Kepler's third law question

Question

According to Kepler's Third Law, if the semi-major axis of an orbit increases by a factor of 4 (in other words, became 4 times larger), what would that indicate about the period of the orbit?

 

Using the simplified version of kepler's third law:

 

p²/a³

 

Thank you. I'm very lost.

Answer 1

Hi Fisher!

 

The upshot of Kepler's Third law, in the simplified form, is that -- for any planet orbiting the Sun, the quantity p²/a³ is a constant.  Hence, if we (in some supernatural way) alter either the period or the semi-major axis of the orbit, the other quantity will also adjust (naturally) so that the period and semi-major axis still had the same value for period squared over semi-major axis cubed.  Thus, if we start with period p and semi-major axis a, and then move to having period p' and semi-major axis a', we can write:

 

(p²/a³) = ((p')²/(a')³)

 

We are also given the relationship that the new semi-major axis, a', is four times larger than the original, a.  Thus, we can write:

 

a' = 4a

 

And the Kepler's Third law relation above becomes:

 

(p²/a³) = ((p')²/(4a)³) = (p')²/(64a³)

 

Rearranging a bit, this becomes:

 

(p')² = (p²/a³)(64a³) --> (p')² = p²(64) -->(taking square root of both sides) p' = p(8) = 8p

 

Thus, the new period p' equals 8 times the original period, once the semi-major axis is increased.  This is the general reason why the outer planets, with much larger semi-major axes of orbit, have much longer periods than the inner planets.

 

Hope this helps!  If you have any other questions about this, just let me know.

Answer 2

Original: p² / a³

Altered: p² / (4a)³, p² / 64a³

 

The period is 1/64.