Orbits — Circular Satellite Motion

Spec mapping: OCR H556 Module 5.4 — Gravitational fields (gravity providing centripetal force for a circular orbit; orbital speed $v = \sqrt{GM/r}$ ; orbital period $T = 2\pi\sqrt{r^{3}/(GM)}$ ; kinetic, potential, and total energy of a satellite in a circular orbit; weightlessness in orbit explained as free fall rather than absence of gravity). Refer to the official OCR H556 specification document for exact wording.

This lesson puts everything from Lessons 7–10 to work. A satellite in a circular orbit is a perfect laboratory: Newton's law of gravitation supplies the inward force, the centripetal-force equation matches it to the orbital motion, and gravitational potential energy and kinetic energy combine to give the total mechanical energy of a bound orbit. From these ingredients we derive the orbital speed $v = \sqrt{GM/r}$ , the period $T = 2\pi\sqrt{r^{3}/(GM)}$ (Kepler's third law again), and the orbital energy $E_{\text{total}} = -GMm/(2r)$ . We then apply the formulae to the ISS, geostationary orbit, and the genuinely counterintuitive case of orbital decay — air resistance increases the orbital speed.

The lesson also lands the single biggest popular misconception in space physics: astronauts on the ISS are not in "zero gravity". The gravitational field at 400 km altitude is only 11% less than at the surface. Weightlessness in orbit is free fall, not absence of gravity. The A* discriminator is the ability to explain the distinction between "gravity acts on the astronaut" (true at all times) and "the astronaut feels weight" (only when a contact force opposes gravity). Lesson 9's escape velocity provides the natural ceiling: $v_{\text{esc}} = \sqrt{2GM/r} = \sqrt{2} \times v_{\text{orbit}}$ , the $\sqrt{2}$ ratio between bound circular motion and just-escape.

The Condition for a Circular Orbit

A satellite of mass $m$ in a circular orbit of radius $r$ around a much larger central body of mass $M$ must obey two constraints:

Gravity provides the centripetal force. The only force on the satellite is the inward gravitational pull from the central body; this must equal the centripetal force required for circular motion at the observed speed:

$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}$

The gravitational force does no work on the orbit. Because gravity points along $\hat{r}$ (inward) and the velocity points along $\hat{\theta}$ (tangential), $\vec{F} \cdot \vec{v} = 0$ at every instant. Gravity bends the trajectory but does not change the satellite's kinetic energy. As a consequence, the orbital speed is constant — circular orbits are uniform circular motion at fixed $v$ .

From the first constraint, $m$ cancels and rearranging gives the orbital speed:

$v = \sqrt{\frac{GM}{r}}$

Three immediate consequences:

The orbital speed depends only on $M$ and $r$ , not on the satellite's mass. Saturn V and a ball-bearing at the same altitude orbit at the same speed.
Higher orbits are slower. $v \propto 1/\sqrt{r}$ , the opposite of the everyday intuition that "faster things go further". Doubling the orbital radius reduces $v$ by a factor of $\sqrt{2}$ .
Low orbits are fast. The ISS at 400 km altitude orbits at ~7.67 km s⁻¹ — fast enough to cross London in 5 seconds.

Worked Check — ISS Orbital Speed

Using $M_{E} = 5.97 \times 10^{24}$ kg and $r = R_{E} + h = 6.77 \times 10^{6}$ m for the ISS:

$v = \sqrt{\frac{GM_{E}}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^{6}}} = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^{6}}} = \sqrt{5.88 \times 10^{7}} = 7.67 \times 10^{3} \text{ m s}^{-1}$

About 7.67 km s⁻¹ — matching the operational ISS speed. The figure depends only on the Earth's mass and the orbital radius, never on the spacecraft's own mass.

Orbital Period — Kepler's Third Law Revisited

From Lesson 10, Kepler's third law (derived for a circular orbit) gives

$T^{2} = \frac{4\pi^{2}}{GM} r^{3}$

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

An equivalent form uses $v = \sqrt{GM/r}$ and the orbital circumference $2\pi r$ :

$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi r \times \sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^{3}}{GM}}$

— the same answer reached by either route. Choose whichever matches your data: if you know $r$ and $M$ , use $T = 2\pi\sqrt{r^{3}/(GM)}$ directly; if you already have $v$ , use $T = 2\pi r / v$ .

Orbital Energy

For a satellite in a circular orbit, the kinetic, potential, and total energies all take clean forms.

Kinetic energy. Using $v^{2} = GM/r$ :

$E_{k} = \tfrac{1}{2} m v^{2} = \tfrac{1}{2} m \cdot \frac{GM}{r} = \frac{GMm}{2r}$

— positive, as KE always is.

Gravitational potential energy (Lesson 9):

$E_{p} = -\frac{GMm}{r}$

— negative, with the conventional reference $V(\infty) = 0$ .

Total mechanical energy (the sum):

$E_{\text{total}} = E_{k} + E_{p} = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

Four observations:

The total energy is negative, like the potential energy alone. Negative total energy is the diagnostic signature of a bound orbit — to escape to infinity (where the energy would be zero or positive), the satellite would need an external energy boost. Bound orbits have $E_{\text{total}} < 0$ ; the escape limit is $E_{\text{total}} = 0$ ; hyperbolic flybys have $E_{\text{total}} > 0$ .
$E_{k} = -\tfrac{1}{2} E_{p}$ , equivalently $E_{\text{total}} = -E_{k} = +\tfrac{1}{2} E_{p}$ . This is the virial theorem for inverse-square forces, applicable to bound orbits (averaged over a period for non-circular cases).
$|E_{\text{total}}| = \tfrac{1}{2}|E_{p}|$ . The depth of the energy well at radius $r$ is half the depth of the potential well at the same radius — the other half is bound up as orbital kinetic energy.
Higher orbits have larger (less negative) total energy. $E_{\text{total}} \propto -1/r$ ; increasing $r$ makes $E_{\text{total}}$ less negative. To boost a satellite to a higher orbit, you must do positive work on it — even though the satellite's KE actually decreases in the higher orbit. The KE drops by a small amount; the PE rises by twice as much; the total rises (becomes less negative).

The fourth point is genuinely surprising. A satellite raised from 300 km to 1000 km altitude moves slower in the higher orbit ( $v_{1000} < v_{300}$ ), but its total energy is higher because the gain in PE outweighs the loss in KE. Boosting outward costs more potential energy than you save in kinetic — hence the need for rocket propulsion.

The Escape-Velocity Connection

From Lesson 9, $v_{\text{esc}} = \sqrt{2GM/r}$ at radius $r$ . The ratio of escape speed to circular orbital speed at the same radius is

$\frac{v_{\text{esc}}}{v_{\text{orbit}}} = \frac{\sqrt{2GM/r}}{\sqrt{GM/r}} = \sqrt{2}$

— a clean factor of $\sqrt{2}$ . Doubling the kinetic energy of a satellite in circular orbit (i.e. multiplying its speed by $\sqrt{2}$ ) raises its total energy to exactly zero, making it just-bound. From a bound orbit to an escape trajectory is one factor of $\sqrt{2}$ in speed, equivalently one factor of 2 in kinetic energy.

Worked Example 1 — Speed and Period at Geostationary Altitude

A geostationary satellite orbits at $h = 36\,000$ km above the Earth's surface. Calculate (a) orbital radius, (b) orbital speed, (c) period.

(a) Orbital radius: $r = R_{E} + h = 6.37 \times 10^{6} + 3.60 \times 10^{7} = 4.24 \times 10^{7}$ m.

(b) Orbital speed:

$v = \sqrt{\frac{GM_{E}}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{4.24 \times 10^{7}}} = \sqrt{9.39 \times 10^{6}} = 3.07 \times 10^{3} \text{ m s}^{-1} \approx 3.07 \text{ km s}^{-1}$

About 3.07 km s⁻¹ — much slower than the ISS's 7.67 km s⁻¹, consistent with the $1/\sqrt{r}$ scaling.

(c) Period:

$T = \frac{2\pi r}{v} = \frac{2\pi \times 4.24 \times 10^{7}}{3.07 \times 10^{3}} = \frac{2.66 \times 10^{8}}{3.07 \times 10^{3}} = 8.67 \times 10^{4} \text{ s} \approx 24.1 \text{ h}$

Just under one sidereal day, as required for a geostationary orbit (Lesson 12 develops this).

Worked Example 2 — Orbital Decay (The Counterintuitive Speedup)

A satellite in low Earth orbit at 300 km altitude decays under atmospheric drag to 200 km altitude. What is the change in its orbital speed? Does it speed up or slow down?

At 300 km ( $r_{1} = 6.67 \times 10^{6}$ m):

$v_{1} = \sqrt{\frac{GM_{E}}{r_{1}}} = \sqrt{\frac{3.98 \times 10^{14}}{6.67 \times 10^{6}}} = \sqrt{5.97 \times 10^{7}} = 7.73 \times 10^{3} \text{ m s}^{-1}$

At 200 km ( $r_{2} = 6.57 \times 10^{6}$ m):

$v_{2} = \sqrt{\frac{GM_{E}}{r_{2}}} = \sqrt{\frac{3.98 \times 10^{14}}{6.57 \times 10^{6}}} = \sqrt{6.06 \times 10^{7}} = 7.78 \times 10^{3} \text{ m s}^{-1}$

Change in speed: $\Delta v = v_{2} - v_{1} = +0.05$ km s⁻¹ = $+50$ m s⁻¹. The satellite speeds up as its orbit decays.

This is the orbital paradox: drag removes energy from the satellite, yet increases its speed. Resolution requires looking at total energy. Drag reduces $E_{\text{total}}$ , which is $-GMm/(2r)$ ; since $E_{\text{total}}$ is negative and getting more negative, $r$ must decrease. In the lower orbit, $E_{p} = -GMm/r$ becomes more negative by twice as much as $E_{k} = GMm/(2r)$ becomes more positive. So:

$E_{p}$ drops (more negative).
$E_{k}$ rises (faster speed).
$E_{\text{total}}$ drops (less than $E_{p}$ does, but still down) — exactly the energy lost to drag.

Energy bookkeeping: the potential energy released by descending in altitude is split between increased kinetic energy and energy dissipated to atmospheric drag. The drag robs energy from the orbit, the satellite responds by descending, the descent releases more PE than the drag has stolen, and the surplus goes into KE — hence the speedup.

A discriminator.* When OCR asks "what happens to the kinetic energy of a satellite whose orbit decays?", the correct response is: KE increases, PE decreases by more, total energy decreases by the difference. Stating "KE decreases because the satellite slows" is the canonical mid-band mistake — it imports horizontal-motion intuition into orbital mechanics where it does not apply.

Worked Example 3 — Energy Needed to Raise a Satellite

A 1000 kg satellite is to be moved from a circular orbit at 300 km altitude to a circular orbit at 1000 km altitude. Calculate the total work needed.

Total energy of a circular orbit: $E_{\text{total}} = -GMm/(2r)$ .

$\Delta E = E_{2} - E_{1} = -\frac{GMm}{2r_{2}} - \left(-\frac{GMm}{2r_{1}}\right) = \frac{GMm}{2}\left(\frac{1}{r_{1}} - \frac{1}{r_{2}}\right)$

At 300 km: $r_{1} = 6.67 \times 10^{6}$ m. At 1000 km: $r_{2} = 7.37 \times 10^{6}$ m.

$\frac{1}{r_{1}} - \frac{1}{r_{2}} = \frac{1}{6.67 \times 10^{6}} - \frac{1}{7.37 \times 10^{6}} = (1.500 - 1.357) \times 10^{-7} = 1.43 \times 10^{-8} \text{ m}^{-1}$

$\Delta E = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000}{2} \times 1.43 \times 10^{-8} = 1.99 \times 10^{17} \times 1.43 \times 10^{-8} = 2.84 \times 10^{9} \text{ J}$

About 2.84 GJ — comparable to the chemical energy in 700 kg of TNT. Launching and manoeuvring in space is energetically expensive. Note that $\Delta E > 0$ — the agent does positive work on the satellite, even though the satellite's kinetic energy falls in the higher orbit. The PE rises by ~ $5.7$ GJ; the KE falls by ~ $2.8$ GJ; the net is the $+2.84$ GJ supplied externally.

Worked Example 4 — Orbit Selection Decision Tree

Which orbital altitude to choose for a given mission? The decision tree summarises the dominant trade-offs.

graph TD
    A["Mission requirement"] --> B{"Need to stay over<br/>one ground point?"}
    B -- "Yes" --> C["Geostationary orbit<br/>r ≈ 42 200 km, T = 24 h<br/>(comms, weather)"]
    B -- "No" --> D{"Need high-res imaging<br/>or polar coverage?"}
    D -- "Yes, polar" --> E["Sun-synchronous LEO<br/>~700 km altitude<br/>(Earth observation)"]
    D -- "Yes, low-altitude imaging" --> F["LEO 300–600 km<br/>T ≈ 90–95 min<br/>(Hubble, Sentinel)"]
    D -- "Global navigation?" --> G["MEO ~20 200 km<br/>T = 12 h<br/>(GPS, Galileo)"]
    F --> H["Watch for drag below 400 km<br/>— orbit decay & speedup"]
    E --> H

A* candidates should be able to reason about why a chosen orbit fits the mission. Geostationary is forced by the fixed-ground-point requirement; GPS at 12 h fits the global-navigation constraint via constellation geometry; Sun-synchronous polar orbits are chosen so that an Earth-observation satellite passes over the same ground point at the same local solar time every day.

Weightlessness in Orbit — the Big Misconception

Students, TV presenters, and many science museum exhibits routinely describe astronauts on the ISS as being in "zero gravity". This is simply wrong. At the ISS altitude of 400 km, the gravitational field strength is

$g_{\text{ISS}} = \frac{GM_{E}}{r^{2}} = \frac{3.98 \times 10^{14}}{(6.77 \times 10^{6})^{2}} = 8.69 \text{ N kg}^{-1}$

— only 11% smaller than the surface value of 9.81 N kg⁻¹. A 70 kg astronaut has a gravitational weight of $70 \times 8.69 = 608$ N at the ISS, compared with 687 N at the surface. The astronaut is decidedly in gravity.

So why do they appear weightless?

Because both the astronaut and the space station are in free fall. Both are accelerating inward toward the Earth's centre at the same rate $g_{\text{ISS}} = 8.69$ m s⁻². They fall together — the station does not "drop out from under" the astronaut, and the astronaut does not "fall behind" the station. There is no contact force between the astronaut and the station's floor, walls, or seat. The astronaut experiences no normal contact force, and therefore no felt sensation of weight.

The crucial distinction: what we call "feeling weight" in everyday life is not the gravitational pull on us. It is the normal contact force from the ground (or chair, or floor) supporting us against gravity. Standing on the Earth's surface, gravity pulls us down at 9.81 m s⁻² and the floor pushes us up at 9.81 m s⁻² — the contact force is what our nervous system registers as "weight". In free fall, there is no contact force; gravity is still acting (and just as strongly), but we do not feel it.

A few familiar terrestrial illustrations:

Parabolic flight ("vomit comet"). An aircraft flies a precisely calculated parabolic arc such that for ~20 seconds the aircraft (and everything inside it) is in free fall. Passengers feel weightless. Real gravity has not changed.
A lift in free fall. If the cable snaps and the lift accelerates downward at $g$ , the passengers float weightlessly inside the cab. Gravity is still pulling them at $g$ ; so is the lift cab; no contact force.
A pebble dropped from a tower. During its fall the pebble is weightless from its own perspective — there is no contact force on it. If you fell with it (matching velocity), you would see the pebble hover stationary beside you.

A discriminator.* For full marks on "explain why an astronaut on the ISS feels weightless" you need both ideas explicitly: (i) astronaut and station are in free fall with the same gravitational acceleration; (ii) consequently there is no contact force between astronaut and station, and weight as a felt sensation is contact-force-mediated, not gravity-mediated. Quoting just "they are in free fall" earns a partial mark; the contact-force explanation is the AO3 discriminator.

Summary Formulas for Circular Orbits

Quantity	Formula	Notes
Orbital speed	$v = \sqrt{GM/r}$	Higher orbits slower; mass-independent
Orbital period	$T = 2\pi\sqrt{r^{3}/(GM)}$	Kepler's third law
Kinetic energy	$E_{k} = GMm/(2r)$	Positive, increasing with proximity
Potential energy	$E_{p} = -GMm/r$	Negative, $V(\infty) = 0$ convention
Total energy	$E_{\text{total}} = -GMm/(2r)$	Negative for bound orbit; $= -E_{k} = E_{p}/2$
Escape velocity	$v_{\text{esc}} = \sqrt{2GM/r}$	$= \sqrt{2} \times v_{\text{orbit}}$ at the same radius

Memorising these six lines is enough for most OCR orbital-mechanics problems.

Orbits — Circular Satellite Motion

Orbits — Circular Satellite Motion

The Condition for a Circular Orbit

Worked Check — ISS Orbital Speed

Orbital Period — Kepler's Third Law Revisited

Orbital Energy

The Escape-Velocity Connection

Worked Example 1 — Speed and Period at Geostationary Altitude

Worked Example 2 — Orbital Decay (The Counterintuitive Speedup)

Worked Example 3 — Energy Needed to Raise a Satellite

Worked Example 4 — Orbit Selection Decision Tree

Weightlessness in Orbit — the Big Misconception

Summary Formulas for Circular Orbits

Synoptic Links

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