Download Problems 4 File

Intermediate Dynamics – Problem Sheet 4
These questions concentrate on gravity as an example of a central force, Kepler’s laws and
planetary motion.
Problem 4.1
(a.) State Kepler’s three laws of planetary motion.
(b.) A planet of mass, m, orbitting the Sun, whose mass is denoted M , is subject to a
central force,
GM m
r
F (r) = −
r3
Show that F conserves angular momentum.
(c.) The eccentricity of an elliptical orbit is defined by
rmax − rmin
=
rmax + rmin
where rmax is the maximum distance from the sun and rmin is the minimum distance from
the sun. An astronomical unit (A.U.) is the distance from the earth to the sun - its semimajor axis, and equals 92,956,000 miles.
Suppose a planet orbits the Sun in an elliptical orbit such that rmin = 4A.U. and has
eccentricity = 34 . Calculate:
(i.) The length of its semi-major axis.
(ii.) The period of the planet’s orbit (in Earth years), τ .
Problem 4.2
”I see myself as a huge fiery comet, a shooting star. Everyone stops, points up and gasps
”Oh look at that!” Then whoosh, and I’m gone... and they’ll never see anything like it ever
again... and they won’t be able to forget me ever.” Jim Morrison
(a) The asteroid 2060 Chiron is 8.5 astronomical units from the Sun at perihelion (when
it is closest to the Sun). Chiron’s elliptical orbit is extremely eccentric: At aphelion (when
it is furthest from the Sun) it is 18.6 A.U. from the sun. What is the period (in Earth years)
of its orbit around the sun?
(b) In 1682 a bright comet appeared. Edmond Halley computed its orbit and found it
very similar to those of comets seen in 1607 and 1531. He predicted the comet would return
in 1758. By then Halley was dead, but Halley’s comet did return on Christmas night 1758.
What is the semi-major axis of Halley’s comet?
(c) L.Kohoutek at Hamburg Observatory discovered a comet in 1973 which has a period
of about 75,000 years. What is the length of its semi-major axis?
Problem 4.3
A planet of mass m is orbiting the Sun. At perihelion, the magnitude L of angular momentum
2
GN M m
is mv(1 − )a and at aphelion it is mv 0 (1 + )a. At perihelion, the energy E is mv
2 − a(1−)
and at aphelion it is
mv 02
2
−
GN M m
a(1+) .
(a) Using conservation of angular momentum deduce that
v=
L
m(1 − )a
and
v0 =
L
m(1 + )a
(b) Substituting these values into the formulas for the energy E, and using conservation of
energy, deduce
L2 = GN M m2 a(1 − 2 )
(c) Substituting the formula for v into the formula for E, and using the formula for L2 ,
deduce that
GN M m
E=−
2a