Physics › Gravitational fields › Orbits and satellites
Orbits and satellites
An orbit is gravity supplying the centripetal force, and everything else follows from that one substitution. Out of it come Kepler's third law, the kinetic, potential and total energy of a satellite, escape velocity, and the geostationary orbit that stays above one point on the equator.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Gravitational potential and Circular motion.
IN THIS TOPIC
- State Kepler's three laws, and use the equal-areas law to compare a planet's speed at the two ends of its orbit.
- Derive T² ∝ r³ from gravity as the centripetal force.
- Find the orbital speed at any radius, and say why lower orbits are faster.
- Account for a satellite's kinetic, potential and total energy.
- Use escape velocity, and explain why the escaping mass drops out of it.
- Describe synchronous, geostationary and low orbits, including the geostationary plane and radius.
COMMON MISCONCEPTION
A satellite needs its engines running to stay in orbit.
Gravity takes the centripetal job
A satellite in a circular orbit is doing circular motion, so something must supply the centripetal force. Gravity does, entirely. No engine and no thrust come into it. The satellite is permanently falling towards the planet while its tangential speed carries it forever past the edge. Setting the two faces of one force equal is the master move of the whole topic,
from which the orbital speed at any radius, v = √(GM/r), falls straight out. Lower orbits are faster orbits.
Kepler's three laws
Most of a century before Newton wrote down a force law, Johannes Kepler was handed Tycho Brahe's naked-eye records of the planets and spent twenty years fitting curves to them. What came out were three laws, and they still frame this whole topic. Kepler described; Newton, later, explained. All three follow from the inverse-square law, and the third one you are about to derive yourself.
The first law is the law of ellipses. Every planet moves in an ellipse with the Sun at one focus. An ellipse has two foci, and the point examiners press on is that the second one is empty: nothing sits there, and the Sun is certainly not at the centre. A circle is the special case in which the two foci coincide, which is exactly why the circular treatment above is worth so much. Planetary orbits are ellipses of small eccentricity, and the Earth's distance from the Sun stays within two per cent of its average all year round.
The second law is the law of equal areas. The line joining a planet to the Sun sweeps out equal areas in equal times. Near the Sun that line is short, so it must swing through a wide angle to cover the area; far out it is long, and a narrow wedge is enough. The planet therefore moves fastest at perihelion, its closest approach, and slowest at aphelion, its furthest. Comets make the point violently, tearing round the Sun in weeks and then crawling through the outer part of the orbit for decades. Underneath it is conservation of angular momentum, which a purely radial pull has no way to change.
The third law is the law of periods. The square of the orbital period is proportional to the cube of the mean radius, with the mean radius being the semi-major axis of the ellipse, and for the near-circular orbits you will be given it is simply the radius. It is the only one of the three that puts one orbit into arithmetic with another, and it is the one the specifications ask you to derive.
GUIDED PRACTICE
A comet at both ends
A comet passes perihelion at 0.50 AU from the Sun travelling at 60 km s−1, and its aphelion lies at 35 AU. Use the second law to find its speed at aphelion.
Show the working
At either end of the orbit the velocity is at right angles to the Sun-comet line, so in a short time the swept area is a thin triangle of height r and base vδt, giving an area ½rvδt. Equal areas in equal times therefore means rv is the same at both ends.
v = 60 × 0.50/35 = 0.86 km s−1. Seventy times slower at the far end, which is why a comet is visible for a few months out of an orbit lasting decades. The step only works at perihelion and aphelion, because nowhere else is the velocity perpendicular to the radius.
The third law, derived
The derivation of the period-radius law is expected, and it runs to three lines. Replace v in the centripetal equation with the circumference over the period, v = 2πr/T, then rearrange.
so T2 ∝ r3, and distant orbits are disproportionately slow. Plot T2 against r3 for any family of satellites and every one sits on a single straight line through the origin, whose gradient measures the central mass. Astronomers weigh planets with exactly this graph.
WORKED EXAMPLE
A satellite's speed and clock
A satellite orbits at r = 7.0 × 106 m from the Earth's centre. Find its orbital speed and period.
Gravity is the centripetal force, so GMm/r2 = mv2/r, and the satellite's own mass cancels immediately.
v = the square root of GM/r = the square root of (6.67 × 10−11 × 5.97 × 1024)/(7.0 × 106) = 7.5 km s−1.
T = 2πr/v = 2π × 7.0 × 106/7540 = 5.8 × 103 s, about 97 minutes. That is the classic low-orbit hour and a half, and about fifteen sunrises a day.
Orbital energy and escape
An orbiting satellite holds kinetic energy and negative gravitational potential energy, and its total energy stays constant while it stays in one stable orbit. For a circular orbit each piece has a clean expression. The potential energy is from the potential lesson; the orbit condition GMm/r2 = mv2/r rearranges to , exactly half the well's depth; and adding them gives the total,
negative, as any bound orbit's total must be, and rising towards zero as r grows. Moving between orbits is where the energy balance has its famous quirk. Drop to a lower orbit and the satellite speeds up, its kinetic energy rising, yet the total falls, because the potential energy fell twice as far. Atmospheric drag on a low satellite plays that story out in slow motion: the drag drains a little total energy, the orbit sinks, and the fall converts enough potential energy to raise the kinetic energy despite the loss, so the satellite ends lower and faster, the opposite of every intuition about friction.
Leaving entirely means paying the whole depth of the well from radius r. Set the kinetic energy equal to that climb, ½mv2 = GMm/r, and out comes the escape velocity,
about 11 km s−1 from the Earth's surface, independent of the escaping mass.
GUIDED PRACTICE
Verify the eleven
Using M = 5.97 × 1024 kg and R = 6.37 × 106 m, verify the quoted escape speed from the Earth's surface.
Show the working
v = the square root of 2GM/R = the square root of (2 × 6.67 × 10−11 × 5.97 × 1024)/(6.37 × 106).
v = 1.1 × 104 m s−1, the 11 km s−1 quoted earlier. The factor of 2 distinguishes escaping from orbiting; forgetting it lands on the orbital speed instead, a wrong answer that looks plausible.
The special orbits
A synchronous orbit has a period equal to its planet's rotation period. The geostationary orbit adds three more conditions. It is circular, it lies in the equatorial plane, and it travels the same way as the Earth's spin, so the satellite hangs over one fixed point on the equator. Kepler's law then fixes its radius at about 4.2 × 107 m from the Earth's centre, roughly 36,000 km up. Every satellite dish that never moves is aimed at that one ring in the sky.
Low orbits, a few hundred kilometres up, trade coverage for closeness. Periods run near 90 minutes, imaging detail is fine and signal delays are small, but each satellite sees any given place only briefly, so low-orbit systems fly in constellations.
INDEPENDENT PRACTICE
Kepler reaches Mars
Mars orbits the Sun at 1.52 times the Earth's orbital radius. Using Kepler's third law as a ratio, find the Martian year in Earth years.
Show the working
T2 ∝ r3, so T = 1.52 raised to the power 3/2 = 1.9 Earth years.
No G, no solar mass, no metres. The ratio form strips the law back to its skeleton, and the skeleton alone dates the Martian calendar.
ASSESSMENT FOCUS
- Kepler's three laws are recall marks, so learn all three in order. An ellipse with the Sun at one focus; the Sun-planet line sweeping equal areas in equal times; and T² ∝ r³. The second is the one candidates paraphrase into nothing, so give the statement first and the consequence second, that the planet is fastest at its closest point.
- The T² ∝ r³ derivation is required. Set gravitational force equal to centripetal force, substitute v = 2πr/T, then rearrange. Practise it until the three lines write themselves.
- Lower orbit, faster satellite, shorter period; but a lower orbit also means lower total energy. Both halves get examined, sometimes in the same question.
- Escape velocity comes from energy. The kinetic energy supplied equals the depth GMm/r, and the escaping mass cancels, so a pebble and a rocket share one escape speed.
- Geostationary needs all four conditions. A circular orbit, a 24-hour period, the equatorial plane, and the same sense as the Earth's rotation, at a radius near 4.2 × 107 m from the centre.
- “Why does the satellite need no engine?” Gravity supplies the centripetal force, and the circular path is an equipotential, so no work is needed. Two sentences, full marks.
- The satellite's own mass cancels out of every speed, period and radius calculation here, so a question that hands it to you is often checking whether you notice. Its mass matters only once you are asked for an energy.
CHECK YOURSELF
Show that the geostationary orbit has a radius of about 4.2 × 107 m. (M = 5.97 × 1024 kg; take the period as 24 hours.)
Show a hint
Kepler's derived law, rearranged for r cubed.
Show the answer
T = 24 × 3600 = 8.64 × 104 s. Rearrange the derived law to r3 = GMT2/4π2.
r3 = (6.67 × 10−11 × 5.97 × 1024 × (8.64 × 104)2) / 4π2 = 7.5 × 1022 m3.
r = 4.2 × 107 m from the Earth's centre, about 36,000 km above the surface, in the equatorial plane. One radius, shared by every geostationary satellite ever flown. Purists use the sidereal day of 23 h 56 min, which trims the radius by under 0.2% and never changes an AQA answer.
Kepler: an ellipse with the Sun at one focus, equal areas in equal times, and T² ∝ r³.
Gravity is the centripetal force; T² grows as r³.
Lower orbit, faster satellite, lower total energy.
Geostationary asks for a circle over the equator, 24 hours round it, and one fixed ring.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the orbits and satellites questions page.
WHERE TO GO NEXT
- Proportional reasoning is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- State Kepler's three laws, and use the equal-areas law to compare a planet's speed at the two ends of its orbit.
- Derive T² ∝ r³ from gravity as the centripetal force.
- Find the orbital speed at any radius, and say why lower orbits are faster.
- Account for a satellite's kinetic, potential and total energy.
- Use escape velocity, and explain why the escaping mass drops out of it.
- Describe synchronous, geostationary and low orbits, including the geostationary plane and radius.
Open the full revision checklist to track your progress across the whole unit.