In short

A satellite of mass m in a circular orbit of radius r around a body of mass M has total mechanical energy E = -\dfrac{GMm}{2r}, negative because it is bound. Its kinetic and potential energies are K = +\dfrac{GMm}{2r} and U = -\dfrac{GMm}{r}, so K = -U/2 and E = U/2 = -K — the kinetic energy is exactly half the magnitude of the potential energy. The binding energy is |E| = \dfrac{GMm}{2r} — the work you must do against gravity to lift the satellite out to infinity and leave it there at rest. Counter-intuitively, a satellite that loses energy to drag speeds up: it drops to a smaller r, and K = GMm/(2r) grows as r shrinks. These relations are the circular-orbit version of a deeper law — the virial theorem — which says that for any stable gravitationally bound system, \langle K \rangle = -\tfrac{1}{2}\langle U \rangle on average.

The Chandrayaan-3 lander, in August 2023, spent its last five kilometres of descent fighting something strange. As it dropped toward the Moon, its engines fired retrograde — against the direction of motion — and yet the lander had to slow down, not speed up. Every Indian watching the landing on TV that night saw the speed number on screen tick slowly toward zero. That number had started the descent near 1.7 km/s. By touchdown, it had to reach essentially 0.

What was strange is this: if an ordinary object falling toward the Moon simply lets gravity do its work, it speeds up all the way down. But a satellite in orbit does the opposite — lose energy, drop to a lower orbit, and your speed increases. The engineers at ISRO, designing Chandrayaan's de-orbit burn, had to fight this effect: they needed to dump kinetic energy fast enough that the lander would stop before hitting the ground, not speed-up-and-crash.

That paradox is the beginning of the story this article tells. Once you understand the energy of a satellite — how its kinetic energy, potential energy, and total energy are linked — you understand why falling orbits speed up, why the atmosphere pulls the ISS slowly downward and faster, why comets move so fast at perihelion and so slowly at aphelion, and why a gravitationally bound system has a universal ratio between kinetic and potential energy. The formulas are two lines. The physics is everywhere.

The three energies of a circular orbit

Start from the two results you already have. For a satellite of mass m in a circular orbit of radius r around a central body of mass M:

Use these to compute all three relevant energies.

Kinetic energy

The kinetic energy is K = \tfrac{1}{2}mv^2. Substitute v^2 = GM/r:

K = \tfrac{1}{2}m \cdot \frac{GM}{r} = \frac{GMm}{2r}

Why: this comes from the force balance for a circular orbit. Gravity pulls inward with magnitude GMm/r^2; circular motion needs a centripetal pull of mv^2/r. Setting the two equal and solving for \tfrac{1}{2}mv^2 gives exactly GMm/(2r). This is not a new result — it is the orbital-velocity formula, one line further along.

Notice that K is positive (as kinetic energies always are) and grows as r shrinks. A satellite in a low orbit moves faster than one in a high orbit. A 7.9 km/s ISS has more kinetic energy per kilogram than a 3.1 km/s geostationary satellite.

Potential energy

The gravitational potential energy of the two-body system, taking U = 0 at infinity, is:

U = -\frac{GMm}{r}

Why: this comes from integrating the gravitational force F = -GMm/r^2 from the satellite's position out to infinity. The negative sign says the system is bound — you would have to do positive work to separate the two masses all the way to infinity, where U = 0. The deeper U dips (more negative), the more tightly bound the pair.

U is negative, and becomes more negative as r shrinks. A low-orbit satellite sits deeper in Earth's gravity well than a high-orbit satellite.

Total mechanical energy

Add them:

E = K + U = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right) = \frac{GMm}{2r} - \frac{2GMm}{2r}
\boxed{\,E = -\frac{GMm}{2r}\,}

Why: put U over the common denominator 2r to make the subtraction easy. The positive GMm/(2r) from kinetic energy is exactly half the magnitude of the negative -GMm/r = -2GMm/(2r) from potential energy, so they partly cancel and leave a net negative result of half the potential-energy magnitude.

The total mechanical energy is negative, exactly half the potential energy, and exactly the negative of the kinetic energy:

E = \tfrac{1}{2}U = -K

Three clean relations, falling out of one line of algebra. The rest of this article is about what they mean.

The kinetic–potential ratio: a universal 1:2

The relation K = -\tfrac{1}{2}U, or equivalently |U| = 2K, is the most important single fact about circular-orbit energy. Let's unpack it.

In magnitudes: the potential energy's magnitude is twice the kinetic energy. At every radius, |U| = 2K.

In signs: U is negative, K is positive, and their sum is the negative total energy E.

As a table for a 1000 kg satellite around Earth (GM_E = 3.986 \times 10^{14} m³/s²):

Orbit Radius r K = GMm/(2r) U = -GMm/r E = -GMm/(2r)
Low (ISS, 408 km up) 6.78 \times 10^6 m +29.4 GJ -58.8 GJ -29.4 GJ
GPS (20 200 km up) 2.66 \times 10^7 m +7.5 GJ -15.0 GJ -7.5 GJ
Geostationary (35 800 km up) 4.22 \times 10^7 m +4.7 GJ -9.4 GJ -4.7 GJ
Moon's orbit 3.84 \times 10^8 m +0.52 GJ -1.04 GJ -0.52 GJ

Every row satisfies K = -U/2 = -E. That is not coincidence; it is the 1:2:1 ratio built into circular gravitational orbits.

The three energies of a circular orbit vs radius Three curves plotted against orbit radius r. Kinetic energy K = +GMm/(2r) is positive and decreases with r. Potential energy U = -GMm/r is negative and rises toward zero with r. Total energy E = -GMm/(2r) is negative, exactly mirroring K across the axis. orbit radius r → 0 energy + 0 − K = +GMm/(2r) U = −GMm/r E = −GMm/(2r) = ½U +K −2K (= U) −K (= E)
The three energies of a circular orbit as functions of $r$. The kinetic energy $K$ is positive and falls off as $1/r$. The potential energy $U = -GMm/r$ is negative and twice as deep. The total energy $E = -GMm/(2r)$ is exactly $K$ reflected across zero. At every radius, the three energies stand in the ratio $K : E : U = +1 : -1 : -2$.

Where does the "half" come from?

A one-line derivation makes the factor of 2 transparent. The orbit condition is:

\underbrace{\frac{mv^2}{r}}_{\text{centripetal need}} = \underbrace{\frac{GMm}{r^2}}_{\text{gravity supplies}}

Multiply both sides by r/2:

\tfrac{1}{2}mv^2 = \tfrac{1}{2}\cdot\frac{GMm}{r}

The left side is K. The right side is \tfrac{1}{2}|U|. So:

K = \tfrac{1}{2}|U|, \quad \text{or equivalently}, \quad 2K = -U

Why: the factor of 2 came from the r/2 multiplication, which was itself forced on us by the structure of the force law. Inverse-square gravity gives a potential energy \propto 1/r and a centripetal requirement \propto 1/r (at fixed v) — the ratio between them is what determines the 1:2 energy split. A different force law (say, F \propto 1/r^3) would give a different ratio. The specific answer K = |U|/2 is a fingerprint of inverse-square attraction.

This is the core identity. Every other result in this article is a consequence of it.

Binding energy — the bill for escape

The binding energy of a system is the energy you would have to add to break it apart and leave all pieces at rest at infinity. For a satellite in a circular orbit:

\text{Binding energy} = \text{energy at infinity} - \text{energy now} = 0 - E = -E

Since E = -GMm/(2r), this is:

\boxed{\,E_\text{bind} = \frac{GMm}{2r}\,}

The binding energy is the magnitude of E. A satellite in a low orbit is more deeply bound (larger E_\text{bind}) than one in a high orbit.

Why not just "the potential energy"?

At first glance, you might expect the binding energy to be |U| = GMm/r — the depth of the potential well. But that would be the work needed to lift a stationary object out to infinity. The satellite is not stationary — it already has kinetic energy K = GMm/(2r) that is helping you. Giving up that kinetic energy buys you half of the potential-energy climb for free. The net energy you must add is:

E_\text{bind} = |U| - K = \frac{GMm}{r} - \frac{GMm}{2r} = \frac{GMm}{2r}

Why: the satellite arrives at infinity with zero kinetic energy (by definition of "barely escape"). Along the way, it started with kinetic energy K, and it had to climb out of a well of depth |U|. Energy conservation says you must supply the difference, which is exactly |U| - K = |U|/2.

This is the same number you would need to speed it up to escape velocity in place: if the satellite were at rest at radius r, you would need to give it \tfrac{1}{2}mv_\text{esc}^2 = GMm/r of kinetic energy to escape. But since it already has GMm/(2r), the top-up is only GMm/(2r). Different accounting, same answer.

Plugging in numbers — the ISS and geostationary binding

Take a 1000 kg satellite.

ISS (r = 6.78 × 10⁶ m):

E_\text{bind} = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{2 \times 6.78 \times 10^6} = \frac{3.986 \times 10^{17}}{1.356 \times 10^7} \approx 2.94 \times 10^{10} \text{ J}

That is 29.4 GJ per tonne — the energy cost of pushing a tonne of ISS to infinity and leaving it there at rest. For comparison, burning a litre of petrol releases about 33 MJ, so escaping the ISS altitude with a tonne of payload would take the chemical energy of roughly 890 000 litres of petrol — before accounting for any rocket-equation inefficiencies.

Geostationary (r = 4.22 × 10⁷ m):

E_\text{bind} = \frac{3.986 \times 10^{17}}{2 \times 4.22 \times 10^7} \approx 4.72 \times 10^{9} \text{ J}

About 4.72 GJ per tonne — six times less, because the satellite is already much higher up the well.

The binding energy is what mission planners pay to break a satellite free. It is also what you get back (as kinetic energy and stress heat) if the satellite falls back. When Skylab re-entered in 1979, all of its binding energy — several tens of gigajoules — had to go somewhere. Most became heat on the way down, a little became kinetic energy of fragments at impact.

The paradox of losing energy: drag makes you faster

Now to the most surprising consequence of E = -GMm/(2r). Solve this equation for r:

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

If E becomes more negative (the satellite loses energy), the magnitude |E| grows, so the denominator 2|E| grows, so r shrinks. The satellite drops to a lower orbit.

What happens to its speed? From the orbital-velocity formula, v = \sqrt{GM/r}, and since r shrank, v grew. The satellite speeds up as it loses energy.

Let's confirm by checking the kinetic energy directly. K = GMm/(2r) grows as r shrinks. So yes — the satellite's kinetic energy increases even as its total energy decreases.

How can this be? The satellite loses, say, 1 J of total energy to atmospheric drag. Where does it go? Not into kinetic energy — the satellite's kinetic energy actually grows by 1 J. The only place left is potential energy, and that is precisely where the accounting works:

\Delta E = \Delta K + \Delta U
-1 \text{ J} = +1 \text{ J} + \Delta U \quad \Rightarrow \quad \Delta U = -2 \text{ J}

The potential energy drops by 2 J — the satellite falls deeper into Earth's well. Of that 2 J drop, 1 J is converted into extra kinetic energy (making the satellite faster) and the other 1 J is lost as heat in the atmosphere. The drag force is doing negative work on the satellite equal to 1 J, but the gravitational force is doing positive work of 2 J, and the net is +1 J of kinetic energy gained.

Why this pattern is universal: any satellite losing energy (drag, gravitational radiation, a retrograde burn that is small enough not to eject it) will drop to a lower orbit and go faster. The rule of thumb: for a circular orbit, a unit of energy removed becomes 2 units of potential-energy drop and 1 unit of kinetic-energy gain. The system cannot escape this 2:1 partition because it is enforced by the virial theorem (next section).

The ISS's slow spiral

Every month, the ISS drops by about 2 km because of faint atmospheric drag (even at 408 km, there is enough residual atmosphere to bite). If nothing were done, it would speed up as it fell, its speed growing faster and faster as it accelerated toward denser air. Once it hit the serious atmosphere (around 100 km), it would break up and burn within minutes.

To prevent that, the ISS's thrusters periodically raise its orbit — a reboost. This is a counter-intuitive thing: to raise a satellite's orbit, you fire in the direction of motion (prograde). Firing prograde adds kinetic energy instantly, but then as the satellite coasts up to the higher orbit, it loses kinetic energy (because at higher r it needs to be slower) and gains potential energy. Net result: more total energy, higher orbit, lower speed.

So firing the engine in the forward direction slows you down, after coasting. This is not a typo. It is the same energy accounting as before, in reverse: \Delta K > 0 right at the burn, but after coasting to the new orbit, \Delta K < 0 (you are moving slower) and \Delta U > 0 (you are higher), with \Delta U = -2\Delta K so the net \Delta E is positive.

Orbital mechanics is full of these sign reversals, and they all come from the K = -U/2 rule for circular orbits.

The virial theorem — why this is universal

The relation \langle K \rangle = -\tfrac{1}{2}\langle U \rangle is not special to circular orbits. It is true, on average, for any gravitationally bound system in equilibrium: a pair of stars orbiting each other, a comet in a highly elliptical orbit (when averaged over a period), the molecules of a gas cloud held together by mutual gravity, the stars of a galaxy. The general statement is called the virial theorem.

Virial theorem for gravity (qualitative). For any stable gravitationally bound system in which the force between each pair of particles is \propto 1/r^2, the time-averaged kinetic and potential energies satisfy:

\langle K \rangle = -\tfrac{1}{2}\langle U \rangle

Equivalently, the time-averaged total energy is:

\langle E \rangle = \langle K \rangle + \langle U \rangle = \tfrac{1}{2}\langle U \rangle = -\langle K \rangle

For a circular orbit, the kinetic and potential energies are constant (not just their averages) — and so the virial relations hold exactly at every instant. For an elliptical orbit, K swings from a maximum at perihelion to a minimum at aphelion, and U mirrors it — but averaged over one full revolution, the virial 1:2 ratio is preserved.

Using the virial theorem to weigh galaxies

Astronomers use the virial theorem in reverse. If you can measure how fast the stars in a galaxy are moving — their velocity dispersion — you can compute \langle K \rangle. The virial theorem says this equals -\tfrac{1}{2}\langle U \rangle, which for a roughly spherical galaxy of radius R and total mass M_\text{galaxy} is of order GM_\text{galaxy}^2/R. Rearranging, you get the total gravitating mass of the galaxy — including the mass you cannot see directly.

This is how Fritz Zwicky, in 1933, first concluded that the Coma galaxy cluster contained far more mass than the visible stars — the discovery now called dark matter. The virial theorem, applied to galaxy motions, said the cluster must have about 400 times the mass of the visible galaxies. That factor of 400 was later refined, but the basic inference — that gravitational systems must obey the virial theorem, and the visible mass was not enough — has never been overturned.

Explore: the energy landscape of an orbit

The interactive figure below lets you drag the radius of a circular orbit (Earth satellite, M = M_E, m = 1 kg) between R_E (just above the surface) and 10 R_E. Watch how K, U, and E respond. The 1:2 ratio between K and |U| is built into the equations — the readouts let you verify it at any radius.

Interactive: K, U, and E versus orbit radius Three curves plotted against orbit radius measured in Earth radii. Kinetic energy per unit mass is positive and falls with r. Potential energy per unit mass is negative and rises toward zero. Total energy per unit mass sits between them, negative and approaching zero. 1 3 6 10 orbit radius r / R_E 0 +30 +60 −30 −60 energy per kg (MJ/kg) K U E drag the red point along the axis
Drag the red point to change the orbit radius (in Earth radii). The three curves are $K$ (red, positive), $U$ (dark, negative, twice as deep), and $E = K + U$ (dashed grey, negative, half of $U$). At every radius you can read off the values on the right; verify for yourself that $K = -U/2 = -E$, and that $|E|$ — the binding energy per kilogram — shrinks as you move to higher orbits. At $r = 1 R_E$ (Earth's surface), $|E|/m \approx 31.25$ MJ/kg — that is the energy cost, per kilogram, of escape from the surface.

Worked examples

Example 1: Energy to lift Chandrayaan-3 out of low Earth orbit

Chandrayaan-3 had a total mass of about 3900 kg at launch. Its initial Earth orbit was a low Earth parking orbit at roughly 170 km altitude. From there, a series of burns raised its apogee step by step, finally sending it toward the Moon. Take R_E = 6371 km, GM_E = 3.986 \times 10^{14} m³/s², and work out (a) the total mechanical energy of the spacecraft in the 170 km parking orbit, (b) its binding energy there, and (c) the minimum extra energy needed to reach a parabolic (just-barely-escape) trajectory from that orbit.

Chandrayaan-3 in low Earth parking orbit Earth in the centre with a dashed circular orbit at 170 km altitude. A small spacecraft symbol sits on the orbit. An outward arrow labelled "add energy to escape" points away, representing the burn needed to leave the parking orbit on a trans-lunar trajectory. Earth parking orbit: 170 km altitude C-3 v +ΔE to escape
Chandrayaan-3 in its initial 170 km parking orbit. Total mechanical energy is $E = -GMm/(2r)$; the binding energy $|E|$ is what you must add to reach a marginally-escape trajectory.

Step 1. Orbital radius.

r = R_E + h = 6371 + 170 = 6541 \text{ km} = 6.541 \times 10^6 \text{ m}

Why: orbital radius is measured from Earth's centre, not its surface.

Step 2. Total mechanical energy.

E = -\frac{GM_E m}{2r} = -\frac{(3.986 \times 10^{14})(3900)}{2 \times 6.541 \times 10^6}
E = -\frac{1.554 \times 10^{18}}{1.308 \times 10^7} \approx -1.19 \times 10^{11} \text{ J} = -119 \text{ GJ}

Why: negative, as expected for a bound satellite. Magnitude in gigajoules gives a sense of the energy scale — 119 GJ is comparable to the chemical energy in about 3500 litres of jet fuel.

Step 3. Binding energy.

E_\text{bind} = -E = 1.19 \times 10^{11} \text{ J} \approx 119 \text{ GJ}

Why: binding energy is just the negative of E — the amount you must add to drag the spacecraft out to infinity at rest.

Step 4. Minimum extra energy for a parabolic trajectory.

At the threshold (E = 0), the spacecraft just barely escapes, so the extra energy needed is exactly the binding energy itself:

\Delta E_\text{min} = 0 - E = 1.19 \times 10^{11} \text{ J}

Step 5. Sanity-check using the velocity increment.

Current orbital speed: v = \sqrt{GM_E/r} = \sqrt{3.986 \times 10^{14}/6.541 \times 10^6} \approx 7807 m/s.

Escape speed at this radius: v_\text{esc} = \sqrt{2}\,v \approx 11\,038 m/s.

Velocity increment needed: \Delta v = 11\,038 - 7807 \approx 3231 m/s.

Checking via energy: \tfrac{1}{2}m(v_\text{esc}^2 - v^2) = \tfrac{1}{2}(3900)((11\,038)^2 - (7807)^2) = \tfrac{1}{2}(3900)(6.10 \times 10^7) \approx 1.19 \times 10^{11} J.

Why: the two numbers match, confirming the energy accounting. Adding kinetic energy equal to the binding energy raises the orbit's total energy from -1.19 \times 10^{11} J to 0, which is the escape threshold.

Result: At its 170 km parking orbit, Chandrayaan-3 had E = -119 GJ. Its binding energy was 119 GJ. A velocity increment of about 3.2 km/s would just barely escape Earth.

What this shows: the total energy of a satellite is a single number that tells you everything — what orbit it is in, how hard it is to escape from, how much kinetic energy it has. ISRO's mission planners do this exact calculation for every burn: tally the \Delta E the thrusters need to supply, divide by the spacecraft mass and the exhaust-velocity-times-burn-time, and you have the burn duration.

Example 2: The ISS slowly spiralling down

Atmospheric drag at the ISS altitude removes about 7.5 GJ of total energy per month from the ISS (mass \approx 4.2 \times 10^5 kg, initial radius r_0 = 6.78 \times 10^6 m). Without a reboost, by how much would its orbital radius shrink in one month, and by how much would its orbital speed increase?

ISS orbital decay under atmospheric drag A diagram of Earth and two circular orbits — a larger original orbit with an ISS symbol, and a slightly smaller lower orbit. An inward-pointing dashed arrow connects them, showing the decay. Labels: initial speed 7.67 km/s, final speed slightly larger. Earth ISS (initial) v₀ ≈ 7.67 km/s lower orbit, faster
The ISS loses energy to drag and spirals inward. Counter-intuitively, it speeds up as it drops — the missing energy comes from $U$, not $K$.

Step 1. Change in total energy.

The ISS loses \Delta E = -7.5 \times 10^9 J in one month.

Step 2. Change in radius.

From E = -GMm/(2r), differentiate with respect to r:

\frac{dE}{dr} = \frac{GMm}{2r^2}

For a small change \Delta E, the corresponding radius change is:

\Delta r = \frac{2r^2}{GMm}\,\Delta E

Why: invert the derivative — if E depends on r through -GMm/(2r), then a small \Delta E is tied to a small \Delta r by this formula. The factor of r^2 on the top means high-altitude orbits respond more dramatically to the same energy loss than low orbits (for a fixed mass).

Plug in:

\Delta r = \frac{2 \times (6.78 \times 10^6)^2}{(3.986 \times 10^{14})(4.2 \times 10^5)} \times (-7.5 \times 10^9)

Numerator: 2 \times 4.596 \times 10^{13} = 9.192 \times 10^{13}.

Denominator: 3.986 \times 10^{14} \times 4.2 \times 10^5 = 1.674 \times 10^{20}.

\Delta r = \frac{9.192 \times 10^{13}}{1.674 \times 10^{20}} \times (-7.5 \times 10^9) = (5.49 \times 10^{-7}) \times (-7.5 \times 10^9)
\Delta r \approx -4120 \text{ m} \approx -4.1 \text{ km}

Why: the ISS drops by about 4 km per month under this drag rate. Real ISS logs show orbital decay of 2–5 km per month, depending on solar-cycle conditions that affect the atmosphere's scale height — the order of magnitude matches.

Step 3. Change in orbital speed.

v = \sqrt{\frac{GM}{r}} \quad\Rightarrow\quad \frac{dv}{dr} = -\frac{1}{2}\sqrt{\frac{GM}{r^3}} = -\frac{v}{2r}

So:

\Delta v \approx -\frac{v}{2r}\,\Delta r = -\frac{7669}{2 \times 6.78 \times 10^6} \times (-4120)
\Delta v \approx -5.66 \times 10^{-4} \times (-4120) \approx +2.33 \text{ m/s}

Why: \Delta r is negative (orbit shrinks), and dv/dr is negative (smaller r means faster), so the two negatives multiply to a positive \Delta v — the ISS speeds up by about 2.3 m/s.

Step 4. Verify the 2:1 partition.

Kinetic energy gained: \Delta K = mv\,\Delta v = (4.2 \times 10^5)(7669)(2.33) \approx +7.5 \times 10^9 J.

Potential energy change: \Delta U = -GMm \cdot (-1/r^2)\,\Delta r = (GMm/r^2)\Delta r = m g_\text{iss} \Delta r.

g_\text{iss} = GM/r^2 = 3.986 \times 10^{14} / (6.78 \times 10^6)^2 \approx 8.68 m/s² (Earth's gravity at ISS altitude).

\Delta U = (4.2 \times 10^5)(8.68)(-4120) \approx -1.50 \times 10^{10} J = -15.0 GJ.

So \Delta E = \Delta K + \Delta U = 7.5 - 15.0 = -7.5 GJ. ✓

Why: exactly what we expected. Of the 15 GJ drop in potential energy, 7.5 GJ becomes extra kinetic energy and the other 7.5 GJ is lost to drag heat. The 2:1 partition is automatic.

Result: In one month of unchecked drag, the ISS drops by 4 km, speeds up by 2.3 m/s, and sheds 7.5 GJ of total energy. Every so often, a Progress cargo vehicle docks, fires its engines prograde, and reboost lifts the station back up.

What this shows: the 2:1 virial partition governs the ISS's slow decay exactly as it governs Chandrayaan's burn: lose 1 unit of total energy, drop 2 units of potential energy, and gain 1 unit of kinetic energy. The same rule applies to the many thousands of tracked objects in low Earth orbit — all are slowly spiralling inward, faster as they go. The tons of satellite debris that burn up in Earth's atmosphere every year do so because this effect has been running for years or decades.

Example 3: Binding energies across the solar system

For a 1000 kg probe, compute the binding energy at the surface of (a) Earth, (b) Moon, and (c) Jupiter, and compare.

Use: G = 6.674 \times 10^{-11} N·m²/kg². M_E = 5.972 \times 10^{24} kg, R_E = 6.371 \times 10^6 m. M_\text{Moon} = 7.35 \times 10^{22} kg, R_\text{Moon} = 1.737 \times 10^6 m. M_J = 1.898 \times 10^{27} kg, R_J = 6.991 \times 10^7 m.

Step 1. Interpret "binding energy at the surface."

A probe sitting on the surface is not in orbit, so the circular-orbit formula E_\text{bind} = GMm/(2r) does not apply directly. Instead, the binding energy of a stationary mass is simply |U| = GMm/R — the full depth of the gravitational well. (A surface-orbit probe, skimming the surface at orbital speed, would have E_\text{bind} = GMm/(2R) — half as much.)

The task asks about binding energy at the surface, so we compute |U|.

Why distinguish: a "satellite at radius r" has kinetic energy on its side and so has only half the binding energy of a "stationary object at radius r." Which number to use depends on whether the object is in orbit or at rest.

Step 2. Earth surface binding.

|U_E| = \frac{GM_E m}{R_E} = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{6.371 \times 10^6} \approx 6.25 \times 10^{10} \text{ J}

About 62.5 GJ for 1000 kg. Per kilogram: 62.5 MJ/kg.

Step 3. Moon surface binding.

|U_M| = \frac{(6.674 \times 10^{-11})(7.35 \times 10^{22})(1000)}{1.737 \times 10^6} \approx 2.82 \times 10^9 \text{ J}

About 2.8 GJ for 1000 kg. Per kilogram: 2.8 MJ/kg. About 22 times less than Earth.

Step 4. Jupiter surface binding.

|U_J| = \frac{(6.674 \times 10^{-11})(1.898 \times 10^{27})(1000)}{6.991 \times 10^7} \approx 1.81 \times 10^{12} \text{ J}

About 1.81 TJ for 1000 kg. Per kilogram: 1.81 GJ/kg. About 29 times more than Earth.

Result: Binding energy per kilogram at the surface — Earth 62.5 MJ, Moon 2.8 MJ, Jupiter 1810 MJ.

What this shows: the Moon's tiny binding energy is why a human could in principle jump off it (if only humans could jump at 2.4 km/s) and why the Moon has no atmosphere — the thermal energy of gas molecules at the Moon's surface is comparable to its binding energy, so molecules escape. Jupiter's enormous binding energy is why it holds vast amounts of hydrogen and helium — even the lightest gases cannot escape. Every atmosphere we see is a balance between thermal energy and binding energy, and the binding-energy formula is where that balance is written.

Common confusions

You have the circular-orbit energy identities and the qualitative virial theorem. What follows is for readers who want the elliptical-orbit generalisation, a derivation of the virial theorem from Newton's laws, and the relativistic correction at very small radii (which is how black holes actually swallow energy).

The elliptical orbit — total energy depends only on a

For a general Keplerian ellipse with semi-major axis a and eccentricity \epsilon, the total mechanical energy of the orbiting body is:

\boxed{\,E = -\frac{GMm}{2a}\,}

This is a remarkable result — the energy depends on the semi-major axis alone, not on the eccentricity. Two orbits with the same a but different shapes (one nearly circular, one highly eccentric) have exactly the same total energy. The kinetic and potential energies swing around wildly over the course of a revolution, but their sum is a conserved constant equal to -GMm/(2a).

Setting a = r (for a circle) recovers the circular-orbit result of this article.

Derivation sketch. Start from energy conservation at perihelion (r_p = a(1-\epsilon)) and aphelion (r_a = a(1+\epsilon)), and from angular-momentum conservation which says v_p r_p = v_a r_a. Set the total mechanical energy equal at both points:

\tfrac{1}{2}mv_p^2 - \frac{GMm}{r_p} = \tfrac{1}{2}mv_a^2 - \frac{GMm}{r_a}

Substitute v_a = v_p r_p/r_a and solve for v_p^2:

v_p^2 = \frac{2GM/r_p}{1 + r_p/r_a} = \frac{2GM r_a}{r_p(r_p + r_a)} = \frac{2GM(1 + \epsilon)}{a(1 - \epsilon^2)(2)} = \frac{GM(1+\epsilon)}{a(1-\epsilon)}

Now plug this back into the total energy at perihelion:

E = \tfrac{1}{2}m v_p^2 - \frac{GMm}{r_p} = \tfrac{1}{2}m \cdot \frac{GM(1+\epsilon)}{a(1-\epsilon)} - \frac{GMm}{a(1-\epsilon)}
E = \frac{GMm}{a(1-\epsilon)}\left(\tfrac{1+\epsilon}{2} - 1\right) = \frac{GMm}{a(1-\epsilon)} \cdot \frac{\epsilon - 1}{2} = -\frac{GMm}{2a}

Why: the algebra telescopes because the ellipse's geometry r_p + r_a = 2a and r_p r_a = a^2(1-\epsilon^2) cancel all the \epsilon dependence from the final answer. The eccentricity disappears — a deep feature of 1/r^2 force.

So for any bound orbit, circular or elliptical, Kepler's third law T^2 = (4\pi^2/GM)a^3 and the energy formula E = -GMm/(2a) hold together. Fix the semi-major axis, and you have fixed both the period and the total energy, regardless of the orbit's shape.

The virial theorem — a one-line derivation

For a bound system of particles interacting gravitationally, define the virial:

\mathcal{V} = \sum_i \vec{r}_i \cdot \vec{F}_i

where the sum is over all particles, \vec{r}_i is the position of particle i from any fixed origin, and \vec{F}_i is the net force on particle i.

For a pair of gravitating particles, the force between them is \vec{F}_{ij} = -Gm_im_j(\vec{r}_i - \vec{r}_j)/|\vec{r}_i - \vec{r}_j|^3, and summing over all pairs gives \mathcal{V} = U (the total potential energy). The proof uses the fact that 1/r^2 is a homogeneous function of degree -2 in position, so Euler's identity for homogeneous functions applies.

Separately, if you write Newton's second law \vec{F}_i = m_i \ddot{\vec{r}}_i and dot with \vec{r}_i, you get:

\vec{r}_i \cdot \vec{F}_i = m_i \vec{r}_i \cdot \ddot{\vec{r}}_i = \frac{d}{dt}(m_i \vec{r}_i \cdot \dot{\vec{r}}_i) - m_i \dot{\vec{r}}_i \cdot \dot{\vec{r}}_i = \frac{d}{dt}(\cdot) - 2K_i

Summing over i and time-averaging over a long time (for a stable bound orbit, the total derivative averages to zero):

\langle \mathcal{V} \rangle = -2\langle K \rangle

Since \mathcal{V} = U for gravity:

\langle U \rangle = -2\langle K \rangle \quad \Leftrightarrow \quad \boxed{\langle K \rangle = -\tfrac{1}{2}\langle U \rangle}

This is the virial theorem for any inverse-square attractive force. For power-law forces F \propto r^{n} (so U \propto r^{n+1}), the corresponding theorem is \langle K \rangle = \tfrac{n+1}{2}\langle U \rangle. Gravity has n = -2, giving the coefficient -1/2.

Why losing energy makes things hotter — the negative heat capacity

The kinetic energy of a gas is, by equipartition, related to its temperature: K = \tfrac{3}{2}Nk_BT for an ideal monatomic gas. The virial theorem says that for a self-gravitating bound system, K = -\tfrac{1}{2}U. So as the system loses total energy (radiating into space, for instance), K increases. Therefore the temperature increases.

This is called negative heat capacity: lose energy, get hotter. It is one of the most startling features of self-gravitating systems, and it is why stars are stable. A star radiates energy from its surface. If it were an ordinary body, losing energy would make it cool, slowing the nuclear burn and putting out the fire. But because it is self-gravitating, losing energy makes it hotter, speeding up the nuclear reactions, keeping the star burning. The virial theorem is the reason stars are thermodynamically stable for billions of years.

The same principle applies to globular clusters and galaxy clusters: they evolve over cosmic timescales by losing kinetic energy (via stars ejected from the core, or dynamical friction), and as they do, they contract and get hotter in the centre. This is the "gravothermal catastrophe" of dense stellar systems, first understood in the 1960s.

The Schwarzschild radius, one more time

In the escape-velocity article, the Schwarzschild radius was defined by v_\text{esc} = c, giving r_s = 2GM/c^2. Using the binding-energy picture, you can say: at r = r_s, the binding energy of a mass m equals its rest energy:

E_\text{bind}(r_s) = \frac{GMm}{2r_s} = \frac{GMm}{2 \cdot 2GM/c^2} = \frac{mc^2}{4}

(A factor of 4 rather than 2 because we're using the binding energy for an orbit at r_s, not the potential-energy magnitude at r_s.) Either way, the binding energy scales up to a significant fraction of the rest energy mc^2 for compact objects — a sign that Newtonian gravity has broken down and general relativity must take over. In the exact Schwarzschild geometry, an orbit at r = r_s does not even exist (the innermost stable circular orbit is at r = 3r_s), and the binding energy of a particle at the innermost stable orbit is about 5.7\% of mc^2 — the energy released per unit mass when matter spirals into a black hole from far away. This is the huge luminosity of quasars and active galactic nuclei: the virial theorem, carried to its relativistic extreme.

Where this leads next