Physics · Class 11

Trace an equal-area orbit

An ellipse has the Sun at a focus. The planet moves faster near the Sun and slower farther away, sweeping equal areas in equal times. Increasing the semi-major axis increases the period according to T² proportional to a³.

Open the interactive experiment →

Trace an equal-area orbit

கோள் நீள்வட்டப் பாதையில் சூரியனை ஒரு குவியத்தில் கொண்டு சுற்றும். சம நேரங்களில் சம பரப்புகளை கடக்கும்.

Key notes, relationships and model limits

Key relationship

r=a(1−e cosE); E−e sinE=M; T²=4π²a³/(GM); dA/dt=½√(GMa(1−e²)).

Remember

Common trap

Uniform angular speed does not describe an eccentric Kepler orbit.

Model limits

Planet around a fixed Sun of mass 1.989×10³⁰ kg. The 10-second animation represents one complete physical orbit; equal animation times represent equal physical times.

What you will learn

Connect elliptic geometry, equal swept areas and period scaling.

Try the experiment

Try both scenarios, change one control, and predict the result before checking the graph and readouts.

Think of it this way

Change one input while keeping the others fixed, as you would in a controlled experiment.

Worked example

For a=1.5×10¹¹ m and e=0.5, perihelion r=7.5×10¹⁰ m and aphelion r=2.25×10¹¹ m; the period is about 3.17×10⁷ s.

Why this matters

The model connects trace an equal-area orbit with an observable reading. Check the equation and assumptions before applying it.

Before you start

SI units · algebra · graph reading

Check your understanding

Where is the planet fastest on an elliptic orbit?

At aphelion, farthest from the Sun
At perihelion, closest to the Sun
Its speed is equal everywhere

Read the explanation

Conserved angular momentum requires faster tangential motion at a smaller radius.

Course connection: Class XI Chapter 7. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.