Energy of Orbiting Satellites

Understanding the energy of satellites in orbit brings together gravitational potential energy and kinetic energy in an elegant way. The relationships between these energies are central to understanding how satellites are launched, how they change orbit, and why they eventually de-orbit.

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Kinetic Energy of a Satellite in Orbit

For a satellite in a circular orbit, we established that:

$v^2 = \frac{GM}{r}$v2=rGM​

The kinetic energy is therefore:

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}m \times \frac{GM}{r} = \frac{GMm}{2r}$Ek​=21​mv2=21​m×rGM​=2rGMm​

Note that kinetic energy is always positive.

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Gravitational Potential Energy in Orbit

The gravitational potential energy at orbital radius r is:

$E_p = -\frac{GMm}{r}$Ep​=−rGMm​

This is always negative (the satellite is in a bound state).

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Total Energy of an Orbiting Satellite

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

$E_{\text{total}} = E_k + E_p = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right) = -\frac{GMm}{2r}$Etotal​=Ek​+Ep​=2rGMm​+(−rGMm​)=−2rGMm​

This gives us three beautifully related results:

Energy	Expression	Sign	Relationship
Kinetic energy E_k	+GMm/(2r)	Positive	—
Potential energy E_p	−GMm/r	Negative	E_p = −2E_k
Total energy E_total	−GMm/(2r)	Negative	E_total = −E_k

The total energy is negative, confirming the satellite is bound (it cannot escape without being given energy). The total energy equals the negative of the kinetic energy, and the potential energy is exactly twice the total energy.

Common exam mistake: Students sometimes add E_k and E_p incorrectly by forgetting the negative sign on E_p. The total energy of a bound orbit is always negative. If you get a positive total energy, the object has enough energy to escape — it is not in a bound orbit.

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Energy Diagram

flowchart TD
    subgraph Energy["Energy vs Orbital Radius"]
        direction TB
        A["As r increases:"]
        B["E_k = GMm/(2r) DECREASES\n(satellite moves slower)"]
        C["E_p = −GMm/r INCREASES\n(becomes less negative)"]
        D["E_total = −GMm/(2r) INCREASES\n(becomes less negative)"]
    end
    subgraph Higher["Moving to Higher Orbit"]
        E["Must ADD energy to satellite"]
        F["KE decreases → satellite slows down"]
        G["PE increases more than KE decreases"]
        H["Total energy increases (less negative)"]
    end
    subgraph Lower["Moving to Lower Orbit"]
        I["Satellite LOSES energy"]
        J["KE increases → satellite speeds up"]
        K["PE decreases more than KE increases"]
        L["Total energy decreases (more negative)"]
    end
    Energy --> Higher
    Energy --> Lower

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The Counterintuitive Result: Higher Orbits Are Slower

This confuses many students. To move a satellite to a higher orbit, you must give it energy (fire rockets forward). Yet the satellite ends up travelling slower than before. How?

The answer lies in the energy distribution. When you boost a satellite to a higher orbit:

• You supply energy, increasing the total energy
• The potential energy increases by a large amount (becomes much less negative)
• The kinetic energy decreases (the satellite slows down)
• The net result is an increase in total energy, even though KE has decreased

Think of it like climbing a hill: you expend energy, gain height (PE increases), but you might be walking more slowly at the top.

Worked Example 1: Energy Budget for Orbit Raising

A 500 kg satellite moves from orbit at r₁ = 7.0 × 10⁶ m to r₂ = 1.4 × 10⁷ m around Earth (M = 5.97 × 10²⁴ kg).

At r₁:

$E_k = \frac{GMm}{2r_1} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{2 \times 7.0 \times 10^6} = \frac{1.992 \times 10^{17}}{1.4 \times 10^7} = 1.423 \times 10^{10} \text{ J}$Ek​=2r1​GMm​=2×7.0×1066.674×10−11×5.97×1024×500​=1.4×1071.992×1017​=1.423×1010 J

$E_p = -\frac{GMm}{r_1} = -2.846 \times 10^{10} \text{ J}$Ep​=−r1​GMm​=−2.846×1010 J

$E_{\text{total},1} = -1.423 \times 10^{10} \text{ J}$Etotal,1​=−1.423×1010 J

At r₂ = 2r₁:

$E_k = \frac{GMm}{2r_2} = \frac{1.992 \times 10^{17}}{2.8 \times 10^7} = 7.114 \times 10^{9} \text{ J}$Ek​=2r2​GMm​=2.8×1071.992×1017​=7.114×109 J

$E_p = -\frac{GMm}{r_2} = -1.423 \times 10^{10} \text{ J}$Ep​=−r2​GMm​=−1.423×1010 J

$E_{\text{total},2} = -7.114 \times 10^{9} \text{ J}$Etotal,2​=−7.114×109 J

Energy that must be supplied:

$\Delta E = E_{\text{total},2} - E_{\text{total},1} = (-7.114 \times 10^9) - (-1.423 \times 10^{10}) = +7.11 \times 10^{9} \text{ J} \approx 7.1 \text{ GJ}$ΔE=Etotal,2​−Etotal,1​=(−7.114×109)−(−1.423×1010)=+7.11×109 J≈7.1 GJ

Confirmation: the speed drops from v₁ = √(GM/r₁) = 7,540 m s⁻¹ to v₂ = √(GM/r₂) = 5,330 m s⁻¹.

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Hohmann Transfer Orbits

In practice, orbit changes are performed using a Hohmann transfer — an elliptical orbit that connects two circular orbits using two engine burns:

1. First burn (at lower orbit): Fire engines forward to increase speed. The satellite enters an elliptical transfer orbit with its perigee at r₁ and apogee at r₂.
2. Coast phase: The satellite follows the transfer ellipse, gaining PE and losing KE as it climbs.
3. Second burn (at higher orbit): Fire engines forward again to circularise the orbit at r₂.

This is the most fuel-efficient way to change between coplanar circular orbits and is used for virtually all geostationary satellite deployments.

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De-Orbiting and Atmospheric Drag

For satellites in low Earth orbit, the thin upper atmosphere creates a small drag force. Over time, this drag:

• Removes energy from the satellite (E_total decreases / becomes more negative)
• Causes the orbital radius to decrease
• Causes the satellite to speed up (E_k increases as r decreases)

This seems paradoxical: drag removes energy, yet the satellite goes faster. The resolution is that the PE decreases (becomes more negative) faster than the KE increases. The net energy is being lost — converted to thermal energy through friction with the atmosphere.

Eventually, the satellite descends low enough for drag to become dominant, and it burns up or crashes — this is re-entry.

Worked Example 2: Speed Increase During Orbital Decay

A satellite's orbital radius decreases from 7.0 × 10⁶ m to 6.6 × 10⁶ m due to atmospheric drag. By how much does its speed increase?

$v_1 = \sqrt{\frac{GM}{r_1}} = \sqrt{\frac{3.984 \times 10^{14}}{7.0 \times 10^6}} = 7,544 \text{ m s}^{-1}$v1​=r1​GM​​=7.0×1063.984×1014​​=7,544 m s−1

$v_2 = \sqrt{\frac{GM}{r_2}} = \sqrt{\frac{3.984 \times 10^{14}}{6.6 \times 10^6}} = 7,769 \text{ m s}^{-1}$v2​=r2​GM​​=6.6×1063.984×1014​​=7,769 m s−1

Speed increase: Δv = 225 m s⁻¹ (the satellite is going faster despite losing total energy).

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Relationship Between Escape Velocity and Orbital Speed

At any orbital radius r, the escape velocity is:

$v_{\text{esc}} = \sqrt{\frac{2GM}{r}} = \sqrt{2} \times \sqrt{\frac{GM}{r}} = \sqrt{2} \times v_{\text{orbital}}$vesc​=r2GM​​=2​×rGM​​=2​×vorbital​

So the escape velocity is always √2 (≈ 1.414) times the circular orbital speed at that radius. To escape from a circular orbit, a satellite must increase its speed by a factor of √2 — that is, by about 41%.

Worked Example 3: Escaping from the ISS Orbit

The ISS orbits at v ≈ 7,665 m s⁻¹. To escape:

$v_{\text{esc}} = \sqrt{2} \times 7665 = 10,840 \text{ m s}^{-1}$vesc​=2​×7665=10,840 m s−1

The required speed increase is: Δv = 10,840 − 7,665 = 3,175 m s⁻¹

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Energy Required for Launch

To place a satellite into orbit from Earth's surface, you must supply energy to:

1. Increase the gravitational PE from the surface to the orbital radius
2. Provide the kinetic energy for orbital speed
3. Overcome atmospheric drag during ascent (not calculable from simple theory)

The minimum energy (ignoring drag and assuming launch from rest) is:

$E_{\text{launch}} = E_{\text{total},\text{orbit}} - E_{\text{total},\text{surface}} = \left(-\frac{GMm}{2r}\right) - \left(-\frac{GMm}{2R}\right) = \frac{GMm}{2}\left(\frac{1}{R} - \frac{1}{r}\right)$Elaunch​=Etotal,orbit​−Etotal,surface​=(−2rGMm​)−(−2RGMm​)=2GMm​(R1​−r1​)

Wait — this is not quite right because the object is at rest on the surface, so E_total at the surface is just E_p (no orbital KE):

$E_{\text{launch}} = \left(-\frac{GMm}{2r}\right) - \left(-\frac{GMm}{R}\right) = GMm\left(\frac{1}{R} - \frac{1}{2r}\right)$Elaunch​=(−2rGMm​)−(−RGMm​)=GMm(R1​−2r1​)

Worked Example 4: Minimum Launch Energy for Geostationary Orbit

For a 2000 kg satellite to geostationary orbit (r = 4.22 × 10⁷ m) from Earth's surface (R = 6.37 × 10⁶ m):

$E_{\text{launch}} = GMm\left(\frac{1}{R} - \frac{1}{2r}\right) = 3.984 \times 10^{14} \times 2000 \times \left(\frac{1}{6.37 \times 10^6} - \frac{1}{8.44 \times 10^7}\right)$Elaunch​=GMm(R1​−2r1​)=3.984×1014×2000×(6.37×1061​−8.44×1071​)

$= 7.968 \times 10^{17} \times (1.570 \times 10^{-7} - 1.185 \times 10^{-8})$=7.968×1017×(1.570×10−7−1.185×10−8)

$= 7.968 \times 10^{17} \times 1.452 \times 10^{-7} = 1.157 \times 10^{11} \text{ J} \approx 116 \text{ GJ}$=7.968×1017×1.452×10−7=1.157×1011 J≈116 GJ

In practice, much more energy is needed due to atmospheric drag and the inefficiency of rocket engines.

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Summary

• E_k = GMm/(2r), E_p = −GMm/r, E_total = −GMm/(2r)
• Total energy is negative for bound orbits; |E_total| = E_k; E_p = −2E_k
• Higher orbits: more total energy, lower speed, lower KE, higher (less negative) PE
• Atmospheric drag removes energy but the satellite speeds up as r decreases
• Escape velocity = √2 × orbital velocity at any radius
• Hohmann transfers use two burns to move between circular orbits efficiently

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A-Level Deep Dive: Energy of Orbiting Satellites

Spec mapping

Edexcel 9PH0 specification Topic 12 — Gravitational fields and astrophysics addresses the energy of bodies in circular orbit, including the link between kinetic energy, gravitational potential energy and total mechanical energy for an orbiting satellite (refer to the official specification document for exact wording). The topic sits in Paper 2 (Advanced Physics II) and is examined synoptically with circular motion (Topic 6) and gravitational fields proper (earlier in Topic 12). Candidates are expected to derive $E_k = GMm/(2r)$Ek​=GMm/(2r) from $v^2 = GM/r$v2=GM/r, combine it with the gravitational potential $V_g = -GM/r$Vg​=−GM/r to obtain $E_{\text{total}} = -GMm/(2r)$Etotal​=−GMm/(2r), and use these to analyse orbit-raising, orbit-lowering, atmospheric drag and escape conditions. The Edexcel data and formulae sheet provides $F = Gm_1 m_2 / r^2$F=Gm1​m2​/r2 and $V_g = -GM/r$Vg​=−GM/r, but the energy expressions for circular orbits are derived results that must be reproduced under exam conditions.

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Worked example with full mark scheme

Question (8 marks): A satellite of mass $m = 1200\,\text{kg}$m=1200kg is in a circular orbit of radius $r_1 = 7.20 \times 10^6\,\text{m}$r1​=7.20×106m around the Earth. It is to be raised to a higher circular orbit of radius $r_2 = 1.20 \times 10^7\,\text{m}$r2​=1.20×107m. Take $GM_E = 3.99 \times 10^{14}\,\text{N m}^2\,\text{kg}^{-1}$GME​=3.99×1014N m2kg−1.

(a) Show that the total mechanical energy of a satellite of mass $m$m in a circular orbit of radius $r$r is $E_{\text{total}} = -GMm/(2r)$Etotal​=−GMm/(2r). (3)

(b) Calculate the minimum energy required to raise the satellite from $r_1$r1​ to $r_2$r2​. (4)

(c) State, with a reason, how the kinetic energy of the satellite changes during the orbit raise. (1)

Solution with mark scheme:

(a) Step 1 — express orbital speed using Newton's second law. For circular motion the gravitational force provides the centripetal force: $GMm/r^2 = mv^2/r$GMm/r2=mv2/r, so $v^2 = GM/r$v2=GM/r.

M1 — equating gravitational force to centripetal requirement and isolating $v^2$v2.

Step 2 — write the kinetic energy.

$E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m \cdot \dfrac{GM}{r} = \dfrac{GMm}{2r}$Ek​=21​mv2=21​m⋅rGM​=2rGMm​

A1 — correct $E_k = GMm/(2r)$Ek​=GMm/(2r).

Step 3 — add the gravitational potential energy. $E_p = -GMm/r$Ep​=−GMm/r (data sheet). Then

$E_{\text{total}} = E_k + E_p = \dfrac{GMm}{2r} - \dfrac{GMm}{r} = -\dfrac{GMm}{2r}$Etotal​=Ek​+Ep​=2rGMm​−rGMm​=−2rGMm​

A1 — correct combination producing the printed expression with the correct sign.

(b) Step 1 — write the energy difference.

$\Delta E = E_2 - E_1 = -\dfrac{GMm}{2r_2} - \left(-\dfrac{GMm}{2r_1}\right) = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ΔE=E2​−E1​=−2r2​GMm​−(−2r1​GMm​)=2GMm​(r1​1​−r2​1​)

M1 — correct difference of total energies, with the sign handled correctly so that $\Delta E > 0$ΔE>0 for $r_2 > r_1$r2​>r1​.

Step 2 — substitute values.

$\dfrac{GMm}{2} = \dfrac{(3.99 \times 10^{14})(1200)}{2} = 2.394 \times 10^{17}\,\text{J m}$2GMm​=2(3.99×1014)(1200)​=2.394×1017J m

M1 — correct numerical evaluation of the prefactor (units of J m because the bracket has units of $\text{m}^{-1}$m−1).

Step 3 — evaluate the bracket.

$\dfrac{1}{r_1} - \dfrac{1}{r_2} = \dfrac{1}{7.20 \times 10^6} - \dfrac{1}{1.20 \times 10^7} = (1.389 - 0.833) \times 10^{-7} = 5.56 \times 10^{-8}\,\text{m}^{-1}$r1​1​−r2​1​=7.20×1061​−1.20×1071​=(1.389−0.833)×10−7=5.56×10−8m−1

A1 — correct bracket value to 3 s.f.

Step 4 — combine.

$\Delta E = 2.394 \times 10^{17} \times 5.56 \times 10^{-8} \approx 1.33 \times 10^{10}\,\text{J} \approx 13.3\,\text{GJ}$ΔE=2.394×1017×5.56×10−8≈1.33×1010J≈13.3GJ

A1 — final answer to 3 s.f. with correct unit.

(c) The kinetic energy decreases. Since $E_k = GMm/(2r)$Ek​=GMm/(2r) falls as $r$r increases, the satellite is moving more slowly in the higher orbit even though work has been done on it.

B1 — correct direction with reason linked to $E_k \propto 1/r$Ek​∝1/r.

Total: 8 marks (M3 A4 B1).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A satellite is in a low circular orbit of radius $r$r around the Earth. Atmospheric drag does work $W$W (negative) on the satellite over many orbits, leaving it in a lower circular orbit of radius $r' < r$r′<r.

(a) Explain why the speed of the satellite increases even though the drag force opposes its motion. (3)

(b) Show that the magnitude of the energy lost to drag equals the gain in kinetic energy. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.2) — stating that $E_k = GMm/(2r)$Ek​=GMm/(2r) for a circular orbit, so $E_k$Ek​ increases as $r$r decreases.
• M1 (AO2.1) — recognising that as the satellite spirals inward, $r$r decreases, hence speed $v = \sqrt{GM/r}$v=GM/r​ rises.
• A1 (AO2.2) — clear conclusion: drag removes total energy, but the lost potential energy is more than the kinetic gain, so KE rises despite the retarding force. Examiners reward the explicit chain "drag → $\Delta E_{\text{total}} < 0$ΔEtotal​<0 → $r$r decreases → $E_k$Ek​ increases".

(b)

• M1 (AO1.1) — $\Delta E_{\text{total}} = -GMm/(2r') - (-GMm/(2r)) = -W_{\text{drag}}$ΔEtotal​=−GMm/(2r′)−(−GMm/(2r))=−Wdrag​ in magnitude (the satellite has lost $|W|$∣W∣ of total energy).
• M1 (AO2.1) — $\Delta E_k = GMm/(2r') - GMm/(2r)$ΔEk​=GMm/(2r′)−GMm/(2r), which equals $|\Delta E_{\text{total}}|$∣ΔEtotal​∣ in magnitude because $E_{\text{total}} = -E_k$Etotal​=−Ek​ for any circular orbit.
• A1 (AO2.2) — conclusion stated explicitly: $|W| = \Delta E_k$∣W∣=ΔEk​, so the kinetic energy gain numerically equals the energy dissipated by drag (the rest comes from the potential well).

Total: 6 marks split AO1 = 2, AO2 = 4. This question rewards qualitative reasoning about signs and the relationship $E_{\text{total}} = -E_k$Etotal​=−Ek​ — a synoptic AO2 task that distinguishes B/A candidates from A* candidates.

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