SOLUTIONS TO EXERCISE 30

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(a) Kepler's Third Law can be expressed as P^2 = K r^3, where P is
the orbital period and r is the distance from the center of the
primary object to the orbiting object.  As discussed in class,
for objects orbiting the Sun, the proportionality constant K is
equal to one when P is in units of years and r is in units of
astronomical units (1 AU = 1.50 * 10^8 km).  This newly discovered
minor planet is orbiting the Sun at a distance of 300,000,000 km.
This is 300,000,000 / 150,000,000 = 2 AU.  So we can say that

P^2 = 1 * (2^3) = 8 ;  so P = sqrt(8) = 2.83 years.

Convert this length of time to seconds:

2.83 years = 2.83 * 365.25 * 24 * 3600 = 8.93 * 10^7 sec

And the distance the minor planet travels in one orbit is the
circumference of its orbit, which is

2 * 3.14 * 3 * 10^8 = 1.88 * 10^9 km

So the minor planet's orbital velocity is

1.88 * 10^9 km
---------------  =  21.1 km/sec
8.93 * 10^7 sec

(b) This solution is left as an exercise to the reader.  The answer turns out to