Interpret elliptical motion and use period–radius data to infer a central mass.
Method and worked examples
Kepler’s first law places the central body at one focus of an elliptical orbit. The second law gives equal swept areas in equal times, so an orbiting body moves faster near the focus and more slowly farther away. Equal areas do not mean equal arc lengths. A circle is a special ellipse.
For one dominant central mass, T2 is proportional to a3, where a is semi-major axis. For a circle, a=r. AT2−versus-r3 graph has gradient GM4π2, so M=G×gradient4π2. Objects orbiting different central masses need not share the same gradient. For one mass, radius ratios give period ratios using T ∝ r3ᐟ2.
Example 1
Satellite B has circular radius four times A around the same planet. Compare periods.
T ∝ r3ᐟ2, so TATB=43ᐟ2=8.
Example 2
AT2 versus r3 graph has gradient 5.92×10−14s2m−3. Use G=6.67×10−11 to estimate mass.
M=G×gradient4π2=1.00×1025kg, using π=3.14159.
Watch out:Equal-area motion requires higher speed near the central body and lower speed farther away.
11. Kepler’s laws and orbital data | Year 12 Physics | Ascent Arcade