http://www.ck12.org Chapter 9. Newton’s Universal Law of Gravity
Illustrative Example 1
A satellite is to be placed in orbit about the Earth halfway between the Earth and the moon. The period of the moon
about the Earth is 27.3 days. What is the period of the satellite?
Answer:
Let us call the distance between the Earth and moonRand the distance from the Earth to the satelliteR 2. Using
Kepler’s Third Law for the Earth and Moon, we first solve fork.
T^2 =kr^3 →( 27. 3 )^2 =kR^3 →k=(^27.^3 )
2
R^3. Now that we knowkwe solve for the periodTsof the satellite.Ts^2 =kR
3
23 →substitute fork→
( 27. 3 )^2
R^3 →Ts(^2) =(^27.^3 )^2
R^3
R^3
23 →solve forTs=
√
(√ 27. 3 )^2
23 =^9.^65 daysCheck Your Understanding:
Two exoplanets (planets that orbit stars other than the sun), Close and Far, orbit their parent sun such that Close is
2.0 AU from its sun and Far is 4.0 AU from its sun. If Close takesnyears to orbit its sun, then Far will take:
a. Less than^12 nyears
b.^12 nyears
c. 2nyears
d. More than 2nyears
Answer:This problem is, essentially, the same as Example 1. In that problem, two bodies orbited the Earth, one
twice as far from the Earth as the other. We saw that the period of the moon which was twice as far away as the
satellite had a period of 27.3 days and the satellite had a period of 9.65 days. Clearly, the relationship (which we
can see from the equation) is not linear. Since the moon’s period is greater than twice the satellite’s period, in fact,
almost three times as much, it will take Far longer than 2nyears to orbit its sun. The answer is therefore, D.
http://demonstrations.wolfram.com/OrbitalSpeedAndPeriodOfASatellite/