NOTE: This page uses Taylor's "c" for our "r0" parameter,
i.e. the "c" on this page is our "r0", NOT the focal distance!
Phys 326 – Problem Set No.8 Solutions
Di Zhou
1
Problem 1 : Satellite I
(a) We are given the satellite’s height hmin = 300km at perigee and hmax = 3000km at apogee. Therefore
the distance from the center of the earth is rmin = Re + hmin and rmax = Re + hmax . Since
rmin =
c
1+
rmax =
c
1−
(1)
Therefore we have,
=
rmax − rmin
= 0.17
rmax + rmin
(2)
Therefore we find
c = (1 + )rmin = 7824km
(3)
(b) when the satellite passes by the y axis,
h ≈ b − Re ≈ 1400km
2
(4)
Problem 2 : Satellite II
(a) The satellite’s height hmin = 250km and speed vmax = 8500m/s at perigee. The distance from the earth’s
center is rmin = Re + hmin = 6650km. Thus the angular momentum is given by
l = mvmax rmin
(5)
Therefore,
c=
(vmax rmin )2
l2
=
= 7960km
γµ
GMe
(6)
(b) Finally we have,
rmin =
c
→ = 0.197
1+
(7)
c
= 9910km
1−
(8)
Similarly,
rmax =
So hmax = rmax − Re =3510 km.
1
Problem 3 : Sputnik ➔ EXPANDED SOLUTION with “When-to-use-U-vs-U*” Guidance
We are given these two pieces of info about the orbit: the perigee height hmin = 181 km and apogee height
hmax = 327 km, both measured above the earth’s surface. The perigee and apogee radii we will need are
r = R⊕ + h with R⊕ = 6.4 × 10 6 m .
(a) The period of the orbit can be found from Kepler’s second law, τ 2 = 4π 2 m a 3 / γ where we have replaced
the reduced mass μ with the satellite’s mass m since the the satellite’s mass is infinitesimal compared with the
mass of the Earth. We need γ and a :
• For Earth’s gravity, the force-strength parameter γ is GM ⊕ m = gR⊕2
• For an elliptical orbit, a simple sketch shows that the semi-major axis a is (rmin + rmax ) / 2
4π 2 m 3 4π 2 m ⎛ rmin + rmax ⎞
π 2 (2R⊕ + hmin + hmax )3
Thus, τ =
a =
=
.
⎜
⎟⎠
γ
gR⊕2 m ⎝
2
2gR⊕2
3
2
When you put in the numbers, be very careful not to mix km and m → we have g = 9.8 m/s2 in meters, not
kilometers, so be sure to put in all those distances in meters as well! The result is:
π [ (12.8 + 0.181 + 0.327) × 10 6 ]3/2
τ=
= 5383 sec = 90 min
2 (9.8) [6.4 × 10 6 ]
(b) The maximum speed in the orbit occurs at the minimum distance : at perigee. To get speed into the game,
let’s combine two of our energy equations: E = −γ / a (in lecture we called this the “E Equation”) and the
definition of energy, E = T + U = 12 mv 2 + U .
VITAL ISSUE to address before we continue: should we use U or U * = U + U cf here?
➔ Remember the origin of Ucf : it only appeared when we simplified our work into a one-dimensional, purely
radial problem. If you don’t remember this, figure it out by the combining the definition of Ucf with our
“L Equation” φ = L / µr 2 :
U cf =
L2
µ 2 r 4φ 2 1 2  2 1 2
=
= µr φ = µ vφ ➔ Ucf is just the angular part of the Kinetic Energy !
2 µr 2
2 µr 2
2
2
To summarize this crucial insight, the total energy can be decomposed in two ways:
• E = 12 µv 2 + U , where v 2 = vr2 + vφ2 = r2 + (rφ )2 is the total speed2 of the particle
• E = 12 µr2 + U cf + U , where we expose only the radial part of v2 and absorb the vφ piece into Ucf
All Clear? Then on we go to find Yuri Gagarin’s maximum speed. This vmax is not just the radial speed, it is the
total speed. In fact we already said that vmax occurs at perigee, and there, the ship’s velocity is entirely angular
→ r = vr is zero at any apse. Now combine the “E Equation” with the definition of energy, E = T+U to get v2 :
E=
1 2
γ
mv + U = −
2
2a
→
1 2 GMm
GMm
mv −
=−
2
r
2a
→
⎛ 2 1⎞
v 2 (r) = GM ⎜ − ⎟
⎝ r a⎠
I learned from our TAs via a couple of students with great memories that this last equation has a historical
name : the vis viva equation, where the Latin words mean “live force”. (It even has a wiki page!) Let’s apply
it at perihelion, rmin, to finish this problem:
⎛ 2
⎞
2
rmax
v 2max = v 2 (rmin ) = gR⊕2 ⎜
−
= 2gR⊕2
→ vmax = 7.8 km/s
⎟
rmin (rmax + rmin )
⎝ rmin rmax + rmin ⎠
3
Sputnik
NOTE: R1 and R2 here mean r_min and r_max
(a) The period of a elliptical orbit is the same as that of a circular orbit which has the same radius. Hence
the period of elliptical orbit is
r
r
R3
(R1 + R2 )3 /8
= 2π
= 88min
(9)
τ = 2π
GM
GM
(b) The maximum velocity is given by perigee, hence we have,
GM m
1
GM m
mv 2 −
=−
⇒ v = 7.8km/s
2
R1
R1 + R2
4
(10)
The Earth Becomes a Star
(a) If the radius of an orbit is r, then we have,
Gm1 m2
m1 v 2
2πr
=
= 2π
→T =
2
r
r
v
s
r3
Gm2
(11)
(b) Now in the CM frame, the equation of motion is
µr̈ = F
(12)
For the relative motion this implies that
µω 2 r =
Gm1 m2
r2
(13)
therefore we find,
2πr3/2
2πr3/2
T =p
(14)
= √
GM
Gm1 m2 /µ
√
(c) If m1 = m2 then M = 2m and the period is 2 times of the original value. Therefore the period is 0.71
years.
5
The Virial Theorem
(a) The newton’s second law is
mv 2
= nkrn−1
r
(15)
Hence we find
T =
1
n
mv 2 = U
2
2
(16)
(b) If G = ~r · p~, then
dG
= ṙ · p + ṗ · r = mv 2 + F · r
dt
If we integrate this from 0 to t, we have,
Z
G(t) − G(0)
G(t) − G(0) = (2T + F · r)dt ⇒
= 2 hT i + hF · ri
t
2
(17)
(18)
(c) if the particles orbit is periodic, which means the satellite’s distance and velocity has an upper limit, this
implies that
r·p
(19)
also has an upper limit. Therefore no matter what time t is, the quantity G(t) has an upper limit, say, Gmax .
Therefore we always have
2 hT i + hF · ri ≤
2Gmax
→0
t
t→∞
(20)
(d) For a potential energy U = krn if the satellite’s orbit is periodic, the above argument is valid. In this
point of view, we can always make the time t arbitrary large. In that case we have
n
1
hT i = − hf i = hU i
2
2
and hf i denotes the average force.
3
(21)