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Educora
Intermediate28 min31 / 68

Universal gravitation and satellites

Kepler’s laws, Newton’s law of gravitation, g at the surface and at a height (GM = gR²), the mass and density of a planet, satellite orbits, the first cosmic velocity, geostationary satellites and changing orbits — the CSCA way.

Check yourself
In this lesson you will learn
  • 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

  1. First law (orbits): every planet moves on an ellipse with the Sun at one focus.
  2. 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).
  3. Third law (periods): for all bodies orbiting the same central body the ratio a³/T² is the same.
a³/T² = ka³/T² = k
where:
  • 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.

Example 1: calculating with Kepler’s laws

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?

Show solution
1) For the Earth a = 1 AU and T = 1 year ⇒ k = 1. T² = a³ = 64 ⇒ T = 8 years.
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.
Interactive
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Planets far from the Sun move more slowly and travel a longer path — that is why T grows quickly with r: T ∝ r^(3/2).

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.

F = G · m₁m₂/r²F = G · m₁m₂/r²
where:
  • 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.

Example 2: how does the force change?

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?

Show solution
1) F′ = F · 3/2² = 3F/4.
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.

g = GM/R² GM = gR² gh = GM/(R + h)² = g · R²/(R + h)²g = GM/R² GM = gR² gh = GM/(R + h)² = g · R²/(R + h)²
where:
  • 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.

Interactive
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gh for the Earth (m/s²) against x = h/R. At x = 1 (h = R) g is already 2.5 m/s², four times smaller.
Example 3: g at a height and on another planet

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?

Show solution
1) The distance from the centre is 3R ⇒ gh = g/9 ≈ 1.1 m/s² (writing g/4 treats h as the distance from the centre).
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.
Example 4 (CSCA format): g on a small moon

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

Show solution
g′ = g · (M′/M) · (R/R′)² = 10 · (1/81) · 16 = 160/81 ≈ 2.0 m/s², answer A.
B forgets to square the radius (10 · 4/81), C ignores the radius (10/81), and D inverts the ratio (10 · 81/16).
M = gR²/G M = 4π²r³/(GT²) ρ = M/(4πR³/3) = 3πr³/(GT²R³); r = R: ρ = 3π/(GT²)M = gR²/G M = 4π²r³/(GT²) ρ = M/(4πR³/3) = 3πr³/(GT²R³); r = R: ρ = 3π/(GT²)
where:
  • 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.

Example 5: the Earth’s mass and a planet’s density

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)?

Show solution
1) M = gR²/G = 10 · 4.1 × 10¹³/(6.67 × 10⁻¹¹) ≈ 6 × 10²⁴ kg.
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.

v = √(GM/r) ω = √(GM/r³) T = 2π√(r³/(GM)) a = GM/r²v = √(GM/r) ω = √(GM/r³) T = 2π√(r³/(GM)) a = GM/r²
where:
  • 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.

Example 6: orbit radius and the satellite’s quantities

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).

Show solution
1) v ∝ 1/√r ⇒ 2 : 1; ω ∝ 1/√(r³) ⇒ 8 : 1; T ∝ √(r³) ⇒ 1 : 8; a ∝ 1/r² ⇒ 16 : 1.
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.
Example 7 (CSCA format): two satellites

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

Show solution
A, in the lower orbit, is faster, has larger ω and a, and a shorter period ⇒ answer C (a = GM/r²).
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.
Definition
First cosmic velocity (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.

v₁ = √(GM/R) = √(gR)v₁ = √(GM/R) = √(gR)
where:
  • 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.

Example 8: the first cosmic velocity

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?

Show solution
1) v₁ = √(10 · 6.4 × 10⁶) = √(6.4 × 10⁷) = √(64 × 10⁶) = 8.0 × 10³ m/s (7.9 km/s with g = 9.8).
2) v₁ = √(2.5 · 3.6 × 10⁶) = √(9 × 10⁶) = 3.0 × 10³ m/s.
Definition
Geostationary satellite (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.

Example 9 (CSCA format): a transfer orbit

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.

Show solution
At P the satellite speeds up to enter orbit 2 ⇒ answer C.
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
The test is taken in English or Chinese — recognise the terms in both.

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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By Kepler’s third law, which quantity is the same for all planets of the Sun?

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