Kepler's laws of planetary motion
The Moon changes shape every month and the planets wander among the stars; people watched these regular patterns for thousands of years. Link to earlier classes : the solar system of earlier classes now gets exact laws of motion.
Kepler found three laws that describe how planets move around the Sun.
In the figure : The planet moves on an ellipse with the Sun at one focus; the two shaded regions have equal areas and are swept in equal times.
Rule 1: The three laws — Law of orbits : every planet moves in an ellipse with the Sun at one focus. Law of areas : the line from the Sun to the planet sweeps out equal areas in equal times, so the planet moves faster when it is nearer the Sun; this is conservation of angular momentum, since the gravitational force is central. Law of periods : T² ∝ a³ (T = time period, a = semi-major axis).
- Easy : A planet moves faster at perihelion (closest) than at aphelion (farthest).
- Medium : A planet orbits at 4 times the Earth's distance from the Sun. T² = 4³ = 64 (in years²) ⇒ T = 8 years.
- Exam : Two satellites of the same planet have orbit radii in the ratio 1 : 4; the ratio of their periods = √(1 : 64) = 1 : 8.
Rule 2: Circular orbits — For a nearly circular orbit of radius R around a mass M : T² = 4π²R³ ÷ (GM).
- Easy : The constant T² ÷ R³ is the same for all planets of the same star.
- Medium : If the radius of the orbit is doubled, T becomes 2√2 = 2.83 times larger.
- Exam : Mercury's orbit radius is 0.39 AU : T = 0.39^(3/2) = 0.24 year ≈ 88 days.
ellipse, equal areas, T² ∝ a³; closest to the Sun means fastest.
Saying that the Sun is at the centre of the ellipse (it is at a focus).
Memory clue : "Orbits, Areas, Periods — O A P." Close to the Sun, fast; far from the Sun, slow.
Spot the difference
| Law | Statement | Reason |
|---|---|---|
| First | elliptical orbit | inverse-square force |
| Second | equal areas in equal times | angular momentum conserved |
| Third | T² ∝ a³ | gravitational force law |
Universal law of gravitation
An apple falls to the ground and the Moon stays in its orbit; Newton saw that both come from the same pull. Link to earlier classes : gravitation of class 9 now has exact rules with the constant G.
Every particle attracts every other particle with a force directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
Rule 1: The law — F = G m₁m₂ ÷ r², G = 6.67 × 10⁻¹¹ N m² kg⁻². The force is attractive, acts along the line joining the particles, and the two forces are equal and opposite (third law). For several masses forces add as vectors (superposition).
- Easy : Two 1 kg masses 1 m apart attract each other with 6.67 × 10⁻¹¹ N, an extremely small force.
- Medium : Two 100 kg masses 2 m apart : F = 6.67 × 10⁻¹¹ × 10⁴ ÷ 4 = 1.67 × 10⁻⁷ N.
- Exam : Doubling the distance reduces the force to one-fourth; doubling both masses makes the force four times larger.
Rule 2: Gravity is very weak but long ranged — It dominates only when at least one mass is huge, such as a planet.
- Easy : The pull between two students is too small to notice.
- Medium : The earth (6 × 10²⁴ kg) pulls a 1 kg mass at distance 6.4 × 10⁶ m with F = 6.67 × 10⁻¹¹ × 6 × 10²⁴ ÷ (6.4 × 10⁶)² = 9.8 N.
- Exam : The Moon pulls the earth with the same force that the earth pulls the Moon.
F = G m₁m₂ ÷ r²; inverse-square; equal and opposite on the two bodies.
Using the distance between surfaces instead of centres, and forgetting to square the distance.
Memory clue : "Mass up, force up; distance up, force down (squared)."
Spot the difference
| Quantity | Symbol | Value |
|---|---|---|
| Universal constant | G | 6.67 × 10⁻¹¹ N m² kg⁻² |
| Acceleration due to gravity | g | 9.8 m s⁻² (varies with place) |
Acceleration due to gravity
You weigh slightly less on a high mountain and also slightly less at the equator than at the poles. Link to earlier classes : the value 9.8 m s⁻² is now derived and its variations are explained.
The acceleration due to gravity g is the acceleration of a freely falling body near the earth. It does not depend on the mass of the falling body.
Rule 1: Value at the surface — g = GM ÷ R² (M = mass and R = radius of the earth). Taking M = 6.0 × 10²⁴ kg and R = 6.4 × 10⁶ m, g ≈ 9.8 m s⁻².
- Easy : g = 6.67 × 10⁻¹¹ × 6 × 10²⁴ ÷ (6.4 × 10⁶)² = 9.77 m s⁻².
- Medium : A planet with the same mass but twice the radius has g ÷ 4.
- Exam : A planet with mass 8 times and radius 2 times that of the earth has g = 8 ÷ 4 = 2 times that of the earth.
Rule 2: Above and below the surface — At height h (h ≪ R) : g_h = g(1 − 2h ÷ R); exactly g_h = g R² ÷ (R + h)². At depth d : g_d = g(1 − d ÷ R); g is zero at the centre and maximum at the surface.
- Easy : At h = 320 km (h ÷ R = 0.05) : g_h = 9.8 × (1 − 0.1) = 8.82 m s⁻² (approximate).
- Medium : At depth d = 3200 km (half the radius) : g_d = 9.8 ÷ 2 = 4.9 m s⁻².
- Exam : At the height h = R, g_h = g ÷ 4 = 2.45 m s⁻².
Rule 3: Why g changes with place — The earth is slightly flattened and it spins. Therefore g is least at the equator and most at the poles. It decreases going up and going down.
- Easy : A body weighs a little more at the poles than at the equator.
- Medium : Weight W = mg changes with g; mass does not change.
- Exam : g decreases with height as 1 ÷ r², and with depth linearly.
g = GM ÷ R²; g_h ≈ g(1 − 2h ÷ R); g_d = g(1 − d ÷ R).
Mixing up mass (constant) and weight (depends on g), and using the height formula for large h.
Memory clue : "Height — twice the fall rate (2h/R); depth — once (d/R)."
Spot the difference
| Place | g |
|---|---|
| Surface | g = GM ÷ R² |
| Height h (small) | g(1 − 2h ÷ R) |
| Depth d | g(1 − d ÷ R) |
| Centre of the earth | 0 |
Gravitational potential energy and escape speed
A rocket needs a huge speed at launch; a ball thrown upwards always falls back. Link to earlier classes : the formula mgh is now seen as a small-height approximation of a more general law.
Gravitational potential energy of two masses is the work needed to bring them from infinite separation to the given distance; it is negative. The escape speed is the minimum launch speed to leave the earth forever.
Rule 1: Potential energy — U = −G m₁m₂ ÷ r (zero at infinity). Near the earth's surface, the change is U = mgh (h ≪ R). The total mechanical energy E = K + U is constant for the motion in a gravitational field.
- Easy : For the earth and a 1 kg mass at the surface : U = −6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1 ÷ 6.4 × 10⁶ = −6.25 × 10⁷ J.
- Medium : Raising 1 kg by 10 m near the surface : ΔU = 1 × 9.8 × 10 = 98 J.
- Exam : The negative sign shows that the earth and the mass form a bound system; energy is needed to separate them.
Rule 2: Escape speed — v_e = √(2GM ÷ R) = √(2gR). For the earth, v_e ≈ 11.2 km s⁻¹; for the Moon ≈ 2.4 km s⁻¹. It does not depend on the mass or the direction of the projected body.
- Easy : v_e = √(2 × 9.8 × 6.4 × 10⁶) = 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹.
- Medium : A planet with the earth's mass but half the radius has v_e = 11.2 × √2 = 15.8 km s⁻¹.
- Exam : The escape speed is √2 times the orbital speed near the surface (7.9 km s⁻¹).
U = −GMm ÷ r; v_e = √(2gR) = 11.2 km s⁻¹ for the earth.
Thinking the escape speed depends on the mass of the body, and dropping the minus sign of U.
Memory clue : "11.2 to escape, 7.9 to orbit — ratio root two."
Spot the difference
| Quantity | Formula | Earth value |
|---|---|---|
| Escape speed | √(2GM ÷ R) | 11.2 km s⁻¹ |
| Orbital speed near the surface | √(GM ÷ R) | 7.9 km s⁻¹ |
Earth satellites
Weather satellites, GPS and TV dishes all depend on bodies that fall around the earth without ever hitting it. Link to earlier classes : circular motion and gravitation now join.
A satellite is a body in orbit around a planet; gravity provides the centripetal force.
Rule 1: Orbital speed and period — For a circular orbit of radius r = R + h : v = √(GM ÷ r); T = 2πr ÷ v = 2π√(r³ ÷ GM). Near the earth's surface v ≈ 7.9 km s⁻¹ and T ≈ 85 minutes. A higher orbit means a lower speed and a longer period.
- Easy : At r = 6.4 × 10⁶ m : v = √(6.67 × 10⁻¹¹ × 6 × 10²⁴ ÷ 6.4 × 10⁶) = 7.9 × 10³ m s⁻¹.
- Medium : For r = 6.8 × 10⁶ m, T = 2π√(r³ ÷ GM) = 2π × √(3.14 × 10²⁰ ÷ 4.0 × 10¹⁴) = 5570 s ≈ 93 min.
- Exam : A heavier satellite has the same orbital speed and period; v and T depend only on the orbit radius.
Rule 2: Energy of a satellite — K = GMm ÷ 2r; U = −GMm ÷ r; E = −GMm ÷ 2r (negative: bound). The energy to be supplied to lift it from the surface into orbit is the change of E; to escape from the orbit, E must become zero.
- Easy : K = −E and U = 2E.
- Medium : If the satellite loses energy (air drag), r decreases and its speed increases.
- Exam : For a 1000 kg satellite in an orbit of radius 7 × 10⁶ m : E = −6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1000 ÷ (2 × 7 × 10⁶) = −2.86 × 10¹⁰ J.
Rule 3: Geostationary satellite — Its period is 24 hours, it orbits above the equator in the direction of the earth's spin, so it appears fixed in the sky. Its radius is about 4.2 × 10⁴ km from the centre (36000 km above the surface) and its orbital speed is about 3.1 km s⁻¹. Used in communication and TV.
- Easy : T = 24 h = 86400 s; r = (GMT² ÷ 4π²)^(1/3) = 4.2 × 10⁷ m.
- Medium : Its height above the surface = 4.2 × 10⁷ − 6.4 × 10⁶ = 3.6 × 10⁷ m.
- Exam : Because the period is fixed, a geostationary satellite can have only one orbit radius.
v = √(GM ÷ r); T² ∝ r³; E = −GMm ÷ 2r; geostationary : T = 24 h, above the equator.
Believing that astronauts float because gravity is absent (gravity is there; they are in free fall).
Memory clue : "Higher means slower." And "E = −K" for a circular orbit.
Spot the difference
| Quantity | Low orbit | High orbit |
|---|---|---|
| Speed | larger | smaller |
| Period | shorter | longer |
| Total energy | more negative | less negative |