwards? Outwards from Earth? At an angle?

The distinction matters because velocity is a vector: it has both magnitude and direction.

A more useful tracking arrangement measures the spacecraft repeatedly after the burn. Radio ranging can help determine distance, while Doppler measurements provide information about motion towards or away from the tracking station.

Combining observations over time with an orbital model allows the trajectory to be reconstructed. A single range or Doppler measurement cannot determine the complete orbit.

For a purely forward burn from a circular orbit, the predicted sequence is:

1. The engine increases the satellite’s tangential speed.
2. Its distance from Earth’s centre is initially unchanged.
3. As the satellite moves onwards, its distance from Earth increases.
4. It reaches a maximum distance, then moves inwards again if the new trajectory is a bound ellipse that does not intersect Earth.

These are predictions from the idealised gravitational model, not measurements from a particular mission.

A backward tangential burn produces the reverse pattern: the satellite begins moving closer to Earth after leaving its original circular path.

03The circular-speed test

To predict the new path, we first need a reference speed: the speed required to maintain a circular orbit at the satellite’s current distance from Earth.

For a circular orbit, Earth’s gravitational acceleration supplies exactly the centripetal acceleration needed to keep the satellite travelling around a circle.

Let:

• \(G\) be the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
• \(M\) be the mass of Earth, approximately \(5.97\times10^{24}\ \mathrm{kg}\).
• \(r\) be the orbital radius, measured from Earth’s centre in metres.
• \(v_c\) be the circular orbital speed, in metres per second.

Equating gravitational acceleration and circular centripetal acceleration gives:

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

Rearranging:

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

We can shorten \(GM\) to \(\mu\), Earth’s gravitational parameter:

\[ \mu=GM \]

For Earth, we will use the rounded value:

\[ \mu=3.986\times10^{14} \ \mathrm{m^3/s^2} \]

The circular-speed formula becomes:

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

This equation applies to circular motion around the stated central mass. It does not give the satellite’s instantaneous speed at an arbitrary point on an elliptical orbit.

Faster or slower than the local circular speed?

Suppose a satellite begins in a circular orbit.

At its current radius, gravity has exactly the strength needed to maintain the original circular motion.

Now the satellite receives a forward tangential kick.

Its speed increases, but Earth’s gravitational acceleration at that location is unchanged. Gravity can no longer bend its path tightly enough to maintain the original circle.

The satellite’s new path initially curves away from the old circle, towards larger orbital radii.

For a backward kick, gravity bends the slower satellite’s path more tightly than the original circle requires. Its path moves towards smaller radii.

The closest point in an Earth orbit is called perigee. The farthest point is apogee.

The distinction isn’t determined by which direction the satellite happens to be facing. It is determined by how its distance from Earth changes along the resulting orbit.

A decision table for tangential burns

The following assumes the satellite initially travels in a circular Earth orbit, the post-burn motion remains tangential and in the original direction, and only Earth’s gravity acts afterwards.

Here \(v\) is the speed immediately after the burn.

The escape speed at distance \(r\), neglecting further thrust and other forces, is:

\[ v_{\mathrm{esc}}=\sqrt{\frac{2\mu}{r}} \]

This comes from requiring enough mechanical energy to reach an indefinitely large distance from Earth. It is not a circular-orbit speed.

If a backward burn reduces the tangential speed all the way to zero, the satellite begins falling radially rather than following an ordinary ellipse.

The table also doesn’t apply unchanged to angled burns. We’ll return to that distinction shortly.

05Why the direction of the kick matters

Consider a satellite moving upwards along the right-hand side of its circular orbit when its engine fires.

There are two ways to give it a velocity change of magnitude 400 m/s:

• Fire the engine so the velocity change points forwards, along the satellite’s existing motion.
• Fire it so the velocity change points radially outwards, directly away from Earth’s centre.

Would these produce the same orbit because the velocity changes have the same magnitude?

No.

In the first case, the engine adds directly to the existing tangential velocity.

In the second, the engine adds a perpendicular velocity component. The satellite now moves partly outwards as well as sideways.

The difference is more than geometry.

The amount of kinetic energy transferred also depends on the direction of the velocity change.

Because gravitational potential energy depends only on position, it is unchanged during an ideal instantaneous burn.

The change in mechanical energy therefore equals the change in kinetic energy.

To compare orbits without needing the satellite’s mass, we can use specific mechanical energy: mechanical energy per kilogram.

\[ \varepsilon=\frac12v^2-\frac{\mu}{r} \]

Here \(\varepsilon\) is in joules per kilogram, and \(v\) is the satellite’s total speed.

While the satellite coasts under Earth’s gravity alone, this quantity stays constant.

06How can we recognise the closest and farthest points of an ellipse?

For a circular orbit, the satellite’s velocity is perpendicular to the radius at every point.

For an elliptical orbit, that is generally not true.

As the satellite travels between perigee and apogee, it has both tangential motion and radial motion. At perigee and apogee themselves, the radial component is zero.

Those two points are especially useful because the velocity is purely tangential.

Two conservation laws then help us work out what happens between these points.

First, mechanical energy is conserved during unpowered motion:

\[ \varepsilon=\frac12v^2-\frac{\mu}{r} \]

Second, gravity points towards Earth’s centre, so it produces no torque about that centre. The satellite’s angular momentum about Earth is conserved.

Angular momentum per kilogram has magnitude:

\[ h=rv_t \]

Here \(v_t\) is the tangential component of velocity, and \(h\) is measured in \(\mathrm{m^2/s}\).

At perigee and apogee, the velocity is entirely tangential. Therefore:

\[ r_pv_p=r_av_a \]

The subscripts \(p\) and \(a\) refer to perigee and apogee.

This tells us that a satellite moves faster at perigee than at apogee.

Notice the limit of the original circular-speed model. We can calculate the circular speed at either radius as a reference, but neither end of an elliptical orbit is automatically moving at that circular speed.

A useful consequence for orbit raising

Suppose a satellite starts in a low circular orbit and wants to reach a higher circular orbit.

A forward burn at the low point places it into an ellipse that reaches the target radius.

When it arrives at the far end, it is moving more slowly than the circular speed required at that higher radius.

It therefore needs another forward burn to circularise the orbit.

That leads to a result that can feel backwards: a satellite can receive two forward pushes and finish in a higher circular orbit where its speed is lower than its original circular speed.

The explanation is that its altitude, gravitational potential energy, and orbital path have changed between the two burns.

For this article, the key rule is simpler: a single tangential speed change normally creates an ellipse, not a new circle.

07Questions and solutions

Use the following rounded Earth data where required:

Unless stated otherwise, assume Earth is spherical, the satellite’s mass is negligible compared with Earth’s, burns are instantaneous, and the only force after each burn is Earth’s gravity. Speeds are measured in an Earth-centred, non-rotating reference frame. Take forward tangential and radially outward velocity components as positive.

Question 1

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

Its engine briefly increases its tangential velocity by 0.150 km/s in the forward direction.

Calculate its speed immediately after the burn, and explain whether the new path is circular or elliptical. Identify whether the firing point becomes perigee or apogee.

Solution 1

The satellite’s new speed is approximately 7.21 km/s. It enters an elliptical orbit, with the burn point becoming perigee.

The initial circular speed is:

\[ \begin{aligned} v_c&=\sqrt{\frac{\mu}{r}}\\ &=\sqrt{\frac{398600}{8000}}\\ &=7.06\ \mathrm{km/s} \end{aligned} \]

The gravitational parameter and radius have been expressed in kilometres and seconds.

The tangential velocity after the burn is:

\[ \begin{aligned} v&=7.06+0.150\\ &=7.21\ \mathrm{km/s} \end{aligned} \]

The satellite now has more tangential speed than is required to maintain a circular orbit at its current radius.

Gravity is unchanged at the moment of the burn, so the satellite moves away from its original circular path and reaches larger orbital radii later.

Its new speed is below escape speed, so the trajectory remains a bound ellipse.

The tempting mistake is to say that the satellite immediately enters a larger circular orbit. It cannot do so from this single burn because it is still at the original radius, but no longer has the circular speed required there.

Question 2

A satellite begins in a circular Earth orbit of radius 7000 km.

A short backward tangential burn reduces its speed to 7.15 km/s.

An engineer argues that the satellite is safe because its current altitude is approximately 630 km and the engine did not push it directly towards Earth.

Using conservation of energy and angular momentum, determine whether the satellite can complete its new orbit without intersecting Earth’s surface.

Solution 2

The satellite cannot complete a full orbit without intersecting Earth. Its idealised trajectory has a calculated perigee radius of approximately 5700 km, which lies inside Earth’s surface radius of 6370 km.

First, identify the burn point.

The initial circular speed is:

\[ \begin{aligned} v_c&=\sqrt{\frac{398600}{7000}}\\ &=7.55\ \mathrm{km/s} \end{aligned} \]

After the burn, the satellite travels tangentially at 7.15 km/s.

Because this is below the required circular speed, the burn point becomes apogee.

Thus:

\[ r_a=7000\ \mathrm{km} \]

and

\[ v_a=7.15\ \mathrm{km/s} \]

Next, use conservation of angular momentum.

At the two turning points, the velocity is tangential:

\[ r_av_a=r_pv_p \]

Therefore:

\[ v_p=\frac{r_av_a}{r_p} \]

Here \(r_p\) and \(v_p\) are the unknown perigee radius and speed.

Now apply conservation of mechanical energy.

For the hypothetical unimpeded ellipse:

\[ \frac12v_a^2-\frac{\mu}{r_a} = \frac12v_p^2-\frac{\mu}{r_p} \]

Substituting the angular momentum relationship and rearranging gives:

\[ r_p=\frac{r_av_a^2}{2\mu/r_a-v_a^2} \]

Using kilometres and seconds:

\[ \begin{aligned} r_p &=\frac{7000(7.15)^2}{113.89-51.12}\\ &\approx5700\ \mathrm{km} \end{aligned} \]

Compare this with Earth’s surface radius:

\[ R_E=6370\ \mathrm{km} \]

The calculated perigee lies approximately 670 km below the solid surface.

The spacecraft would therefore encounter Earth before reaching that mathematical turning point.

The engineer’s argument confuses an initial position with a complete future trajectory. The satellite does not need an engine pointing downwards to fall closer to Earth; reducing its sideways speed changes how gravity curves its path.

The calculation also reveals why simply identifying a new ellipse isn’t always enough. We must check whether that ellipse clears the central body’s surface.

Question 3

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

Immediately after a brief manoeuvre, tracking data indicates:

• Its total speed is unchanged from the original circular speed.
• Its radial velocity is now 1.00 km/s outwards.
• Its motion remains in the original orbital plane and retains the original tangential direction.

A technician concludes that the manoeuvre was purely radial and that, because the speed is unchanged, the orbit must still be circular.

Determine the new tangential velocity component and the magnitude of the engine’s velocity change. Explain both errors in the technician’s conclusion.

When the satellite next passes through the same orbital radius while travelling inwards, what will its total speed and radial velocity be?

Solution 3

The new tangential velocity is approximately 6.99 km/s, and the magnitude of the engine’s velocity change is approximately 1.003 km/s. The orbit is not circular, and the manoeuvre cannot have been purely radial.

First, calculate the original circular speed.

\[ \begin{aligned} v_c&=\sqrt{\frac{398600}{8000}}\\ &=7.059\ \mathrm{km/s} \end{aligned} \]

The post-burn total speed remains 7.059 km/s.

Next, resolve the new velocity into components.

Let \(v_r\) be the radial component and \(v_t\) the tangential component.

Because these components are perpendicular:

\[ v^2=v_t^2+v_r^2 \]

Hence:

\[ \begin{aligned} v_t&=\sqrt{v^2-v_r^2}\\ &=\sqrt{7.059^2-1.00^2}\\ &=6.99\ \mathrm{km/s} \end{aligned} \]

The tangential component has decreased, even though total speed has not changed.

Now determine the velocity change produced by the engine.

Before the burn, the satellite has no radial velocity.

Its tangential velocity is 7.059 km/s.

After the burn:

\[ \begin{aligned} \Delta v_t&=6.987-7.059\\ &=-0.072\ \mathrm{km/s}\\ \Delta v_r&=+1.00\ \mathrm{km/s} \end{aligned} \]

The velocity-change magnitude is:

\[ \begin{aligned} |\Delta\vec v| &=\sqrt{(\Delta v_t)^2+(\Delta v_r)^2}\\ &=\sqrt{0.072^2+1.00^2}\\ &\approx1.003\ \mathrm{km/s} \end{aligned} \]

The engine must have changed both velocity components.

A purely outward radial kick would have left the tangential component unchanged at 7.059 km/s. Adding radial velocity to that would have increased the total speed.

Finally, consider the orbit and the return crossing.

The satellite’s mechanical energy per kilogram is unchanged at the instant of the burn because its total speed and distance from Earth are unchanged.

But its angular momentum about Earth’s centre has decreased because its tangential velocity has decreased.

Its orbit is therefore different.

The satellite is also moving outwards immediately after the burn, so it cannot be following a circle of constant radius.

As it coasts, mechanical energy and angular momentum are conserved.

At the next inward crossing of the original radius, its total speed must again be 7.059 km/s, because both its specific mechanical energy and gravitational potential energy are the same as at the outward crossing.

Its tangential speed is again 6.987 km/s, but its radial velocity has reversed:

\[ v_r=-1.00\ \mathrm{km/s} \]

This is the key trap: identical speeds at the same orbital radius do not establish identical orbits.

The direction of motion is essential.

Question 4

A satellite is initially in a circular Earth orbit of radius 7000 km.

A mission requires it to enter a higher circular orbit of radius 14,000 km using two instantaneous tangential burns.

The first burn occurs in the original circular orbit. The satellite then coasts along an elliptical transfer orbit whose apogee is at the target radius. The second burn occurs at that apogee.

Determine:

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

b) The direction and magnitude of each burn.

c) The satellite’s final circular speed, and why it is lower than its original circular speed even though both burns increase its velocity in the forward tangential direction.

Use conservation of energy and angular momentum for the elliptical transfer.

Solution 4

The satellite needs a forward burn of approximately 1.17 km/s at the lower orbit and another forward burn of approximately 0.979 km/s at the higher point. Its final circular speed is 5.34 km/s, lower than the original 7.55 km/s.

First, calculate the speeds in the starting and target circular orbits.

Let the initial radius be \(r_0=7000\ \mathrm{km}\), and the target radius be \(r_1=14000\ \mathrm{km}\).

\[ \begin{aligned} v_0&=\sqrt{\frac{398600}{7000}}\\ &=7.55\ \mathrm{km/s} \end{aligned} \]

For the higher circular orbit:

\[ \begin{aligned} v_1&=\sqrt{\frac{398600}{14000}}\\ &=5.34\ \mathrm{km/s} \end{aligned} \]

These are circular speeds, not transfer-ellipse speeds.

Next, find the transfer speeds using angular momentum conservation.

The transfer perigee is at \(r_0\), and the transfer apogee is at \(r_1\).

\[ r_0v_p=r_1v_a \]

Because \(r_1=2r_0\):

\[ v_a=\frac12v_p \]

Now apply energy conservation along the transfer ellipse.

\[ \frac12v_p^2-\frac{\mu}{r_0} = \frac12v_a^2-\frac{\mu}{r_1} \]

Substituting the radius and speed ratios:

\[ \begin{aligned} \frac12v_p^2-\frac{\mu}{r_0} &=\frac18v_p^2-\frac{\mu}{2r_0}\\ \frac38v_p^2&=\frac{\mu}{2r_0} \end{aligned} \]

Therefore:

\[ v_p=\sqrt{\frac{4\mu}{3r_0}} \]

Substitute the values:

\[ \begin{aligned} v_p&=\sqrt{\frac{4(398600)}{3(7000)}}\\ &=8.71\ \mathrm{km/s} \end{aligned} \]

At apogee:

\[ v_a=\frac{8.71}{2}=4.36\ \mathrm{km/s} \]

Calculate both burns.

At the original lower orbit, the satellite must accelerate from 7.55 km/s to 8.71 km/s:

\[ \begin{aligned} \Delta v_1&=v_p-v_0\\ &=8.71-7.55\\ &=+1.17\ \mathrm{km/s} \end{aligned} \]

At apogee, the satellite has slowed to 4.36 km/s. But the circular speed at that radius is 5.34 km/s.

So:

\[ \begin{aligned} \Delta v_2&=v_1-v_a\\ &=5.34-4.36\\ &=+0.979\ \mathrm{km/s} \end{aligned} \]

Both burns are prograde.

Finally, explain the apparently contradictory speeds.

The satellite initially travels at 7.55 km/s in its lower circular orbit.

Its first burn increases its speed to 8.71 km/s and places it into an ellipse.

As it climbs towards apogee, its gravitational potential energy increases and its kinetic energy decreases. Its speed falls to 4.36 km/s.

The second burn raises that speed to 5.34 km/s, just enough for a circular orbit at the higher radius.

The common mistake is comparing only the initial and final speeds and concluding that at least one engine burn must have slowed the satellite.

That ignores the long interval between burns, during which gravity changes both its speed and its altitude.

Question 5

An idealised spherical planet has mass \(M\), surface radius \(R\), and no atmosphere.

A point-sized satellite initially travels in a circular orbit at radius \(2R\).

Its original circular speed is:

\[ v_0=\sqrt{\frac{GM}{2R}} \]

Mission control is considering two alternative instantaneous manoeuvres at the same orbital position.

• Manoeuvre A rotates the satellite’s velocity by \(30^\circ\) radially outwards from its original tangential direction.
• Manoeuvre B rotates its velocity by \(30^\circ\) radially inwards from its original tangential direction.

In both cases, the satellite’s total speed immediately after the burn remains exactly \(v_0\). Its tangential motion stays in the original direction.

A controller makes three claims:

1. Both satellites must remain on circular orbits because their speeds are unchanged.
2. Manoeuvre A is safe because the satellite initially moves away from the planet.
3. Manoeuvre B is unsafe because the satellite initially moves towards the planet.

Determine the actual minimum and maximum orbital radii for both manoeuvres, using only conservation of mechanical energy and angular momentum.

Then consider the following rescue plan: wait until the first apogee and apply a tangential burn to circularise the orbit there.

Determine the required velocity change as a multiple of \(v_0\), and decide which manoeuvre allows the rescue to occur before the satellite contacts the planet.

Ignore all forces except the planet’s gravity between burns. Treat the surface as solid.

Solution 5

Both manoeuvres produce non-circular, bound trajectories with a minimum radius of \(R\) and a maximum radius of \(3R\). Without further intervention, both paths reach the planet’s surface.

However, Manoeuvre A reaches \(3R\) before encountering the surface, so it can be rescued by a forward circularisation burn of approximately \(0.239v_0\). Manoeuvre B reaches the surface first and cannot follow that same waiting strategy safely.

The central surprise is that a satellite can initially move away from a planet yet still be on an orbit that later intersects its surface.

Step 1: Resolve the new velocities.

Both manoeuvres produce a velocity at \(30^\circ\) to the local tangent.

The tangential component is:

\[ v_t=v_0\cos30^\circ \]

Therefore:

\[ v_t=\frac{\sqrt3}{2}v_0 \]

The radial component is:

\[ v_r=\pm v_0\sin30^\circ \]

So:

\[ v_r=\pm\frac12v_0 \]

The plus sign applies to Manoeuvre A, which moves outwards. The minus sign applies to B, which moves inwards.

Both satellites have the same tangential speed and the same total speed after their manoeuvres.

Step 2: Calculate their mechanical energy.

Let \(\mu=GM\) for the planet.

At radius \(2R\), the total specific mechanical energy is:

\[ \begin{aligned} \varepsilon &=\frac12v_0^2-\frac{\mu}{2R}\\ &=\frac{\mu}{4R}-\frac{\mu}{2R}\\ &=-\frac{\mu}{4R} \end{aligned} \]

Both manoeuvres have the same energy because neither changes the total speed or the orbital radius at the instant of the burn.

Their negative mechanical energy also tells us that, in the absence of a solid planetary surface, their trajectories would remain gravitationally bound.

But equal energy does not mean a circular orbit.

Step 3: Calculate the angular momentum.

The specific angular momentum is:

\[ h=rv_t \]

At the moment of the manoeuvre:

\[ \begin{aligned} h&=2R\left(\frac{\sqrt3}{2}v_0\right)\\ &=\sqrt3Rv_0 \end{aligned} \]

Square this expression:

\[ h^2=3R^2v_0^2 \]

Using \(v_0^2=\mu/(2R)\):

\[ h^2=\frac{3\mu R}{2} \]

Both manoeuvres have the same angular momentum magnitude, which is smaller than the original circular-orbit value.

That is the missing information in the controller’s first claim.

Step 4: Find the turning radii.

At any closest or farthest point, the radial velocity is zero.

The satellite’s speed there is entirely tangential:

\[ v=\frac{h}{r} \]

Its specific mechanical energy at a turning point is therefore:

\[ \varepsilon=\frac{h^2}{2r^2}-\frac{\mu}{r} \]

Substitute the conserved energy and angular momentum:

\[ -\frac{\mu}{4R} = \frac{3\mu R}{4r^2}-\frac{\mu}{r} \]

Multiply through by \(4Rr^2/\mu\):

\[ -r^2=3R^2-4Rr \]

Rearrange:

\[ r^2-4Rr+3R^2=0 \]

Factorise:

\[ (r-R)(r-3R)=0 \]

The two turning radii are:

\[ r_{\min}=R,\qquad r_{\max}=3R \]

These are the same for both manoeuvres.

The ideal mathematical paths both just touch the planet’s surface at their closest approach.

The unchanged speed at the manoeuvre point did not preserve the original circle.

Step 5: Determine which event happens first.

Manoeuvre A has positive radial velocity immediately after the burn.

It is moving outwards from \(2R\), so it reaches the maximum radius \(3R\) first.

If it receives no further burn, it will eventually return inwards and reach \(R\), touching the surface.

Manoeuvre B has negative radial velocity immediately after the burn.

It is already moving inwards from \(2R\), so it reaches the minimum radius \(R\) before it can reach the farthest point.

Thus the initial radial direction does not decide whether a surface encounter is possible. It decides the order in which the satellite encounters the turning points.

Step 6: Calculate the rescue burn for Manoeuvre A.

At apogee, the satellite is at radius \(3R\).

Its speed is found from angular momentum conservation:

\[ \begin{aligned} v_a&=\frac{h}{3R}\\ &=\frac{\sqrt3Rv_0}{3R}\\ &=\frac{v_0}{\sqrt3} \end{aligned} \]

To maintain a circular orbit at \(3R\), it would need:

\[ v_c=\sqrt{\frac{\mu}{3R}} \]

We know:

\[ v_0=\sqrt{\frac{\mu}{2R}} \]

Therefore:

\[ v_c=\sqrt{\frac23}v_0 \]

The required tangential velocity change is:

\[ \begin{aligned} \Delta v&=v_c-v_a\\ &=\left(\sqrt{\frac23} -\frac1{\sqrt3}\right)v_0\\ &=\frac{\sqrt2-1}{\sqrt3}v_0\\ &\approx0.239v_0 \end{aligned} \]

This burn is prograde because the satellite’s speed at apogee is below the circular speed for that radius.

After the burn, it travels in a circular orbit of radius \(3R\), safely outside the planet.

Manoeuvre A can therefore be rescued before the surface encounter.

Manoeuvre B cannot safely wait until its first apogee because it reaches the solid surface at \(R\) first.

The controller’s original reasoning fails because speed, mechanical energy, angular momentum, and radial direction play different roles.

The two manoeuvres have identical speeds and mechanical energies immediately after firing. They even have the same minimum and maximum orbital radii.

Yet their immediate directions determine whether the proposed rescue opportunity occurs before or after the surface encounter.

08What to check when a satellite’s speed changes

The first decision is whether the engine changes tangential speed, radial speed, or both.

For a purely tangential burn from a circular orbit, compare the new speed with the circular speed at the unchanged radius. That identifies whether the satellite begins an inward-reaching ellipse, an outward-reaching ellipse, or an escape trajectory.

For an angled burn, resolve the velocity into components. The satellite’s position, total mechanical energy, and angular momentum together determine the subsequent path.

If the path passes close to a planet, check its minimum radius against the planet’s surface radius. Being above the surface when the engine fires is not enough.

The next useful calculation is deciding what happens after the satellite has followed the new ellipse to another point in its orbit. At that stage, the distinction between instantaneous speed, gravitational potential energy, and total orbital energy becomes essential. That relationship is developed in Energy Changes When Satellites Change Orbit: HSC Physics.