Kepler$$s Laws $2$

Equipment

Ruler, Protractor, Sharp Pencil

Kepler$$s $3$rd Law

The squares of the sidereal periods of the planets are proportional to the cubes of their semi$-$major

axes.

$23$ T kr $=$

where T is the sidereal period of the planet and r is the semi$-$major axis of the planet.

$1.$ We can derive Kepler$$s $3$rd law from Newton$$s law of gravitation:

a$.$ For a satellite of mass m in orbit around a planet $($or moon$)$ of mass M$,$ begin with

F ma g

$=$ centripetal

$.$ Then substitute for

Fg

$($Newton$$s law of gravitation$)$ and for

centripetal a

$($for

an approximately circular orbit$),$ and then solve for

$2$ T $.$

b$.$ You should get that

$23$ T r $$

$,$ just like in Kepler$$s $3$rd law!

$2.$ Now take the square root to get T$.$