12 ORBITAL MOTION 12.6 Planetary orbits

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The satellite’s orbital period is simply

2 π r

T =

v

2 × π × (1.107 × 107 ) (^)
6000
= 3.22 hours.
Worked example 12.4: Halley’s comet
Question: The distance of closest approach of Halley’s comet to the Sun is 0.57 AU.
(1 AU is the mean Earth-Sun distance.) The greatest distance of the comet from
the Sun is 35 AU. The comet’s speed at closest approach is 54 km/s. What is its
speed when it is furthest from the Sun?
Answer: At perihelion and aphelion, the comet’s velocity is perpendicular to its
position vector from the Sun. Hence, at these two special points, the comet’s
angular momentum (around the Sun) takes the particularly simple form
l = m r u.
Here, m is the comet’s mass, r is its distance from the Sun, and u is its speed.
According to Kepler’s second law, the comet orbits the Sun with constant angular
momentum. Hence, we can write
r 0 u 0 = r 1 u 1 ,
where r 0 and u 0 are the perihelion distance and speed, respectively, and r 1 and
u 1 are the corresponding quantities at aphelion. We are told that r 0 = 0.57 AU,
r 1 = 35 AU, and u 0 = 54 km/s. It follows that
u =
u 0 r 0
(^1) r
1
=^
54 × 0.57
35
= 0.879 km/s.
Worked example 12.5: Mass of star
Question: A planet is in circular orbit around a star. The period and radius of the
orbit are T = 4.3 107 s and r = 2.34 1011 m, respectively. Calculate the mass
of the star.