Why Higher Circular Satellites Move More Slowly

Two satellites circle Earth. One is 400 km above the surface, while the other is 35,800 km above it. The higher satellite has a much longer distance to travel around Earth. Which one would you expect to move faster?

Surprisingly, the lower satellite travels at about 7.7 km/s, while the higher one moves at only 3.1 km/s. These are calculated speeds for ideal circular orbits, not measurements of particular satellites.

Higher circular satellites move more slowly because Earth’s gravitational pull becomes weaker with distance. A satellite farther from Earth needs a lower sideways speed for gravity to keep its path circular. It also takes longer to complete each orbit.

The important word is circular. Comparing two established circular orbits is different from asking what happens when a spacecraft accelerates to change orbit.

01Why weaker gravity means a lower orbital speed

Imagine swinging a ball around on a string. The ball’s velocity points along the tangent to its circular path, but the string continually pulls it towards the centre.

Without that inward pull, the ball would travel in a straight line rather than continue around the circle.

An orbiting satellite behaves similarly, except there is no string. Gravity supplies the inward force that continually changes the direction of its velocity.

Here is the key question.

Suppose you move a satellite farther from Earth but try to keep its orbital speed unchanged. Would the weaker gravity still bend its trajectory into a circle?

Not necessarily. Gravity gets weaker faster than the required inward acceleration decreases when you increase the radius while holding speed constant.

For circular motion, the required inward, or centripetal, acceleration is

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

where:

• \(a_c\) is centripetal acceleration in metres per second squared (\(\mathrm{m\,s^{-2}}\)).
• \(v\) is orbital speed in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(r\) is the orbital radius in metres (\(\mathrm{m}\)).

If speed stays constant, doubling the circular path’s radius halves the acceleration required.

But gravity follows a different relationship. Its strength decreases with the square of the distance from Earth’s centre. Doubling the radius reduces gravitational acceleration to one-quarter of its original value.

The available gravity has fallen more sharply than the acceleration needed at the original speed. To restore a circular orbit, the satellite must move more slowly.

This is where the string analogy stops being exact. You can adjust the tension in a string independently. For a satellite in an ideal gravitational orbit, the central mass and distance determine the available gravitational force.

Deriving the circular orbital speed equation

Newton’s law of universal gravitation gives the attractive force between a spherical central body and a satellite outside it:

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

Here:

• \(F_g\) is gravitational force in newtons (\(\mathrm{N}\)).
• \(G\) is the universal gravitational constant, approximately \(6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\).
• \(M\) is the central body’s mass in kilograms (\(\mathrm{kg}\)).
• \(m\) is the satellite’s mass in kilograms.
• \(r\) is the distance between their centres in metres.

For an ideal circular orbit, gravity provides the entire required centripetal force:

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

Equating these expressions gives

\[ \begin{aligned} \frac{GMm}{r^2}&=\frac{mv^2}{r}\\ v^2&=\frac{GM}{r}\\ v&=\sqrt{\frac{GM}{r}} \end{aligned} \]

We take the positive square root because \(v\) represents speed, not a signed velocity component.

This is the relationship that explains the puzzle:

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

For circular orbits around the same central mass, increasing orbital radius decreases orbital speed.

Notice that satellite mass \(m\) cancels. A 100 kg satellite and a 1,000 kg satellite in ideal circular orbits of equal radius around Earth have the same orbital speed.

The heavier satellite experiences more gravitational force, but its greater inertia exactly compensates.

Orbital radius is not altitude

This is a common source of calculation errors.

A satellite’s altitude \(h\) is measured above the surface. Its orbital radius \(r\) is measured from the centre of the central body.

For a spherical planet,

\[ r=R+h \]

where \(R\) is the planet’s radius.

For example, a satellite 400 km above Earth does not have an orbital radius of 400 km.

Using Earth’s approximate mean radius of 6,370 km,

\[ \begin{aligned} r&=6370+400\\ &=6770\ \mathrm{km} \end{aligned} \]

In metres, this is \(6.77\times10^6\,\mathrm{m}\).

Using 400 km directly in the orbital speed equation would give a physically incorrect result.

What happens if the orbital radius becomes four times larger?

For two circular orbits around the same central mass, we can compare speeds without knowing the planet’s mass:

\[ \frac{v_2}{v_1} =\sqrt{\frac{r_1}{r_2}} \]

Let the first orbital radius be \(r\) and the second be \(4r\).

Then

\[ \begin{aligned} \frac{v_2}{v_1} &=\sqrt{\frac{r}{4r}}\\ &=\frac12 \end{aligned} \]

The higher satellite travels at half the speed.

Meanwhile, gravitational acceleration is only one-sixteenth as large, because gravity follows the inverse-square law.

That matches the circular-motion requirement: halving the speed reduces \(v^2\) to one-quarter, and increasing the radius by four reduces \(v^2/r\) by another factor of four.

The required acceleration is therefore one-sixteenth of its original value.

A weaker gravitational field and a lower orbital speed are not competing effects. They are exactly what allows the larger circle to work.

Why the slower satellite takes longer

One complete circular orbit covers a distance of \(2\pi r\). If \(T\) is the orbital period in seconds, then

\[ T=\frac{2\pi r}{v} \]

Substituting the circular-orbit speed equation gives

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

For the same central mass,

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

Return to the satellite at four times the original orbital radius.

Its path is four times as long, but its speed is only half as large. Its orbital period must therefore be eight times as long.

This is why a satellite can have both a lower speed and a much longer orbital period.

02How Newton tested whether gravity could explain an orbit

Artificial satellites didn’t exist when Isaac Newton developed his theory of universal gravitation. But a natural satellite was already available: the Moon.

The unanswered question was whether the force that makes objects fall near Earth’s surface could also be responsible for the Moon’s orbital motion.

If Earth’s gravity only acted near the ground, something else would have to keep the Moon in orbit.

If gravity extended all the way to the Moon, Newton needed to explain why its effect was so much weaker there.

This became part of what is commonly called Newton’s Moon test.

The measurements needed for the Moon test

The comparison requires two independent kinds of information.

First, measurements of falling motion or pendulum motion near Earth’s surface provide an estimate of gravitational acceleration at the surface.

Second, astronomical observations provide the Moon’s distance and orbital period.

Picture an observer on Earth pointing a telescope towards the Moon. By following its changing position against distant stars, the observer can estimate how long the Moon takes to complete an orbit relative to that star background.

Astronomical geometry provides an estimate of the distance from Earth’s centre to the Moon.

These measurements allow the Moon’s required centripetal acceleration to be calculated without first assuming a particular gravitational force law.

There is a subtle measurement problem here.

Measuring the time from one full Moon to the next sounds like a good way to find the Moon’s orbital period. That interval is about 29.5 days.

But the Moon takes only about 27.3 days to return to approximately the same direction relative to distant stars.

Why are these periods different?

Earth travels around the Sun while the Moon orbits Earth. After the Moon completes one revolution relative to the stars, the Earth-Moon-Sun alignment has changed. The Moon must travel farther before the same phase appears again.

The 29.5-day interval is the synodic month, or phase cycle. The 27.3-day interval is the sidereal month, measured relative to distant stars.

For a simplified orbital acceleration calculation, the sidereal period is the appropriate choice.

Using the phase cycle without correcting for Earth’s motion would underestimate the calculated lunar centripetal acceleration by about 14%.

This is an example of a measurement that looks reasonable but answers a slightly different question.

04Why observations from Earth can give the wrong impression

Imagine tracking a satellite with a camera fixed to the ground.

The camera records the satellite moving across the sky. You could estimate how quickly its direction changes, but there is a problem: Earth itself is rotating.

A ground-based observer is not stationary in an inertial, Earth-centred reference frame.

An orbiting satellite might move eastward around Earth while appearing to drift westward relative to Earth’s surface, if Earth’s eastward rotation is faster than the satellite’s angular motion.

A better measurement method is to determine the satellite’s orbital motion in an approximately inertial, Earth-centred frame.

Observations can be corrected for Earth’s rotation, or positions can be determined relative to a suitable inertial reference frame. The measured orbital period and radius can then be used to calculate

\[ v=\frac{2\pi r}{T} \]

This assumes the orbit is circular and \(T\) is the inertial orbital period.

The distinction matters because orbital speed, orbital angular speed, and apparent movement across the sky are not interchangeable.

For more on the difference between appearing stationary and actually being stationary, see Geosynchronous vs Geostationary Orbits: How to Tell Them Apart.

05The important exception: a higher position is not always a slower-moving object

Suppose a satellite is travelling in a circular orbit and briefly fires its engine in the direction of motion.

Its speed increases immediately.

Does that mean it has entered a new, higher circular orbit?

No. At the instant of the burn, its distance from Earth’s centre has not changed. The gravitational force is essentially unchanged, but its speed is now greater than the speed required for a circular orbit at that radius.

The spacecraft generally enters a different trajectory, such as an elliptical orbit.

If the burn is sufficiently large, it could instead escape Earth’s gravitational field.

On an elliptical orbit, the spacecraft’s speed changes as its distance from Earth changes. It generally moves fastest near its closest point to Earth and slowest near its farthest point.

You cannot substitute its instantaneous orbital radius into \(v=\sqrt{GM/r}\) and claim to have calculated its actual speed on the ellipse. That equation gives the speed of a circular orbit at that radius.

The useful decision is:

• If you are comparing two circular orbits around the same central mass, the larger-radius orbit has the lower speed.
• If the central mass changes, compare both \(M\) and \(r\).
• If a spacecraft is accelerating or following an elliptical trajectory, its instantaneous speed requires additional information about its motion and energy.

The effect of a sudden speed change on a satellite’s path follows directly from this distinction.

06Questions and solutions

For Questions 1-5, assume central bodies are spherical, satellite masses are negligible compared with central masses, and gravity is the only force between any specified engine burns. Ignore atmospheric drag and gravitational disturbances from other bodies. All circular-orbit speeds are measured in an inertial frame centred on the relevant central body.

Question 1

Satellite A follows a circular orbit of radius \(8.0\times10^6\,\mathrm{m}\) around a planet, travelling at \(6.0\,\mathrm{km\,s^{-1}}\).

Satellite B follows a circular orbit of radius \(1.8\times10^7\,\mathrm{m}\) around the same planet.

Calculate:

a) Satellite B’s orbital speed.

b) The ratio of satellite B’s orbital period to satellite A’s orbital period.

Solution 1

Satellite B travels at \(4.0\,\mathrm{km\,s^{-1}}\), and its orbital period is \(3.375\) times satellite A’s.

For circular orbits around the same central mass,

\[ \frac{v_B}{v_A} =\sqrt{\frac{r_A}{r_B}} \]

Substitute the given radii:

\[ \begin{aligned} v_B&=6.0\sqrt{\frac{8.0}{18.0}}\\ &=4.0\ \mathrm{km\,s^{-1}} \end{aligned} \]

Satellite B moves more slowly because its circular orbit has a larger radius.

For orbital periods, use \(T=2\pi r/v\):

\[ \begin{aligned} \frac{T_B}{T_A} &=\frac{r_B}{r_A}\frac{v_A}{v_B}\\ &=\frac{18}{8}\times\frac{6}{4}\\ &=3.375 \end{aligned} \]

This ratio is dimensionless.

The tempting mistake is to assume that the orbital period is proportional to orbital radius. That would give a factor of \(18/8=2.25\).

But the larger satellite’s speed is also lower, so its period increases by a greater factor.

Question 2

An idealised Earth rotates eastward once every 24.0 hours relative to distant stars.

Two satellites follow prograde, equatorial circular orbits around Earth:

• Satellite A has an inertial orbital period of 8.0 hours.
• Satellite B has an inertial orbital period of 30.0 hours.

A tracking station records how the longitude directly beneath each satellite changes relative to Earth’s surface.

a) Calculate the angular drift rate and direction of each satellite relative to Earth.

b) Determine which satellite has the greater inertial orbital speed and calculate the ratio of their speeds.

c) Explain how a satellite can appear to drift westward even though it is travelling eastward around Earth.

Solution 2

Satellite A drifts eastward relative to Earth’s surface at \(30^\circ\,\mathrm{h^{-1}}\). Satellite B drifts westward at \(3^\circ\,\mathrm{h^{-1}}\), even though both satellites orbit eastward. Satellite A’s actual orbital speed is approximately 1.55 times B’s.

Part a: Angular drift rates

Earth’s assumed angular rotation rate is

\[ \omega_E=\frac{360^\circ}{24.0\ \mathrm{h}} =15^\circ\,\mathrm{h^{-1}} \]

Satellite A’s inertial angular speed is

\[ \omega_A=\frac{360^\circ}{8.0\ \mathrm{h}} =45^\circ\,\mathrm{h^{-1}} \]

Its eastward drift relative to Earth is

\[ 45-15=30^\circ\,\mathrm{h^{-1}} \]

Satellite B’s inertial angular speed is

\[ \omega_B=\frac{360^\circ}{30.0\ \mathrm{h}} =12^\circ\,\mathrm{h^{-1}} \]

Its Earth-relative drift is

\[ 12-15=-3^\circ\,\mathrm{h^{-1}} \]

Taking eastward as positive, the negative sign means westward drift.

Part b: Compare orbital speeds

For circular orbits around the same central mass,

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

and

\[ v\propto r^{-1/2} \]

Combining the relationships gives

\[ v\propto T^{-1/3} \]

Therefore,

\[ \begin{aligned} \frac{v_A}{v_B} &=\left(\frac{T_B}{T_A}\right)^{1/3}\\ &=\left(\frac{30.0}{8.0}\right)^{1/3}\\ &\approx1.55 \end{aligned} \]

Satellite A has the greater orbital speed.

Part c: Resolve the apparent contradiction

Satellite B moves eastward in the inertial frame at \(12^\circ\) per hour, but Earth rotates eastward at \(15^\circ\) per hour.

Earth effectively rotates underneath B faster than B moves around the planet, so B’s longitude shifts westward relative to the surface.

The tempting argument is that westward drift means westward orbital velocity. It doesn’t.

The measurements use different reference frames. Earth’s rotation must be accounted for before using an apparent angular rate to infer a satellite’s true orbital speed.

Question 3

Two satellites follow circular orbits around different planets.

Planet P has mass \(M\). Satellite X orbits P at radius \(r\).

Planet Q has mass \(9M\). Satellite Y orbits Q at radius \(4r\).

Assume both planets are sufficiently compact that the specified orbits lie outside their surfaces.

a) Determine the ratio \(v_Y/v_X\).

b) Determine the ratio \(T_Y/T_X\).

c) Compare the gravitational accelerations experienced by the satellites.

d) A student says, “Y must travel more slowly because its orbit is four times larger.” Explain why the claim fails.

Solution 3

Satellite Y moves 1.5 times faster than X, has an orbital period \(8/3\) times as long, and experiences only \(9/16\) of X’s gravitational acceleration.

This is possible because Y orbits a much more massive planet.

Part a: Orbital speeds

For circular orbits,

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

For satellite X,

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

For satellite Y,

\[ \begin{aligned} v_Y&=\sqrt{\frac{G(9M)}{4r}}\\ &=\frac32\sqrt{\frac{GM}{r}} \end{aligned} \]

Therefore,

\[ \frac{v_Y}{v_X}=\frac32=1.5 \]

Part b: Orbital periods

Since \(T=2\pi r/v\),

\[ \begin{aligned} \frac{T_Y}{T_X} &=\frac{4r}{r}\frac{v_X}{v_Y}\\ &=4\times\frac23\\ &=\frac83 \end{aligned} \]

Although Y is faster, its orbital path is four times as long. It therefore still takes longer to complete each revolution.

Part c: Gravitational accelerations

For a spherical central body,

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

Thus,

\[ \begin{aligned} \frac{g_Y}{g_X} &=\frac{G(9M)/(4r)^2}{GM/r^2}\\ &=\frac9{16} \end{aligned} \]

Y experiences less gravitational acceleration even though its orbital speed is higher.

That is consistent with \(a_c=v^2/r\): Y’s speed squared is \(9/4\) times X’s, but its orbital radius is four times as large.

Part d: Explain the misconception

The student’s rule assumes both satellites orbit the same central mass.

Here, the central mass is not held constant. Increasing mass by a factor of nine strengthens gravity enough to more than compensate for the fourfold increase in orbital radius when determining circular speed.

The general comparison is governed by \(M/r\), not radius alone.

This also explains why different planets or stars can support very different orbital speeds at the same distance.

Question 4

Two satellites, A and B, travel anticlockwise in circular orbits around the same planet.

Their orbital radii are \(r\) and \(4r\), respectively. Both follow the same orbital plane.

Satellite A completes one orbit every 96.0 minutes.

At time \(t=0\), both satellites lie on the same radial line extending from the planet’s centre, on the same side of the planet.

a) Determine B’s orbital period.

b) Find the first positive time when the satellites again lie on the same radial line on the same side of the planet.

c) Calculate their separation at that moment in terms of \(r\).

d) An operator says that the faster inner satellite should eventually catch and rendezvous with the outer satellite without changing either orbit. Explain the error.

Solution 4

Satellite B’s period is 768 minutes. The satellites next have the same planet-centred direction after approximately 109.7 minutes, but they remain \(3r\) apart. They cannot rendezvous while remaining on these different circular orbits.

Part a: Orbital period

Using Kepler’s third-law relationship for circular orbits around the same mass,

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

Therefore,

\[ \begin{aligned} T_B&=96.0(4)^{3/2}\\ &=96.0\times8\\ &=768\ \mathrm{min} \end{aligned} \]

Part b: Next alignment

Orbital angular speed is

\[ \omega=\frac{2\pi}{T} \]

For the satellites to align again on the same side, A must gain one complete revolution relative to B.

If \(t\) is the elapsed time in minutes,

\[ (\omega_A-\omega_B)t=2\pi \]

Substituting their periods,

\[ \begin{aligned} t&=\frac{1}{1/96.0-1/768}\\ &\approx109.7\ \mathrm{min} \end{aligned} \]

The minute units follow because the orbital periods were supplied in minutes.

Part c: Their separation

When they next align on the same side, both satellites point along the same radial direction from the planet’s centre.

Their distances from the centre remain \(r\) and \(4r\).

Therefore,

\[ d=4r-r=3r \]

where \(d\) is their separation, in the same length units as \(r\).

Part d: Why they cannot rendezvous

Satellite A is indeed faster. In fact, its linear orbital speed is twice B’s, while its angular speed is eight times B’s.

But a rendezvous requires both satellites to occupy the same position at the same time.

They cannot do that while remaining in concentric circular orbits of different radii. Their radial separation is always at least \(3r\), even when their angular positions match.

The operator has confused matching angular position with matching physical position.

Actual rendezvous would require at least one satellite to change its trajectory.

Question 5

A spacecraft initially travels in a circular orbit of radius \(r\) around a planet of mass \(M\). Its initial circular orbital speed is

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

The spacecraft must reach a new circular orbit of radius \(2r\).

A controller plans a two-burn transfer. The first burn occurs tangentially at radius \(r\), placing the spacecraft on an elliptical orbit with its closest point at \(r\) and farthest point at \(2r\).

The spacecraft then coasts without thrust to the farthest point, where a second tangential burn changes its path into the higher circular orbit.

Assume the burns are instantaneous, the gravitational field is central and inverse-square, and mechanical energy and angular momentum are conserved during the unpowered transfer.

Let \(v_p\) be the spacecraft’s speed just after the first burn, and \(v_a\) its speed just before the second burn.

a) Using conservation of angular momentum and mechanical energy, determine \(v_p\) and \(v_a\) in terms of \(u\).

b) Determine whether each burn must increase or decrease the spacecraft’s speed. Express the required changes in speed in terms of \(u\).

c) Explain why the spacecraft’s final circular orbital speed is lower than its initial speed, even though both burns increase its speed.

d) At the farthest point of the transfer ellipse, a student calculates the inward acceleration as \(v_a^2/(2r)\). Explain why this does not give the gravitational acceleration at that point, and state the correct magnitude and direction of the acceleration.

Solution 5

Both burns must increase the spacecraft’s speed, even though its final circular orbital speed is lower than its original speed.

The speeds on the transfer ellipse are

\[ v_p=\frac{2u}{\sqrt3}, \qquad v_a=\frac{u}{\sqrt3} \]

The gravitational acceleration at the farthest point of the ellipse has magnitude \(GM/(4r^2)\) and points towards the planet.

The crucial distinction is that the transfer ellipse and the final circular orbit pass through the same outer position but require different instantaneous speeds there.

Part a: Derive the transfer speeds

At the closest and farthest points of the ellipse, the spacecraft’s velocity is tangential to its path.

Because gravity acts towards the planet’s centre, it produces no torque about that centre. Angular momentum is conserved during the coast.

Therefore,

\[ mv_pr=mv_a(2r) \]

Cancelling \(mr\),

\[ v_p=2v_a \]

The spacecraft must move twice as fast at the closest point as at the farthest point.

Now use conservation of mechanical energy.

At distance \(r\), the gravitational potential energy is \(-GMm/r\). At distance \(2r\), it is \(-GMm/(2r)\).

Kinetic energy is \(mv^2/2\).

Equating the total energies at the two ends of the ellipse and dividing by \(m\),

\[ \frac{v_p^2}{2}-\frac{GM}{r} = \frac{v_a^2}{2}-\frac{GM}{2r} \]

Substitute \(v_p=2v_a\):

\[ \begin{aligned} \frac{4v_a^2-v_a^2}{2} &=\frac{GM}{2r}\\ \frac32v_a^2&=\frac{GM}{2r}\\ v_a^2&=\frac{GM}{3r} \end{aligned} \]

Since \(u^2=GM/r\),

\[ v_a=\frac{u}{\sqrt3} \]

and therefore,

\[ v_p=\frac{2u}{\sqrt3} \]

Numerically,

\[ v_p\approx1.155u, \qquad v_a\approx0.577u \]

These speeds are proportional to \(u\), so they have the same units as \(u\), such as \(\mathrm{m\,s^{-1}}\).

Part b: Determine both burn directions

Before the first burn, the spacecraft’s circular speed is \(u\).

Immediately after the burn, it must have the transfer speed \(2u/\sqrt3\).

The change in speed is

\[ \begin{aligned} \Delta v_1 &=\left(\frac2{\sqrt3}-1\right)u\\ &\approx0.155u \end{aligned} \]

This is positive, so the first burn is forward, or prograde.

Now consider the farthest point of the transfer ellipse.

The spacecraft arrives there at speed

\[ v_a=\frac{u}{\sqrt3}\approx0.577u \]

But the circular orbital speed at radius \(2r\) is

\[ \begin{aligned} v_c&=\sqrt{\frac{GM}{2r}}\\ &=\frac{u}{\sqrt2}\\ &\approx0.707u \end{aligned} \]

The transfer spacecraft is moving too slowly to follow the desired circular orbit at that radius.

The second burn must therefore increase its speed by

\[ \begin{aligned} \Delta v_2 &=\left(\frac1{\sqrt2}-\frac1{\sqrt3}\right)u\\ &\approx0.130u \end{aligned} \]

This burn is also prograde.

Both \(\Delta v\) values are positive speed increases, measured in the same units as \(u\).

Part c: Explain the lower final speed

The spacecraft begins at circular speed \(u\) and finishes at circular speed

\[ v_c\approx0.707u \]

The final speed is lower, despite both engine burns increasing the speed at the instant they occur.

There is no contradiction.

The first burn increases the spacecraft’s kinetic energy and places it onto an elliptical transfer orbit. During the outward coast, gravity does negative work on the spacecraft. Its kinetic energy decreases while its gravitational potential energy increases.

By the time the spacecraft reaches the farthest point of the ellipse, its speed has fallen to approximately \(0.577u\).

The second forward burn increases that speed to approximately \(0.707u\), just enough for the higher circular orbit.

The tempting argument is that the second burn must be backwards because the final orbit is slower than the original orbit.

But the speed that matters for the second burn is the spacecraft’s speed immediately before that burn, not its speed in the original lower orbit.

Part d: Explain the acceleration error

At the farthest point, the spacecraft is at distance \(2r\) from the planet.

Newton’s law gives the gravitational acceleration magnitude

\[ \begin{aligned} g_a&=\frac{GM}{(2r)^2}\\ &=\frac{GM}{4r^2} \end{aligned} \]

The acceleration points towards the planet’s centre.

If outward radial acceleration is defined as positive, its radial component is

\[ a_r=-\frac{GM}{4r^2} \]

By contrast, the student’s calculation gives

\[ \begin{aligned} \frac{v_a^2}{2r} &=\frac{u^2/3}{2r}\\ &=\frac{GM}{6r^2} \end{aligned} \]

The two results differ.

The student’s error is applying the circular-orbit formula \(a_c=v^2/r\) to an elliptical trajectory while using the distance from the planet as though it were the radius of a circular path.

For circular motion, the orbital radius is also the radius of curvature of the path. For an ellipse, those quantities generally differ.

Gravity still determines the spacecraft’s actual acceleration through \(GM/r^2\). The shape of the trajectory determines how that acceleration relates to speed and curvature.

This is the deeper lesson behind the original question.

A higher circular orbit requires a lower speed, but being farther from Earth does not uniquely determine a spacecraft’s speed unless its trajectory is also specified.

Once that distinction is clear, the next useful question is why a spacecraft can finish a transfer with less kinetic energy but more total orbital energy. The relationship is explained in Energy Changes When Satellites Change Orbit: HSC Physics.