ASTRONOMY FUNDAMENTALS - Chapter 3, Exercise 1 Solution
==========================================================
Applying Kepler's Third Law

PROBLEM
-------
A hypothetical planet orbits the Sun with a semi-major axis of 4 AU.
Using Kepler's third law (T^2 proportional to a^3, with Earth as the
reference where a = 1 AU and T = 1 year), calculate this planet's real
orbital period in years.

SOLUTION
--------
Kepler's third law, using Earth as the reference point (a = 1 AU,
T = 1 year), simplifies to:

    T^2 = a^3

where T is measured in years and a is measured in AU.

Step 1 - substitute a = 4:
    T^2 = 4^3
    T^2 = 64

Step 2 - solve for T by taking the square root:
    T = sqrt(64)
    T = 8

ANSWER: The planet's orbital period is 8 years.

----
WHY THIS WORKS AS AN ANSWER

Kepler's third law states that the square of the orbital period is
proportional to the cube of the semi-major axis. Using Earth's own real
orbit (1 AU, 1 year) as the reference point conveniently makes the
proportionality constant exactly 1, which is why the relationship
simplifies to the clean equation T^2 = a^3 when T is in years and a is
in AU - no extra unit conversion or constant is needed.

This result also demonstrates something genuinely important about the
relationship: a planet doesn't need to be 4 times farther out to take 4
times as long to orbit - it takes considerably longer than that. Going
from 1 AU to 4 AU (4 times farther) increases the orbital period from 1
year to 8 years (8 times longer), because the period scales with the
CUBE of the distance, not the distance itself. This is a real, direct
consequence of a planet's own orbital speed slowing down significantly
the farther out it orbits, which is itself a real consequence of
gravity's own inverse-square weakening with distance.