Gravitational fields and orbits

Curriculum links

Quoted word for word from the official documents.

Step by step

The caption is what the class reads on screen; the talking points are in your notes drawer (N) when you present. Steps marked Pupils only appear only in the pupil lesson.

1. 01

   Field strength falls with distance

   Presented and for pupils

   Caption

   Earth's gravitational field strength at a distance r from its centre is g = GM/r². Twice as far out, the field is a quarter as strong.

   Talking points

   • NASA's Earth Fact Sheet gives GM = 0.39860 × 10⁶ km³ s⁻², which is 3.9860 × 10¹⁴ m³ s⁻², and a mean radius of 6,371.000 km.
   • At the mean radius g = 3.9860 × 10¹⁴ ÷ (6.371 × 10⁶)² = 9.820 N/kg, the figure NASA lists as Earth's mean surface gravity.
   • OCR's specification models the mass of a spherical object as a point mass at its centre, which is why r is measured from Earth's centre: about 6,800 km for the ISS, whose height above the ground is only about 420 km.

   Ask the class

   The ISS is about 6.80 × 10⁶ m from Earth's centre. Using GM = 3.986 × 10¹⁴ m³ s⁻², what is g there, in N/kg?

   Answer8.62 N/kg. Answers from 8.57 to 8.67 are marked right.

   Whyg = GM/r² = 3.986 × 10¹⁴ ÷ (6.80 × 10⁶)² = 3.986 × 10¹⁴ ÷ 4.624 × 10¹³ = 8.62 N/kg. That is 88 per cent of the 9.82 N/kg at the surface: gravity is still strong at the station's height.

   Sources 1 3 7Present from this step →

2. 02

   Speed in a circular orbit

   Presented and for pupils

   Caption

   In a circular orbit gravity provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). The closer the orbit, the faster it must go.

   Talking points

   • NASA's trajectory file of 7 October 2026 predicts the station's path every four minutes to 22 October. It keeps the ISS between 6,794 and 6,802 km from Earth's centre, about 420 km above the equator's radius of 6,378 km.
   • The same predictions give the station's speed as 7.65 to 7.66 km/s.
   • The label marks where the ISS is now. The station itself is far too small to see at true scale from this distance.

   Ask the class

   Taking r = 6.80 × 10⁶ m and GM = 3.986 × 10¹⁴ m³ s⁻², what is the ISS's orbital speed in km/s?

   Answer7.66 km/s. Answers from 7.62 to 7.7 are marked right.

   Whyv = √(GM/r) = √(3.986 × 10¹⁴ ÷ 6.80 × 10⁶) = √(5.862 × 10⁷) = 7,656 m/s = 7.66 km/s. NASA's trajectory data give 7.65 to 7.66 km/s.

   Sources 1 3Present from this step →

3. 03

   The ISS, now

   Presented and for pupils

   Caption

   This is the International Space Station where it is right now, flown on its latest orbital elements at real speed.

   Talking points

   • The viewer works out the station's position from CelesTrak's orbital elements. Those of 8 October 2026 give 15.488 orbits a day, one every 92.98 minutes.
   • NASA's predicted trajectory counts 231 orbits, ascending node to ascending node, from 13:30:48 UTC on 7 October to 11:13:18 UTC on 22 October 2026: 92.91 minutes each.
   • These figures are for October 2026. On another date the station's height, and so its period, can be a little different.

   Ask the class

   Using r = 6.80 × 10⁶ m and v = 7,656 m/s, find the ISS's orbital period in minutes.

   Answer93 minutes. Answers from 92.5 to 93.5 are marked right.

   WhyT = 2πr/v = 2π × 6.80 × 10⁶ ÷ 7,656 = 5,581 s = 93.0 minutes. NASA's trajectory data give 92.91 minutes from one ascending node to the next.

   Sources 3 4Present from this step →

4. 04

   Time the ISS yourself

   Pupils only

   Caption

   Your turn. The clock runs at one minute a second. Time one orbit of the ISS.

   Talking points

   • Any point on the path works: the moment the label comes back into view from behind the Earth, or the moment it passes straight above the Earth's centre on the screen. The ISS reaches that point once per orbit, so the time between two visits is one period.
   • The camera holds its direction relative to the Sun, which moves 360° in 365.256 days, about 0.06° in one orbit of the ISS. The picture is effectively fixed in space while the class times.
   • Answers from 90 to 96 minutes are marked right.

   Pupil question

   How many minutes did your one orbit of the ISS take?

   Answer93 minutes. Answers from 90 to 96 are marked right.

   WhyThe calculation gave 93.0 minutes and NASA's trajectory data 92.91 minutes. With each time read to the nearest minute, a careful timing lands within a minute or two of 93.

   Pupil task

   Choose a point on the ISS's path, such as where its label comes back into view from behind the Earth, or where it passes straight above the Earth's centre on your screen. Pause when the ISS is there and note the time. Play, and pause again when it next reaches the same point. Leave the view where it is while you time.

   HintAt one minute a second an orbit takes about a minute and a half of real time. Subtract the first time from the second, then type the answer in the box above.

   The viewer checks the task as the pupil works and says when it is done.

   Sources 1 3Open this step as a pupil →

5. 05

   Geostationary orbit

   Presented and for pupils

   Caption

   Every active satellite is drawn here. The gold ring far out is the geostationary belt: those satellites turn with the Earth and stay above the same points on the equator.

   Talking points

   • NASA: a satellite in a circular geosynchronous orbit directly over the equator, with zero eccentricity and inclination, "will have a geostationary orbit that does not move at all relative to the ground."
   • NASA gives the period of a geosynchronous weather satellite as 23 hours, 56 minutes and 4 seconds. That is Earth's sidereal rotation period, 23.9345 hours in NASA's Earth Fact Sheet and 86,164.09 s in JPL's constants.
   • The clock runs at an hour a second, so the Earth turns once in about 24 seconds, and the gold ring turns with it. The low satellites close to the Earth go round many times in that while.

   Ask the class

   What must the period of a geostationary orbit be?

   1. A24 hours exactly, one solar day
   2. B23 h 56 min 4 s, one sidereal dayRight answer
   3. CAbout 93 minutes
   4. D27.3 days

   WhyThe satellite must turn with the Earth, which turns once relative to the stars in a sidereal day, 23 h 56 min 4 s (86,164 s). NASA's notes define the 24-hour day as the time from noon to noon, measured against the Sun.

   Sources 1 2 5 6Present from this step →

6. 06

   The radius of geostationary orbit

   Presented and for pupils

   Caption

   Putting one sidereal day into T² = (4π²/GM) r³ fixes the radius. Every geostationary satellite orbits at the same distance from Earth's centre.

   Talking points

   • r = (GMT²/4π²)^(1/3) = (3.986 × 10¹⁴ × 86,164² ÷ 39.48)^(1/3) = 4.2164 × 10⁷ m. NASA gives the same figure: 42,164 km from the centre of the Earth, about 36,000 km above its surface.
   • Above the equator's radius of 6,378 km that is 35,786 km. The orbital speed there is 2πr/T = 3.07 km/s.
   • The solar day, 86,400 s, would give 42,241 km, 77 km too far out, so the question asks for four significant figures.

   Ask the class

   Using GM = 3.986 × 10¹⁴ m³ s⁻² and the sidereal day, T = 86,164 s, calculate the radius of geostationary orbit in km, to 4 significant figures.

   Answer42,160 km. Answers from 42,130 to 42,190 are marked right.

   Whyr³ = GMT²/4π² = 3.986 × 10¹⁴ × 7.4242 × 10⁹ ÷ 39.478 = 7.4960 × 10²² m³, so r = 4.216 × 10⁷ m = 42,160 km. NASA gives 42,164 km. With the 24-hour solar day the answer would be 42,240 km, which is marked wrong.

   Sources 1 5 6Present from this step →

7. 07

   Low orbits and the geostationary belt

   Presented and for pupils

   Caption

   Low orbits are close and fast. Geostationary orbit is distant and slow, and fixed over the equator. Each suits different work.

   Talking points

   • NASA: most scientific satellites, including NASA's Earth Observing System fleet, are in low Earth orbit. NASA's Aqua satellite takes about 99 minutes to orbit at about 705 km up.
   • NASA: geostationary satellites "provide a constant view of the same surface area", which makes them valuable for weather monitoring and useful for communication.
   • NASA: because they are parked over the equator, geostationary satellites "don’t work well for far northern or southern locations".

   Ask the class

   Which of these are true of a geostationary orbit? Choose all that are.

   More than one answer is right.

   1. AIt lies in the plane of the equatorRight answer
   2. BIt goes round in the same direction as Earth spinsRight answer
   3. CA geostationary satellite can stay directly above London
   4. DIts orbital speed is about 7.7 km/s

   WhyNASA: geostationary satellites "rotate with the Earth directly above the equator". London is far north of the equator, so no geostationary satellite can be overhead there. At 42,164 km the speed is 3.07 km/s; 7.7 km/s is the speed of a low orbit like the ISS's.

   Sources 6Present from this step →

Sources

Every fact in this lesson comes from these sources, and each step lists the ones it uses.

1. Earth Fact Sheet, NASA NSSDCA
2. Notes on the Planetary Fact Sheets, NASA NSSDCA
3. ISS trajectory data (Orbital Ephemeris Message from NASA Johnson Space Center), linked from this NASA page
4. ISS (ZARYA) orbital elements, NORAD 25544, CelesTrak
5. Astrodynamic Parameters, JPL Solar System Dynamics
6. Catalog of Earth Satellite Orbits, NASA Science (Earth Observatory)
7. OCR A Level Physics A (H556) specification, version 3.0