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General physics
🎯 Learning Objectives
  • Analyse circular orbits by relating gravitational force to the centripetal acceleration it provides
  • Derive orbital speed (v = sqrt{dfrac{GM}{r}}) for a small mass orbiting a large mass
  • Calculate orbital period (T = dfrac{2pi r}{v}) and understand its dependence on radius
  • Sketch force vs radius graphs and discuss stability of circular orbits
🗣️ Language Objectives
  • Use terms “centripetal acceleration,” “orbital speed,” “orbital period,” “inverse-square law” correctly
  • Explain relationships between force, mass, velocity, and radius in clear English
  • Interpret and discuss derivations and graphs of circular motion
  • Describe stability and perturbations of orbits using precise vocabulary
📚 Key Terms and Translations
English TermRussianKazakh
Centripetal accelerationЦентростремительное ускорениеОрталыққа тартатын үдеу
Orbital speedОрбитальная скоростьОрбиталдық жылдамдық
Orbital periodПериод обращенияОрбита периоды
Gravitational constant (G)Гравитационная постояннаяГравитациялық тұрақты
Inverse-square lawЗакон обратных квадратовКвадраттық кері заң
Radius (r)РадиусРадиус
🃏 Vocabulary Study Cards

Centripetal Acceleration

Definition: (a_c = dfrac{v^2}{r}), acceleration toward centre of circle

Use: Required to maintain circular motion

Orbital Speed

Definition: (v = sqrt{dfrac{GM}{r}})

Context: Speed for a stable circular orbit

Orbital Period

Definition: (T = dfrac{2pi r}{v})

Implication: Depends on radius and central mass

Inverse-Square Law

Definition: Gravitational force ∝ (1/r^2)

Effect: Strength decreases rapidly with distance

📖 Glossary of Terms

Centripetal Acceleration

The acceleration of a body moving in a circle of radius (r) at speed (v), directed toward the centre.

Translation
Russian: Ускорение тела, движущегося по окружности радиуса (r) со скоростью (v), направленное к центру.
Kazakh: Радиусы (r) және жылдамдығы (v) бар доға бойымен қозғалатын дененің орталыққа бағытталған үдеуі.
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Orbital Speed

The constant speed a small mass must have to maintain a circular orbit under gravity.

Translation
Russian: Постоянная скорость, необходимая малой массе для круговой орбиты под действием гравитации.
Kazakh: Гравитация әсерінен сақиналық орбитаны ұстап тұру үшін қажетті тұрақты жылдамдық.
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🔬 Theory: Circular Orbits in Gravitational Fields

Relating Gravity to Centripetal Force

For a mass (m) orbiting a much larger mass (M) at radius (r), the gravitational force
(F_g = G dfrac{M m}{r2}
must equal the centripetal force
(F_c = m dfrac{v^2}{r}).

Equate them:
(;G dfrac{M m}{r^2} = m dfrac{v^2}{r})

Cancel (m) and solve for orbital speed:

(v = sqrt{dfrac{G M}{r}})

Then the orbital period is:

(T = dfrac{2pi r}{v} = 2pi sqrt{dfrac{r^3}{G M}})

Translation
Russian: Приравнивая (G frac{Mm}{r^2}) и (m frac{v^2}{r}), получаем (v = sqrt{frac{GM}{r}}) и (T = 2pisqrt{frac{r^3}{GM}}).
Kazakh: (G frac{Mm}{r^2}) және (m frac{v^2}{r}) теңестіріп, (v = sqrt{frac{GM}{r}}) және (T = 2pisqrt{frac{r^3}{GM}}) алынады.
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Theory Questions

Easy: Write the equation that equates gravitational and centripetal forces for circular orbit.
Answer
(G dfrac{M m}{r^2} = m dfrac{v^2}{r}).
Medium: Derive the expression for orbital speed (v) from (G dfrac{M m}{r^2} = m dfrac{v^2}{r}).
Answer
Cancel (m), multiply by (r): (Gfrac{M}{r}=v^2), so (v=sqrt{frac{GM}{r}}).
Medium: Show that (T = 2pisqrt{dfrac{r^3}{GM}}) by substituting (v) into (T=2pi r/v).
Answer
Substitute (v): (T=2pi r/sqrt{frac{GM}{r}}=2pisqrt{frac{r^3}{GM}}).
Hard (Critical Thinking): Discuss how non-uniform mass distributions or perturbations affect circular orbit stability.
Answer
Deviations from spherical symmetry cause precession and orbital decay; perturbations (e.g. third bodies) lead to elliptical or unstable orbits.
💪 Memorization Exercises

Complete the Formulas

  1. Centripetal acceleration: (a_c = dfrac{v^2}{__}).
  2. Orbital speed: (v = sqrt{dfrac{__,M}{r}}).
  3. Orbital period: (T = 2pisqrt{dfrac{r^3}{__,M}}).
  4. Gravitational force: (F = G dfrac{M m}{r^__}).

Answer
1. (r)
2. (G)
3. (G)
4. (2)
🎥 Video Lesson

Additional Video Resources:

Orbital Motion & Kepler’s Laws

Gravity and Circular Motion

📐 Worked Examples

Example 1: Low Earth Orbit Speed

Calculate speed of satellite orbiting Earth (M=5.97×1024 kg) at r=7.0×106 m.

Orbital speed diagram

Solution
Answer
(v=sqrt{frac{6.67times10^{-11}times5.97times10^{24}}{7.0times10^6}}approx7.5times10^3text{ m/s}.)

Example 2: Geostationary Orbit Period

Find period for satellite at r=4.23×107 m around Earth.

Orbital period plot

Solution
Answer
(T=2pisqrt{frac{(4.23times10^7)^3}{6.67times10^{-11}times5.97times10^{24}}}approx8.64times10^4text{ s}approx24text{ h}.)
🧪 Interactive Investigation

Explore orbits and Kepler’s laws with PhET:


Investigation Answers
Demonstrate (vpropto r^{-1/2}) and (Tpropto r^{3/2}) by measuring orbital parameters.
👥 Collaborative Group Activity

Use a Quizizz challenge on circular orbits:


📝 Individual Assessment

Solve these structured questions:

  1. Derive (v=sqrt{frac{GM}{r}}) from force balance.
  2. Calculate speed for Mars satellite at r=4.0×106 m (M=6.42×1023 kg).
  3. Show how period scales with radius and mass of central body.
  4. Discuss effect of small radial perturbation on circular orbit stability.
  5. Compare energy (kinetic + potential) of circular vs elliptical orbit at same perigee.
🤔 Lesson Reflection
  • How does radius affect orbital speed and period?
  • Which assumption (negligible mass, perfect circle) is most limiting?
  • What real-world factors (atmosphere, perturbations) alter ideal orbits?
  • How can you apply circular orbit concepts to satellite design?