Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Orbits and the Wider Universe
Using orbital data to find the mass of the central object
- 3 things to know
- Component 2
Overview
Recall questions on rearranging to M = 4π²r³/GT² and to M = v²r/G, choosing the right form for the data given, worked determinations of the mass of the Sun, of a planet and of the Galaxy, and checking the result by working forwards.
What you need to know
Answer each one before you reveal it.
State the equation used to find the mass of a central body from the orbital speed and radius of a body orbiting it.
Why does an orbital measurement give the mass of the central body but not the mass of the orbiting body?
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the equation used to find the mass of a central body from the period and radius of an orbit about it.
Model answer
Rearranging T² = 4π²r³/GM gives M = 4π²r³/GT², where M is in kilograms, r is the orbital radius in metres and T is the period in seconds.
- 2
State the equation used to find the mass of a central body from the orbital speed and radius of a body orbiting it.
Model answer
From v² = GM/r the central mass is M = v²r/G, where v is the orbital speed in m s⁻¹ and r is the orbital radius in metres.
- 3
Calculate the mass of the Sun from the Earth's orbit, taking the orbital radius as 1.50 × 10¹¹ m and the period as 3.16 × 10⁷ s.
Model answer
The mass is M = 4π² × (1.50 × 10¹¹)³ ÷ (6.67 × 10⁻¹¹ × (3.16 × 10⁷)²) = 2.0 × 10³⁰ kg. Substituting back, T = √(4π² × 3.375 × 10³³ ÷ (6.67 × 10⁻¹¹ × 2.0 × 10³⁰)) returns 3.16 × 10⁷ s.
- 4
Calculate the mass of the Earth from the Moon's orbit, taking the orbital radius as 3.84 × 10⁸ m and the period as 2.36 × 10⁶ s.
Model answer
The mass is M = 4π² × (3.84 × 10⁸)³ ÷ (6.67 × 10⁻¹¹ × (2.36 × 10⁶)²) = 6.0 × 10²⁴ kg. Substituting back, T = √(4π² × 5.66 × 10²⁵ ÷ (6.67 × 10⁻¹¹ × 6.02 × 10²⁴)) returns 2.36 × 10⁶ s.
- 5
Calculate the mass of Jupiter from the orbit of Io, taking the orbital radius as 4.22 × 10⁸ m and the period as 1.53 × 10⁵ s.
Model answer
The mass is M = 4π² × (4.22 × 10⁸)³ ÷ (6.67 × 10⁻¹¹ × (1.53 × 10⁵)²) = 1.9 × 10²⁷ kg. Substituting back, T = √(4π² × 7.52 × 10²⁵ ÷ (6.67 × 10⁻¹¹ × 1.90 × 10²⁷)) returns 1.53 × 10⁵ s.
- 6
Why does an orbital measurement give the mass of the central body but not the mass of the orbiting body?
Model answer
The orbiting body's mass cancels out of the centripetal condition, so nothing in the period or the radius depends on it.
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