- State Kepler’s three laws and use a³/T² = k for bodies orbiting the same centre.
- Use F = Gm₁m₂/r² and GM = gR² to find g at a height and the mass and density of a planet.
- Compare satellites in different circular orbits (v, ω, T, a) and calculate the first cosmic velocity.
- Explain geostationary satellites and how speed and period change when a spacecraft changes orbit.
China’s Tiangong space station circles the Earth about every hour and a half at a height of roughly 400 km, while the geostationary satellites of the BeiDou navigation system seem to “hang” above one point of the equator. The same force — universal gravitation — holds all of them; only the radius of the orbit differs. This lesson is the second half of the CSCA syllabus line “Circular motion and universal gravitation”: you will apply F = mv²/r from the lesson “Circular motion” to the force of gravity.
Kepler’s laws
- First law (orbits): every planet moves on an ellipse with the Sun at one focus.
- Second law (areas): the line from the Sun to the planet sweeps equal areas in equal times. So the planet is fastest at perihelion (
近日点) and slowest at aphelion (远日点). - Third law (periods): for all bodies orbiting the same central body the ratio a³/T² is the same.
- asemi-major axis of the ellipse (the radius r for a circular orbit), m
- Torbital period, s
- ka constant that depends only on the mass of the central body: k = GM/(4π²)
Compare only bodies that orbit the same centre: planets of the Sun with each other, moons of Jupiter with each other — never the Moon with the Earth.
1) An asteroid’s orbit has a semi-major axis of 4 AU (1 AU = the radius of the Earth’s orbit). What is its period?
2) A comet has a period of 125 years. What is the semi-major axis of its orbit?
3) A comet’s nearest and farthest distances from the Sun are in the ratio 1 : 3. What is the ratio of its speeds at these points?
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2) a³ = T² = 125² ⇒ a = 125^(2/3) = 5² = 25 AU.
3) At the ends of the major axis the velocity is perpendicular to the radius, and the second law gives v₁r₁ = v₂r₂ ⇒ vnear : vfar = 3 : 1.
Newton’s law of universal gravitation
Any two bodies attract each other. By Newton’s third law the forces are equal and opposite: the Earth pulls the Moon exactly as hard as the Moon pulls the Earth. The formula holds for point masses and uniform spheres; for spheres r is the distance between the centres.
- Ggravitational constant, 6.67 × 10⁻¹¹ N · m²/kg²
- m₁, m₂masses of the bodies, kg
- rdistance between the centres, m
The force is proportional to the product of the masses and inversely proportional to the square of the distance: double r and F becomes four times smaller.
1) One mass is tripled and the distance is doubled. How does F change?
2) Two students of 50 kg each stand 1 m apart. Estimate the gravitational force between them.
3) The Earth’s mass is about 81 times the Moon’s. How do the forces with which they attract each other compare, and their accelerations?
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2) F = 6.67 × 10⁻¹¹ · 50 · 50/1² ≈ 1.7 × 10⁻⁷ N — far too small to feel; gravity matters only for planet-sized masses.
3) The forces are equal (third law). Since a = F/m, the Moon’s acceleration is 81 times larger than the Earth’s.
g at the surface and at a height; mass and density of a planet
At the surface of a planet the weight of a body equals the gravitational force (we ignore the planet’s rotation): mg = GMm/R². This gives g = GM/R² and the most useful equality in CSCA: GM = gR². Chinese textbooks call it the “golden substitution” (黄金代换) — when a problem gives g and R instead of M and G, replace GM by gR².
- M, Rmass and radius of the planet
- hheight above the surface; the distance from the centre is R + h
- ghfree-fall acceleration at height h
g falls with the square of the distance from the centre: g/4 at 2R, g/9 at 3R.
1) At the surface g = 10 m/s². What is g at a height h = 2R?
2) A body weighs 900 N on the Earth. What is its weight at a height h = R/2?
3) A planet has twice the Earth’s mass and twice its radius. What is g at its surface?
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2) The distance from the centre is 1.5R ⇒ the weight is multiplied by 1/1.5² = 4/9: 900 · 4/9 = 400 N.
3) g ∝ M/R² ⇒ g′ = 10 · 2/2² = 5 m/s².
A moon has 1/81 of the Earth’s mass and 1/4 of its radius. On the Earth g = 10 m/s². What is the free-fall acceleration at the surface of the moon, approximately?
A) 2.0 m/s² B) 0.5 m/s² C) 0.12 m/s² D) 51 m/s²
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B forgets to square the radius (10 · 4/81), C ignores the radius (10/81), and D inverts the ratio (10 · 81/16).
- r, Tradius and period of the orbit of a moon or satellite
- ρmean density of the planet, kg/m³
The satellite’s own mass cancels — only the central body can be “weighed” this way. For an orbit just above the surface, the density needs only T.
1) g = 10 m/s², R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N · m²/kg². Find the Earth’s mass.
2) A probe orbits just above the surface of a planet with a period T = 5.0 × 10³ s. What is the planet’s mean density (π ≈ 3.14)?
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2) ρ = 3π/(GT²) = 9.42/(6.67 × 10⁻¹¹ · 2.5 × 10⁷) ≈ 9.42/(1.67 × 10⁻³) ≈ 5.7 × 10³ kg/m³ — about the density of the Earth.
Satellites: orbits, cosmic velocities, geostationary satellites
In a circular orbit only gravity acts on the satellite, and it plays the role of the centripetal force: GMm/r² = mv²/r = mω²r = m · 4π²r/T² = ma. Each equality gives one formula, and the satellite’s mass m cancels in all of them. An astronaut on board feels weightless (完全失重) because he and the station “fall” with the same acceleration.
- rorbit radius r = R + h (from the centre of the planet)
- Mmass of the central body (GM = gR²)
The higher the orbit, the smaller v, ω and a, and the longer T.
1) Two satellites have orbit radii in the ratio 1 : 4. What are the ratios of v, ω, T and a?
2) A satellite orbits at a height h = R. How many times slower is it than a satellite just above the surface?
3) Find the period of an orbit just above the surface (g = 10 m/s², R = 6.4 × 10⁶ m, π ≈ 3.14).
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2) r = 2R ⇒ v = √(GM/2R) = v₁/√2 — √2 times slower (≈ 5.6 km/s).
3) T = 2π√(R/g) = 2π√(6.4 × 10⁵) = 2π · 800 ≈ 5.0 × 10³ s ≈ 84 min — the shortest possible period of an Earth satellite.
Satellites A and B move around the Earth in circular orbits with rA < rB. Which statement is correct?
A) vA < vB B) TA > TB C) aA > aB D) ωA < ωB
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A and D come from the error “bigger orbit — bigger speed” (using v = ωr as if ω were fixed); B turns T ∝ r^(3/2) upside down.
第一宇宙速度)The speed of a circular orbit just above the surface of a planet (r ≈ R). For the Earth ≈ 7.9 km/s. It is the minimum launch speed for a satellite and the maximum speed of any circular orbit. The second cosmic velocity (11.2 km/s) escapes the Earth’s gravity, the third (16.7 km/s) the Solar System.
- Rradius of the planet, m
- gfree-fall acceleration at the planet’s surface
mg = mv²/R ⇒ v₁ = √(gR): at the surface gravity supplies the centripetal force.
1) g = 10 m/s², R = 6.4 × 10⁶ m. Find v₁ for the Earth.
2) A planet has a radius of 3.6 × 10⁶ m and g = 2.5 m/s² at its surface. What is its first cosmic velocity?
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2) v₁ = √(2.5 · 3.6 × 10⁶) = √(9 × 10⁶) = 3.0 × 10³ m/s.
地球同步卫星)A satellite that moves in the plane of the equator, in the direction of the Earth’s rotation, with the Earth’s rotation period (≈ 24 h), so it always stays above the same point of the equator. Since T is fixed, r is fixed too: all geostationary satellites share one orbit — r ≈ 4.2 × 10⁴ km ≈ 6.6R (height ≈ 3.6 × 10⁴ km), v ≈ 3.1 km/s. Their masses may differ.
Changing orbits. To move to a higher orbit a spacecraft fires its engine to speed up: now mv²/r is larger than gravity, and the craft rises along an elliptical transfer orbit. At the far point it speeds up again to enter the new circular orbit — where its speed is lower and its period longer than before. To come down, it slows down. At a given point the acceleration a = GM/r² does not depend on which orbit the craft is on, but the speed does. A craft docking with a station starts in a lower orbit and speeds up; speeding up in the station’s own orbit would lift it higher and make it fall behind.
A satellite moves from circular orbit 1 to elliptical orbit 2 at point P, then to circular orbit 3 at point Q (r₃ > r₁; P is the nearest and Q the farthest point of orbit 2). Which statement is correct?
A) At Q the acceleration on orbit 2 is smaller than on orbit 3.
B) The speed on orbit 3 is greater than on orbit 1.
C) At P the speed on orbit 2 is greater than on orbit 1.
D) The period on orbit 2 is longer than on orbit 3.
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A: at the same point a = GM/r² is the same. B: the higher orbit is slower. D: the ellipse’s semi-major axis is smaller than r₃, so by Kepler’s third law its period is shorter.
How CSCA asks about this
Gravitation items in CSCA are usually two steps: one equation (GMm/r² = …) and one ratio. You have about 75 seconds per item. The typical patterns:
- Kepler’s laws: ellipse and focus, speed at perihelion/aphelion, a period or radius from a³/T² = k;
- how F changes when a distance or mass changes; who measured G (Cavendish);
- g at a height, g on another planet, a planet’s mass and density (GM = gR², ρ = 3π/(GT²));
- comparing two satellites (v, ω, T, a) — often from a table or a graph;
- the first cosmic velocity, “which statement is correct?” about geostationary satellites, changing orbits and docking.
| Term | 中文 | Pinyin |
|---|---|---|
| universal gravitation | 万有引力 | wànyǒu yǐnlì |
| gravitational constant | 引力常量 | yǐnlì chángliàng |
| Kepler’s laws | 开普勒定律 | Kāipǔlè dìnglǜ |
| ellipse / focus | 椭圆 / 焦点 | tuǒyuán / jiāodiǎn |
| semi-major axis | 半长轴 | bàn chángzhóu |
| perihelion / aphelion | 近日点 / 远日点 | jìnrìdiǎn / yuǎnrìdiǎn |
| satellite / orbit | 卫星 / 轨道 | wèixīng / guǐdào |
| first cosmic velocity | 第一宇宙速度 | dì-yī yǔzhòu sùdù |
| geostationary satellite | 地球同步卫星 | dìqiú tóngbù wèixīng |
| weightlessness | 完全失重 | wánquán shīzhòng |
| density | 密度 | mìdù |
| torsion balance | 扭秤 | niǔchèng |
| changing orbit | 变轨 | biànguǐ |
Key points
- Kepler: ellipses, equal areas (fastest at perihelion), a³/T² = k for one central body.
- F = Gm₁m₂/r² with r between the centres; Cavendish measured G.
- g = GM/R², GM = gR²; at a height gh = gR²/(R + h)².
- Mass: M = 4π²r³/(GT²); for an orbit just above the surface ρ = 3π/(GT²).
- Satellite: v = √(GM/r), T = 2π√(r³/GM); higher orbit — smaller v, ω, a, longer T; v₁ = √(gR) ≈ 7.9 km/s.
- Geostationary: above the equator, T = 24 h, all at one radius; to move up a craft speeds up, yet in the new orbit it is slower.
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