Escape Velocity and Orbit Types Revision Notes | A Level AQA Physics

1. The Big Question: How Fast to Break Free?

Picture throwing a ball upward. The harder you throw, the higher it goes before gravity pulls it back. Now ask: is there a speed at which it never comes back? That speed is the escape velocity — and working out what sets it is the first challenge of this lesson.

This topic sits at the heart of AQA's gravitational fields unit. Escape velocity, synchronous orbits, and the distinction between low Earth and geostationary orbits all appear regularly in calculation and extended-answer questions. Mastering the energy argument behind escape velocity will also sharpen your understanding of orbital mechanics more broadly.

2. Escape Velocity — The Energy Argument

Gravitational potential energy near a planet is negative: it equals $-\frac{GMm}{r}$, where $G$ is the gravitational constant, $M$ is the planet's mass, $m$ is the object's mass, and $r$ is the distance from the planet's centre. The negative sign reflects the fact that you must do work on the object to move it away from the planet.

For an object to escape to infinity, its total mechanical energy must be at least zero. At infinity, both kinetic energy and gravitational potential energy are zero. So the minimum launch condition is: kinetic energy at the surface plus gravitational potential energy at the surface equals zero.

Notice that $m$, the mass of the escaping object, cancels completely. Escape velocity is the same for a grain of sand and a spacecraft launched from the same point — it depends only on the planet's mass and radius. This is a direct consequence of the equivalence of gravitational and inertial mass.

Worked example: Calculating Earth's Escape Velocity

Calculate the escape velocity from Earth's surface. Use $G = 6.67 \times 10^{-11}$ N m² kg⁻², $M_{Earth} = 5.97 \times 10^{24}$ kg, $R_{Earth} = 6.37 \times 10^{6}$ m.

1. $v_{esc} = \sqrt{\frac{2GM}{R}}$ — This comes from setting total mechanical energy to zero: $\frac{1}{2}mv^2 = \frac{GMm}{R}$. [M1]
2. $v_{esc} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^{6}}}$ — All values are in SI units (m, kg, N m² kg⁻²), so no conversion is needed. [M1]
3. $2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} = 2 \times 3.982 \times 10^{14} = 7.964 \times 10^{14}$ N m² kg⁻¹ — $6.67 \times 5.97 = 39.82$, so $6.67 \times 10^{-11} \times 5.97 \times 10^{24} = 39.82 \times 10^{13} = 3.982 \times 10^{14}$. Multiply by 2 to get $7.964 \times 10^{14}$.
4. $\frac{7.964 \times 10^{14}}{6.37 \times 10^{6}} = 1.250 \times 10^{8}$ m² s⁻² — $7.964 \div 6.37 \approx 1.250$ and $10^{14} \div 10^{6} = 10^{8}$.
5. $v_{esc} = \sqrt{1.250 \times 10^{8}} = \sqrt{1.250} \times \sqrt{10^{8}} = 1.118 \times 10^{4}$ m s⁻¹ — $\sqrt{10^{8}} = 10^{4}$ and $\sqrt{1.250} \approx 1.118$. [A1]
6. $v_{esc} \approx 1.12 \times 10^{4}$ m s⁻¹ $= 11.2$ km s⁻¹ — Sanity check: Earth's escape velocity is well-known to be approximately 11.2 km/s. This matches.

Answer: $v_{esc} \approx 1.12 \times 10^{4}$ m s⁻¹ (approximately 11.2 km s⁻¹)

3. Orbital Motion and the Idea of a Synchronous Orbit

Now that you have the energy picture for escape, consider what happens at speeds below escape velocity. An object launched horizontally at the right speed will follow a circular orbit: gravity provides the centripetal force needed to keep it moving in a circle rather than flying off into space or crashing back down.

For a circular orbit of radius $r$ around a planet of mass $M$, equating gravitational force to centripetal force gives $\frac{GMm}{r^2} = \frac{mv^2}{r}$, which simplifies to $v = \sqrt{\frac{GM}{r}}$. The orbital period $T$ follows from $v = \frac{2\pi r}{T}$, leading to Kepler's third law: $T^2 \propto r^3$.

The key insight is that the orbital radius for a synchronous orbit is fixed by the planet's mass and rotation rate — there is only one radius at which the orbital period matches the planet's rotation. For Earth, this radius works out to approximately $4.23 \times 10^{7}$ m (about 42,300 km) from Earth's centre, placing the satellite roughly 35,900 km above the equatorial surface.

Worked example: Deriving the Geostationary Orbital Radius

Show that the geostationary orbital radius for Earth is approximately $4.23 \times 10^{7}$ m. Use $G = 6.67 \times 10^{-11}$ N m² kg⁻², $M_{Earth} = 5.97 \times 10^{24}$ kg, $T = 24 \times 3600$ s.

1. $\frac{GMm}{r^2} = \frac{4\pi^2 m r}{T^2}$ — Centripetal acceleration for circular motion is $\frac{4\pi^2 r}{T^2}$. The satellite mass $m$ cancels. [M1]
2. $r^3 = \frac{GMT^2}{4\pi^2}$ — Multiply both sides by $r^2$ and divide by $\frac{4\pi^2}{T^2}$. [M1]
3. $T = 24 \times 3600 = 86400$ s — Always convert hours to seconds before substituting.
4. $T^2 = 86400 \times 86400 = 7.465 \times 10^{9}$ s² — $86400^2$: $8.64 \times 10^4$ squared gives $(8.64)^2 \times 10^8 = 74.65 \times 10^8 = 7.465 \times 10^9$.
5. $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.465 \times 10^{9}$ — $6.67 \times 5.97 \approx 39.82$, so $GM = 3.982 \times 10^{14}$. Then $3.982 \times 10^{14} \times 7.465 \times 10^{9} = 29.72 \times 10^{23} = 2.972 \times 10^{24}$.
6. $r^3 = \frac{2.972 \times 10^{24}}{4\pi^2} = \frac{2.972 \times 10^{24}}{39.48} = 7.527 \times 10^{22}$ m³ — $4\pi^2 \approx 4 \times 9.870 = 39.48$.
7. $r = (7.527 \times 10^{22})^{\frac{1}{3}} \approx 4.23 \times 10^{7}$ m — $(10^{22})^{1/3} \approx 10^{7.33}$; $(7.527)^{1/3} \approx 1.96$; combined gives $\approx 4.23 \times 10^{7}$ m. [A1]

Answer: $r \approx 4.23 \times 10^{7}$ m from Earth's centre (altitude $\approx 35{,}900$ km above the equatorial surface)

4. Geostationary Orbits — Plane, Radius, and Applications

A geostationary satellite must satisfy three simultaneous conditions: its orbit must be circular, it must lie in the equatorial plane (directly above the equator), and its period must be exactly 24 hours. These three constraints together fix both the altitude and the direction of travel (west to east, matching Earth's rotation).

Because the radius is fixed at $\approx 4.23 \times 10^{7}$ m from Earth's centre, all geostationary satellites share the same orbital ring. This means there is a finite number of slots available — a practical limitation that international bodies regulate carefully to avoid signal interference between satellites.

• Orbital plane: equatorial (0° inclination)
• Orbital radius: approximately $4.23 \times 10^{7}$ m from Earth's centre
• Altitude above surface: approximately 35,900 km
• Orbital period: 24 hours (matching Earth's rotation)
• Direction: west to east (prograde, matching Earth's spin)
• Appears stationary from the ground — always above the same point on the equator

The stationary appearance makes geostationary satellites ideal for applications requiring a continuous, fixed link between ground stations and the satellite. Television broadcasting, weather monitoring over a fixed region, and long-range telecommunications all exploit this property — a dish antenna on the ground can point at a fixed position in the sky without any tracking mechanism.

There is, however, a significant drawback: the large orbital radius means signals take approximately 0.24 seconds to travel one-way from the ground to the satellite (a one-way distance of roughly $4.23 \times 10^{7}$ m at the speed of light), giving a round-trip delay of approximately 0.28 seconds. This latency makes geostationary satellites unsuitable for real-time two-way communication such as video calls or internet browsing where delay is noticeable.

5. Low Earth Orbits — Characteristics and Uses

At the other end of the altitude spectrum sit low Earth orbit (LEO) satellites, typically operating between 200 km and 2000 km above the surface. At these altitudes, the orbital period is roughly 90 minutes to 2 hours — the satellite completes many orbits per day and is visible from any given ground station for only a few minutes at a time.

LEO satellites are not restricted to the equatorial plane. Many are placed in polar orbits (inclination close to 90°), which means the satellite passes over every latitude on Earth as the planet rotates beneath it. Over time, a polar-orbit satellite builds up complete global coverage — a property that geostationary satellites, fixed above the equator, cannot provide for high-latitude regions.

The proximity of LEO satellites to Earth's surface has two important consequences. First, the gravitational field strength is only slightly less than at the surface, so the centripetal acceleration required is large — LEO satellites travel at approximately 7,800 m s⁻¹ (about 7.8 km s⁻¹). Second, residual atmospheric drag at the lowest LEO altitudes gradually slows satellites, causing orbital decay; periodic boosts are needed to maintain altitude.

6. Exam Technique and Common Pitfalls

Gravitational fields questions at A-level frequently combine escape velocity and orbital mechanics in multi-part questions. The most reliable approach is to identify which energy or force balance applies, write the relevant equation, and substitute carefully in SI units throughout. Mixing km and m is the single most common source of arithmetic errors in this topic.

When a question asks you to explain why a geostationary orbit must be equatorial, the mark scheme expects a chain of reasoning: if the orbit were inclined, the satellite would not remain above the same point on the equator, so it would not appear stationary to a ground observer — defeating the purpose of a geostationary orbit. State the condition, then justify it with the consequence.

For compare questions on orbit types, structure your answer as a direct parallel: state a feature for LEO, then the corresponding feature for GEO, then move to the next feature. Examiners award marks for explicit comparisons, not for two separate descriptions that happen to be placed near each other.

Across this lesson, three specification requirements have been addressed in full: the derivation and calculation of escape velocity using energy conservation; the definition and conditions for synchronous and geostationary orbits including the fixed equatorial radius; and the comparison of LEO and geostationary satellites in terms of orbital plane, altitude, period, and application. Each of these can appear as a calculation, a descriptive, or an evaluative question — and the worked examples and recall tasks above have given you practice across all three command-word styles.