Satellite Orbits

Satellites — both natural and artificial — orbit planets and stars because of the balance between gravitational attraction and the satellite's tendency to travel in a straight line (inertia). Understanding satellite orbits connects gravitational fields with circular motion, and is a favourite topic for Edexcel exam questions.

Orbital Motion as Circular Motion

For a satellite in a circular orbit, the gravitational force provides the centripetal force needed to maintain circular motion:

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

where:

M is the mass of the central body (e.g. Earth)
m is the mass of the satellite
r is the orbital radius (measured from the centre of M)
v is the orbital speed

Crucially, the satellite's mass m cancels:

$\frac{GM}{r^2} = \frac{v^2}{r}$

$v^2 = \frac{GM}{r}$

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

This is a profound result: the orbital speed depends only on the mass of the central body and the orbital radius, not on the satellite's mass. A 1 kg probe and a 400,000 kg space station at the same orbital radius travel at exactly the same speed.

Common exam mistake: Using r as the altitude above the surface instead of the distance from the centre. Always remember: r = R_planet + h, where h is the altitude.

Deriving Orbital Period

Since the satellite travels a circular path of circumference 2πr at speed v, the orbital period T is:

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

Squaring:

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

This is Kepler's third law for circular orbits: T² is proportional to r³. Double the orbital radius and the period increases by a factor of 2√2 ≈ 2.83.

Worked Example 1: ISS Orbital Speed and Period

The ISS orbits at 408 km altitude. Calculate its orbital speed and period.

r = 6.37 × 10⁶ + 4.08 × 10⁵ = 6.778 × 10⁶ m

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.778 \times 10^6}} = \sqrt{\frac{3.984 \times 10^{14}}{6.778 \times 10^6}} = \sqrt{5.878 \times 10^7} = 7665 \text{ m s}^{-1}$

That is about 7.67 km s⁻¹ — roughly 27,600 km h⁻¹.

$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.778 \times 10^6}{7665} = \frac{4.259 \times 10^7}{7665} = 5557 \text{ s} \approx 92.6 \text{ min}$

The ISS completes roughly 15.5 orbits per day.

Kepler's Third Law

Kepler's third law states that for any system of orbiting bodies around the same central mass:

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

This constant depends only on the central mass M. For all objects orbiting the Sun, T²/r³ is the same. For all objects orbiting Earth, T²/r³ is the same (but a different constant from the Sun's system).

Satellite	Altitude (km)	r (×10⁶ m)	v (m s⁻¹)	T (hours)
Low Earth orbit (LEO)	200	6.57	7,790	1.47
ISS	408	6.78	7,665	1.54
Hubble	547	6.92	7,590	1.59
GPS	20,200	26.6	3,870	11.97
Geostationary	35,786	42.2	3,070	24.00
Moon	384,400	384.4	1,022	655 (27.3 days)

Notice: as orbital radius increases, speed decreases and period increases. Higher orbits are slower.

Worked Example 2: Using Kepler's Third Law as a Ratio

The Moon orbits Earth at r₁ = 3.84 × 10⁸ m with period T₁ = 27.3 days. A satellite orbits at r₂ = 6.78 × 10⁶ m. Find its period.

$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$

$T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2} = 27.3 \times \left(\frac{6.78 \times 10^6}{3.84 \times 10^8}\right)^{3/2}$

$T_2 = 27.3 \times (0.01766)^{3/2} = 27.3 \times 2.347 \times 10^{-3} = 0.0641 \text{ days} = 92.3 \text{ min}$

This matches our earlier ISS calculation, confirming Kepler's third law works beautifully as a ratio method.

Geostationary Orbits

A geostationary satellite has a very specific set of orbital characteristics:

Orbital period = exactly 24 hours (one sidereal day = 86,164 s)
Orbits in the equatorial plane (0° inclination)
Orbits from west to east (same direction as Earth's rotation)
Remains above a fixed point on the equator
Orbital radius ≈ 42,200 km from Earth's centre (altitude ≈ 35,800 km)

flowchart TD
    subgraph Geostationary["Geostationary Orbit"]
        G1["T = 24 hours exactly"]
        G2["Equatorial plane only"]
        G3["West-to-east rotation"]
        G4["Fixed above one point"]
        G5["r ≈ 42,200 km"]
    end
    subgraph Uses["Applications"]
        U1["TV broadcasting"]
        U2["Weather monitoring"]
        U3["Communications relay"]
    end
    subgraph Low["Low Earth Orbit (LEO)"]
        L1["T ≈ 90 minutes"]
        L2["Any inclination"]
        L3["h = 200–2000 km"]
        L4["Moves across sky"]
    end
    subgraph LowUses["LEO Applications"]
        LU1["Earth observation"]
        LU2["ISS / crewed missions"]
        LU3["Spy satellites"]
    end
    Geostationary --> Uses
    Low --> LowUses

Deriving the Geostationary Orbital Radius

From T² = 4π²r³/(GM):

$r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times (86164)^2}{4\pi^2}$

$r^3 = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.424 \times 10^9}{39.48}$

$r^3 = \frac{2.958 \times 10^{24}}{39.48} = 7.494 \times 10^{22}$

$r = (7.494 \times 10^{22})^{1/3} = 4.216 \times 10^7 \text{ m} \approx 42,200 \text{ km}$

There is only one possible radius for a geostationary orbit. The altitude above the surface is 42,200 − 6,370 = 35,800 km.

Exam tip: The term geosynchronous means period = 24 hours but does not require the equatorial plane. A geosynchronous satellite in an inclined orbit traces a figure-of-eight pattern in the sky. Only geostationary satellites remain truly fixed above one point.

Polar and Low Earth Orbits

Low Earth Orbits (LEO) have altitudes of roughly 200–2000 km. Satellites in LEO move fast (about 7–8 km s⁻¹) and complete an orbit in roughly 90 minutes.

Polar orbits are a special type of LEO inclined at approximately 90° to the equator. As the satellite orbits pole-to-pole, the Earth rotates beneath it. Over the course of a day, a polar-orbit satellite can image the entire surface of the Earth, making it ideal for weather observation, Earth monitoring, and reconnaissance.

Weightlessness in Orbit

Astronauts in orbit appear to be "weightless," but this is not because gravity has disappeared. As we calculated earlier, g at ISS altitude is still about 8.7 N kg⁻¹. The astronaut and the ISS are both in free fall — they accelerate towards Earth at the same rate. With no contact force between the astronaut and the floor, there is no sensation of weight. This is more accurately called apparent weightlessness or microgravity.

Summary

For circular orbits: gravitational force = centripetal force, giving v = √(GM/r)
Orbital speed decreases with increasing radius; period increases with radius
Kepler's third law: T² = 4π²r³/(GM), so T² ∝ r³ for a given central mass
Geostationary orbits: T = 24 h, equatorial, west-to-east, r ≈ 42,200 km
Orbital radius r is measured from the centre, not the surface: r = R + h
"Weightlessness" in orbit is apparent — caused by free fall, not absence of gravity

A-Level Deep Dive: Satellite Orbits

Spec mapping

Edexcel 9PH0 specification Topic 12 — Gravitational Fields covers the use of Newton's law of gravitation, the modelling of circular orbital motion under gravity, and the application of these ideas to satellites including geostationary orbits (refer to the official specification document for exact wording). Within Topic 12, satellite orbits sit alongside gravitational field strength, gravitational potential and gravitational potential energy, and exam questions routinely demand confidence in linking circular motion (Topic 5) to the inverse-square gravitational law. The same skill set is reused in Topic 13 (Electric and magnetic fields) where charged particles orbit under Coulomb attraction, and Topic 14 (Nuclear and particle physics) when modelling cyclotron motion. The Edexcel formula booklet provides Newton's law $F = Gm_1m_2/r^2$ , the centripetal acceleration $a = v^2/r = \omega^2 r$ , and gravitational potential $V_{grav} = -GM/r$ — but Kepler's third law $T^2 = 4\pi^2 r^3/(GM)$ is not listed and must be derived in the answer.

Worked example with full mark scheme

Question (8 marks):

(a) By considering the gravitational force on a satellite in a circular orbit of radius $r$ around a planet of mass $M$ , derive an expression for the orbital period $T$ in terms of $G$ , $M$ and $r$ . (4)

(b) A geostationary satellite orbits the Earth with a period of one sidereal day, $T = 86\,164\text{ s}$ . Using $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$ and $M_E = 5.97 \times 10^{24}\text{ kg}$ , calculate the altitude of the satellite above the Earth's surface. Earth's radius $R_E = 6.37 \times 10^6\text{ m}$ . (4)

Solution with mark scheme:

(a) Step 1 — equate gravitational force to centripetal force.

For circular motion, the net inward force is $F = mv^2/r$ . The only force is gravitational attraction, $F = GMm/r^2$ . Setting them equal:

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

M1 — equating gravitational force to centripetal force, with $r$ as the orbital radius (not altitude). A common candidate error is to write the centripetal expression with the surface radius of the planet rather than the orbital radius — this loses the M1 immediately.

Step 2 — solve for orbital speed.

The mass $m$ of the satellite cancels (a key physical point): $v^2 = GM/r$ , so $v = \sqrt{GM/r}$ .

A1 — correct expression for $v$ with the satellite mass eliminated and the dependence shown clearly.

Step 3 — relate period to speed.

The satellite traverses a circumference $2\pi r$ each period, so $T = 2\pi r / v$ . Substituting:

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

M1 — correctly relating $T$ to $v$ and $r$ via the circumference.

Step 4 — final form.

Squaring both sides gives the explicitly Keplerian form:

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

A1 — correct final expression in either form.

(b) Step 1 — rearrange for $r^3$ .

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

M1 — correct rearrangement.

Step 2 — substitute values.

$r^3 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(86\,164)^2}{4\pi^2}$

$T^2 = (86\,164)^2 = 7.424 \times 10^9\text{ s}^2$ . Numerator: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.424 \times 10^9 = 2.957 \times 10^{24}$ . Denominator: $4\pi^2 = 39.48$ . So $r^3 = 7.49 \times 10^{22}\text{ m}^3$ .

M1 — correct substitution including using $T$ in seconds, not hours.

Step 3 — take the cube root.

$r = (7.49 \times 10^{22})^{1/3} = 4.22 \times 10^7\text{ m}$ .

A1 — correct orbital radius (around 42,200 km).

Step 4 — convert to altitude.

$h = r - R_E = 4.22 \times 10^7 - 6.37 \times 10^6 = 3.58 \times 10^7\text{ m} \approx 35\,800\text{ km}$ .

A1 — correct subtraction of Earth's radius. Forgetting this step (giving 42,200 km as the answer) is the single most common error.

Total: 8 marks (M3 A4, with the structural M1 in (a) for force balance).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A satellite of mass $m = 1200\text{ kg}$ orbits the Earth in a circular path at altitude $h = 8.0 \times 10^5\text{ m}$ .

(a) Show that the orbital speed is approximately $7.4\text{ km s}^{-1}$ . (3)

(b) Calculate the orbital period in minutes and comment on whether this satellite could provide continuous coverage of a single point on Earth. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1) — $r = R_E + h = 6.37 \times 10^6 + 8.0 \times 10^5 = 7.17 \times 10^6\text{ m}$ .
M1 (AO2) — applying $v = \sqrt{GM/r}$ with correct values.
A1 (AO2) — $v = \sqrt{(6.67 \times 10^{-11})(5.97 \times 10^{24})/7.17 \times 10^6} \approx 7.45 \times 10^3\text{ m s}^{-1}$ , consistent with the printed result.

(b)

M1 (AO1) — $T = 2\pi r/v = 2\pi(7.17 \times 10^6)/7450 \approx 6045\text{ s}$ .
A1 (AO2) — $T \approx 101\text{ min}$ .
B1 (AO3) — comment: the period is far less than 24 hours, so the satellite moves rapidly across the sky relative to a fixed observer and cannot give continuous coverage of a single point. Continuous coverage requires either a geostationary orbit or a constellation of LEO satellites.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This is a typical Paper 2 structure: a calculation (part a) followed by a calculation plus interpretive comment (part b). The interpretive AO3 mark is what separates A from A* candidates — a numerical answer alone is incomplete.

Synoptic links

Connects to:

Topic 12 — Gravitational potential energy: the orbital radius determines $V_{grav} = -GM/r$ , and combining this with kinetic energy gives total mechanical energy $E = -GMm/(2r)$ for a circular orbit. Increasing orbital radius increases (less negative) total energy, which is why raising a satellite requires energy input.
Topic 12 — Escape velocity: $v_{esc} = \sqrt{2GM/r}$ at the surface is exactly $\sqrt{2}\times$ the surface-skimming orbital speed. A satellite released from a circular orbit with no extra speed remains bound; doubling its kinetic energy (multiplying speed by $\sqrt{2}$ ) is the threshold for escape.
Topic 12 — Kepler's laws of planetary motion: Kepler's first law (elliptical orbits with the central body at one focus) generalises the circular case, and Kepler's second law (equal areas in equal times) is conservation of angular momentum. Edexcel reduces to the circular special case but expects awareness of the broader laws.
Topic 5 — Circular motion: the entire derivation rests on $F = mv^2/r = m\omega^2 r$ . Confidence in switching between angular and linear forms is essential.
GPS positioning (synoptic / applied): GPS satellites at $r \approx 26\,600\text{ km}$ and $T \approx 12\text{ h}$ rely on precise timing — a worked example for Kepler's third law in a real-world context, also touching on relativistic time dilation as Year-2 enrichment.

Mark-scheme literacy

Satellite-orbit questions on 9PH0 Paper 2 split AO marks as follows: