How is escape speed different from orbital speed?

Escape speed is the minimum to leave for ever and is root two times the circular speed at the same radius.

Why are low orbits so fast?

A small r makes the square root of GM over r large, so low satellites run near 7.7 km/s.

S6 Physics Gravitation, Paper 2: orbits, escape speed and binaries

2026-11-02

Ten graded items reversing g for mass, then circular orbit period, escape speed, binary star force and geostationary speed.

Focus　Measure every orbital radius from the centre, use T equals 2 pi root r cubed over GM for orbits and root 2gR for escape speed.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S6 Physics: Gravitation
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationTaking g as 9.8 m/s^2, find the weight of a 60 kg student.
FoundationTwo 2 kg spheres are 0.5 m apart. Find the gravitational force between them.
FoundationA planet has radius 3.2 x 10^6 m and surface g of 5.0 m/s^2. Find its mass.
FoundationAt twice the Earth radius from the centre, what is g if the surface value is 9.8 m/s^2?
IntermediateA satellite circles at 8.0 x 10^6 m from Earth centre. Find its period, with Earth mass 6.0 x 10^24 kg.
IntermediateMars has surface g of 3.7 m/s^2 and radius 3.4 x 10^6 m. Find its escape speed.
IntermediateA planet has Earth mass but four times the radius. Compare its escape speed with the Earth value of 11.2 km/s.
ChallengeTwo stars of 2.0 x 10^30 kg orbit 1.0 x 10^11 m apart. Find the gravitational force between them.
ChallengeA geostationary satellite has radius 4.2 x 10^7 m and period 86400 s. Find its speed.
Error redoA station orbits 400 km above the ground and a student uses r = 4.0 x 10^5 m to get 31.6 km/s. Find the mistake and the correct speed.

03　Answers and one-line notes

#	Answer	One-line note
1	588 N	Weight is mass times g, so 60 kg gives 588 newtons at the Earth surface.
2	About 1.07 x 10^-9 N	Substituting into the inverse square law gives about 1.07 times ten to the minus ninth newtons.
3	About 7.7 x 10^23 kg	Rearranging g equals GM over R squared gives a mass of about 7.7 times ten to the twenty-third kilograms.
4	2.45 m/s^2	Doubling the distance from the centre quarters g, giving about 2.45 metres per second squared.
5	About 7.1 x 10^3 s, roughly 118 minutes	The period works out near 7100 seconds, about 118 minutes, which is typical for a low orbit.
6	About 5.0 km/s	Escape speed is the square root of twice g times R, giving about 5.0 kilometres per second on Mars.
7	Half, about 5.6 km/s	Escape speed falls as the square root of the radius, so four times the radius halves it to about 5.6 kilometres per second.
8	About 2.67 x 10^28 N	Squaring the masses and dividing by the square of the separation gives about 2.67 times ten to the twenty-eighth newtons.
9	About 3.1 km/s	Circumference over period gives about 3.05 kilometres per second, slower than a low orbit.
10	r must be measured from the centre, 6.8 x 10^6 m, giving about 7.7 km/s	He used the altitude instead of the distance from the Earth centre, which gives about 7.7 kilometres per second.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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