Edexcel A Level (IAL) Physics-5.36 Orbital Motion- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -5.36 Orbital Motion- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -5.36 Orbital Motion- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
Applying Newton’s Laws of Motion and Universal Gravitation to Orbital Motion
Orbital motion occurs when an object moves in a curved path under the influence of a central force. For satellites and planets, this central force is provided by gravity.
Orbital Motion and Newton’s First Law
- An object moving in a straight line at constant speed will continue to do so unless a force acts.
- In orbit, the object does not move in a straight line.
- This means a force must be acting on it.
Conclusion: A satellite remains in orbit because a force continuously changes its direction of motion.
Centripetal Motion and Newton’s Second Law
An object moving in a circular orbit undergoes centripetal acceleration.
Centripetal acceleration: \( a = \dfrac{v^2}{r} \)
Using Newton’s second law:
Centripetal force: \( F = m\dfrac{v^2}{r} \)
- The force is always directed towards the centre of the orbit.
- The speed may be constant, but velocity changes direction.
Gravitational Force as the Centripetal Force
For orbital motion, the centripetal force is provided by gravity.
Newton’s law of universal gravitation:
\( F = \dfrac{G M m}{r^2} \)
Equating centripetal force and gravitational force:
\( \dfrac{G M m}{r^2} = m\dfrac{v^2}{r} \)
Cancel \( m \):
\( v^2 = \dfrac{G M}{r} \)
Meaning of the Result
- Orbital speed depends on the mass of the central body.
- Orbital speed decreases with increasing orbital radius.
- Satellite mass does not affect orbital speed.
Key idea: Gravity provides exactly the force needed to keep the object in circular motion.
Orbital Motion and Newton’s Third Law
- The satellite exerts a gravitational force on the planet.
- The planet exerts an equal and opposite force on the satellite.
- The satellite accelerates more because it has much smaller mass.
Condition for Circular Orbit
- The gravitational force must act perpendicular to velocity.
- The object must have sufficient tangential speed.
- If speed is too low → object falls back.
- If speed is too high → object escapes orbit.
Example (Easy)
Why does a satellite in orbit not fall straight down to Earth?
▶️ Answer / Explanation
It has sufficient horizontal velocity, so gravity causes it to follow a curved path rather than falling vertically.
Example (Medium)
State the force that provides the centripetal acceleration for a satellite in circular orbit.
▶️ Answer / Explanation
The gravitational force between the satellite and the planet.
Example (Hard)
Explain why a satellite closer to Earth must move faster than one further away.
▶️ Answer / Explanation
- Gravitational force is stronger at smaller radius.
- Greater centripetal force is required.
- Therefore a higher orbital speed is needed.
Escape Velocity and Orbital Period (Kepler’s Third Law)
Gravitational fields determine both whether an object can escape from a planet and how objects move in stable orbits.
Escape Velocity
Escape velocity is the minimum speed an object must have to escape completely from a gravitational field without further propulsion.
Key condition:
- Final kinetic energy at infinity = 0
- Gravitational potential energy at infinity = 0
Derivation using energy conservation:
Initial KE = increase in gravitational potential energy
\( \dfrac{1}{2}mv^2 = \dfrac{GMm}{r} \)
Cancel \( m \):
\( v = \sqrt{\dfrac{2GM}{r}} \)
Escape velocity equation:
- Independent of mass of object
- Depends on mass and radius of planet
Meaning of Escape Velocity
- Object continues slowing but never returns
- Speed tends to zero at infinity
- No further force or thrust required
Important: Escape velocity is not the same as escape acceleration.
Orbital Motion and Orbital Period
For a satellite in a circular orbit, gravitational force provides centripetal force.
Gravitational force: \( F = \dfrac{GMm}{r^2} \)
Centripetal force: \( F = m\dfrac{v^2}{r} \)
Equating:
\( \dfrac{GMm}{r^2} = m\dfrac{v^2}{r} \)
\( v = \sqrt{\dfrac{GM}{r}} \)
Orbital Period
Orbital period \( T \) is the time taken for one complete orbit.
\( v = \dfrac{2\pi r}{T} \)
Substitute orbital speed:
\( \dfrac{2\pi r}{T} = \sqrt{\dfrac{GM}{r}} \)
Rearranging:
\( T^2 = \dfrac{4\pi^2 r^3}{GM} \)
Kepler’s Third Law
Kepler’s third law states:
The square of the orbital period is proportional to the cube of the orbital radius
\( T^2 \propto r^3 \)
- Applies to planets and satellites
- Same central mass → same constant
Comparison: Orbit vs Escape
- Orbital speed: \( v = \sqrt{\dfrac{GM}{r}} \)
- Escape speed: \( v = \sqrt{\dfrac{2GM}{r}} \)
- Escape speed is √2 times orbital speed
Example (Easy)
Does escape velocity depend on the mass of the object?
▶️ Answer / Explanation
No. The mass cancels out during derivation.
Example (Medium)
If orbital radius increases, what happens to orbital period?
▶️ Answer / Explanation
Orbital period increases since \( T^2 \propto r^3 \).
Example (Hard)
Explain why a satellite must have less speed than escape velocity to remain in orbit.
▶️ Answer / Explanation
- Escape velocity gives zero KE at infinity
- Lower speed keeps object bound
- Gravity bends path into orbit