Physics › Gravitational fields › Orbits and satellites
Orbits and satellites
An orbit is Year 13's two big ideas shaking hands: gravity supplies the centripetal force, and everything follows, the T squared proportional to r cubed law, the energy books of a satellite, escape velocity, and the one special orbit that hangs over a fixed point on the equator.
Builds on Gravitational potential and Circular motion.
IN THIS TOPIC
- Derive T² ∝ r³ from gravity as the centripetal force.
- Account for a satellite's kinetic, potential and total energy, and use escape velocity.
- Describe synchronous, geostationary and low orbits, including the geostationary plane and radius.
WHAT YOU PROBABLY THINK
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, and something must supply the centripetal force. Gravity does, entirely. No engine, no thrust: the satellite is permanently falling toward the planet while its tangential speed carries it forever past the edge. Setting the two faces of the same force equal is the topic's master move:
from which the orbital speed at any radius, v = √(GM/r), falls straight out: lower orbits are faster orbits.
Kepler's third law, derived
AQA requires the derivation of the period-radius law, and it is three lines. Replace v with the circumference over the period, v = 2πr/T, in the equation above, and rearrange:
so T2 ∝ r3: 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 hands you the central mass. Astronomers weigh planets with exactly this graph.
The energy books, and the exit price
An orbiting satellite holds kinetic energy and (negative) gravitational potential energy, and its total energy is constant in a stable orbit. The bookkeeping has a famous quirk: drop to a lower orbit and the satellite speeds up, kinetic energy rising, yet the total falls, because the potential energy fell further. Atmospheric drag on a low satellite therefore makes it faster while draining it, the opposite of every intuition about friction.
Leaving entirely means paying the well's whole depth from radius r. Setting kinetic energy equal to the climb, ½mv2 = GMm/r, gives the escape velocity:
about 11 km s−1 from the Earth's surface, independent of the escaping mass.
The special orbits
A synchronous orbit has a period equal to its planet's rotation period. The geostationary orbit adds two conditions: it lies in the equatorial plane and travels the same direction as the Earth's spin, so the satellite hangs over one fixed point on the equator. Kepler's law fixes its radius at about 4.2 × 107 m from the Earth's centre, roughly 36,000 km up, and 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: short periods around 90 minutes, fine detail for imaging and sensible signal delays, at the price of each satellite seeing any given place only briefly, which is why low-orbit systems fly in constellations.
THE EXAM BIT
- The T² ∝ r³ derivation is required: gravitational force equals centripetal force, substitute v = 2πr/T, rearrange. Practise until the three lines write themselves.
- Lower orbit, faster satellite, shorter period; but lower orbit also means lower total energy. Both halves get examined, sometimes in one question.
- Escape velocity comes from energy, kinetic supplied equals well depth GMm/r, and the mass cancels. A pebble and a rocket share the same escape speed.
- Geostationary needs all three conditions: 24-hour period, equatorial plane, same sense as the Earth's rotation. Radius about 4.2 × 107 m from the centre.
- “Why does the satellite need no engine?”: gravity supplies the centripetal force, and along its circular path, an equipotential, no work is required. Two sentences, full marks.
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. Rearranging the derived law: 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.
Gravity is the centripetal force; T² grows as r³.
Geostationary: 24 hours, equatorial, one fixed ring.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.