Download 8th Ed【CH13】

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8th Ed【CH13】Newtonian Gravitation
9th Ed【CH13】Gravitation
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
課本查表可以得知：地球質量 M e = 5.98 ×1024 kg
太陽質量 M S = 1.99 × 1030 kg
8th Ed：Homework of Chapter 13：
7, 11, 13, 21, 23, 27, 31, 35, 39, 45, 49, 53, 56, 61, 65
太陽與地球距離 d S −e = 1.5 × 1011 m
8th Ed【Sample Problem 13-5】
An asteroid, headed directly tower Earth, has a speed of 12km / s relative to the planet when the
asteroid is 10 Earth radii from Earth’s center. Neglecting the effects of earth’s atmosphere on the
asteroid, find the asteroid’s speed v f when it reaches earth’s surface.
8th Ed【Problem 13-8】：9th Ed【Problem 13-8】★
In Fig 13-35, a square of edge length 20.0cm is formed by four spheres of masses m1 = 5.00 g ,
<解>： RE = 6.37 ×10 m
force from them on a central sphere with mass m5 = 2.5 g ?
6
m2 = 3.00 g , m3 = 1.00 g , and m4 = 5.00 g . In unit-vector notation, what is the net gravitational
M E = 5.98 ×1024 kg
1 2 GMm 1 2 GMm
= mvi −
mv f −
RE
2
2
10 RE
v 2f = vi2 +
⎛ 2(6.67 ×10−11 m3 / kg ⋅ s 2 )(5.98 × 1024 kg ) ⎞ 9
2GM
1
(1 − ) = (1.2 ×103 m / s) 2 + ⎜
⎟
RE
10
6.37 × 106 m
⎝
⎠ 10
= 2.567 × 108 m 2 / s 2
v f = 1.6 × 104 m / s = 16km / s
8th Ed【Problem 13-7】：9th Ed【Problem 13-？】
How far from Earth must a space probe be along a line toward the Sun so that the Sun’s
（圖 13-35）
gravitational pull on the probe balances Earth’s pull?
<解>：The gravitational forces on m5 from the two 5.00g masses m1 and m4 cancel each other.
一艘太空船在地球與太陽的連線上，問距離地球多遠時，這艘太空船受到的太陽引力與地
球引力剛好一樣。
Contributions to the net force on m5 come from the remaining two masses:
Fnet =
( 6.67 ×10
−11
N ⋅ m 2 /kg 2 )( 2.50 ×10−3 kg )( 3.00 ×10−3 kg − 1.00 × 10−3 kg )
(
2 × 10−1 m
)
2
= 1.67 × 10−14 N.
<解>： 受到的引力一樣，所以： F =
⇒
r=
d Me
Ms + Me
=
The force is directed along the diagonal between m2 and m3, towards m2. In unit-vector
notation, we have
G
Fnet = Fnet (cos 45°?i + sin 45° j) = (1.18 × 10−14 N) ?i + (1.18 × 10−14 N) j
GM e m GM S m
=
r2
(d − r ) 2
Me
MS
=
r 2 (d − r )2
(150 ×109 m ) 5.98 ×1024 kg
1.99 × 1030 kg + 5.98 × 1024 kg
8th Ed【Problem 13-9】：9th Ed【Problem 11-5】
Miniature black holes. Left over from the big-bang beginning of the universe. Tiny black holes
might still wander through the universe. If one with a mass 1×1011 kg (and a radius of only
= 2.60 ×108 m.
1×10−16 m ) reached Earth, at what distance from your head would its gravitational pull on you
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match that of Earth’s?
微型黑洞。大爆炸之初的宇宙所遺留下來的。小型黑洞仍可能徘徊在宇宙之中。如果其中
−16
一個具有質量 1×10 kg （且半徑僅有 1× 10 m ）的黑洞來到地球，問在什麼樣的距離下，
你的腦袋被它的引力拉扯與地球引力相當？
11
<解>：地球重力對你（體重為 m） Fg =
GM E m
GM E
= mg 。其中 g =
= 9.8 m / s 2 。
RE2
RE2
如果你和微型黑洞的距離為 r。重力對你 Fg =
比較 2 個式子： mg =
GM b m
= mg.
r2
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中心位置到頂點離為 r =
L
L
=
2 cos 300
3
鉛直方向重力： Fm −m4 = 2
Gmm4
Gmm
sin 300 = 3 2 4
r2
L
FM − m4 =
GMm4
GMm4
=3
r2
L2
鉛直方向淨重力為 0，所以 Fm− m4 = FM − m4
可以得知： M = m
(b) Since m4 cancels in that last step, then the amount of mass in the center sphere is not
relevant to the problem. The net force is still zero.
GM E m GM b m
=
RE2
r2
GM b
(6.67 × 10−11 m3 /kg ⋅ s 2 )(1× 1011 kg)
⇒ r=
=
≈ 0.8 m.
g
9.8 m/s 2
8th Ed【Problem 13-13】：9th Ed【Problem 13-13】
Figure 13-39 shows a spherical hollow inside a lead sphere of radius R = 4cm ; the surface of the
8th Ed【Problem 13-11】：9th Ed【Problem 13-11】
hollow passes through the center of the sphere and “touches” the right side of the sphere. The mass
of the sphere before hollowing was M = 2.95kg . With what gravitational force does the
hollowed-out lead sphere attract a small sphere of mass m = 0.431kg that lies at a distance
As seen in Fig. 13-37, two spheres of mass m and a third sphere of mass M form an equilateral
triangle, and a fourth sphere of mass m4 is at the center of the triangle. The net gravitational force
spheres and of the hollow?
on that central sphere from the three other spheres is zero. (a) What is M in terms of m? (b) If we
double the value of m4 , what then is the magnitude of the net gravitational force on the central
sphere?
d = 9cm from the center of the lead sphere, on the straight line connecting the centers of the
圖 13-39 顯示了鉛球半徑 R = 4cm ，內部挖空了一個球型空腔，球型空腔表面與鉛球表面相
切，且通過球心。未挖空前鉛球質量 M = 2.95kg 。鉛球外有一顆小球質量 m = 0.431kg ，離
鉛球球心距離 d = 9cm ，問挖空後的鉛球作用在小球上的重力為何？
如圖 13-37，兩顆球（質量 m）
，第三顆球（質量 M）
，位於正三角形的頂點位置。第四顆球
質量 m4 位在三角形中心，所受的淨重力為 0。（a）問 M 和 m 的關係？（b）把 m4 加倍，問
中心球的淨重力大小為何？
（圖 13-39）
GMm
<解>：實心鉛球對小球重力： F1 =
d2
4 R 3
π( )
R
M
空腔半徑 r = ，所以被挖掉的空腔質量： M c = 3 2 M =
（指的是跟空腔一樣
4
2
8
3
πR
3
（圖 13-37）
體積的鉛球質量。想像成有 2 個鉛球對小球引力。）
<解>：(a)正三角形邊長為 L
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空腔對小球重力： F2 =
Fundamentals of Physics, 8th Ed
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M
m
8
R
(d − ) 2
2
G
小球 m 受到的重力為：
⎛ 1
⎞ GMm ⎛
⎞
1
1
F = F1 − F2 = GMm ⎜ 2 −
⎟=
⎜1 −
⎟
2
2
2
⎜d
⎟
⎜
⎟
d
8 ( d − R/2 ) ⎠
⎝
⎝ 8 (1 − R/2d ) ⎠
=
<解>：(a) The gravitational acceleration is ag =
⎞
(6.67 × 10−11 m3 /s 2 ⋅ kg)(2.95 kg)(0.431 kg) ⎛
1
⎜1 −
−2
−2
2 ⎟
−
×
⋅
×
(9.00 × 10−2 m) 2
8[1
(4
10
m)
/(2
9
10
m)]
⎝
⎠
GM
= 7.6 m/s 2 .
R2
(b) Note that the total mass is 5M. Thus, ag =
= 8.31×10−9 N
8th Ed【Problem 13-17】：9th Ed【Problem 13-17】
( 3R )
2
= 4.2 m/s 2 .
8th Ed【Problem 13-23】：9th Ed【Problem 13-21】
2
At what altitude above Earth’s surface would the gravitational acceleration be 4.9m / s ?
<解>：The acceleration due to gravity is given by ag =
G ( 5M )
GM
, where M is the mass of Earth and r
r2
Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a
star has a radius of 20 km, what must be its minimum mass so that material on its surface remains in
place during the rapid rotation?
某一中子星自轉速率 1 rev/s。假設有一中子星半徑 20km，該中子星快速轉動時，材料仍在表
面，問中子星的最小質量為何？（提醒：重力抵銷向心力，粒子是在無重力狀態。）
is the distance from Earth’s center. We substitute r = R + h, where R is the radius of Earth
and h is the altitude, to obtain ag =
GM
GM
−R.
. We solve for h and obtain h =
ag
( R + h) 2
According to Appendix C, R = 6.37 × 106 m and M = 5.98 × 1024 kg, so
h=
( 6.67 ×10
−11
m3 / s 2 ⋅ kg )( 5.98 ×1024 kg )
( 4.9m / s )
2
<解>： 1rev / s = 2π rad / s
g =0=
M=
GM
− Rω 2
R2
R 3ω 2 (2000)3 (2π )2
=
= 4.7 ×1024 kg
G
6.673 ×10−11
− 6.37 ×106 m = 2.6 ×106 m.
8th Ed【Problem 13-27】：9th Ed【Problem 13-25】
A solid uniform sphere has a mass of 1×104 kg and a radius of 1m. What is the magnitude of the
th
th
8 Ed【Problem 13-21】：9 Ed【Problem 13-23】
One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of
inner radius R, outer radius 2R, and mass 4M. If M = 4.1× 1024 kg and R = 6 × 106 m , what is the
gravitational force due to the sphere on a particle of mass m located at a distance of (a) 1.5 m and (b)
0.5m from the center of the sphere? (c) Write a general expression for the magnitude of the
gravitational force on the particle at a distance r ≤ 1m from the center of the sphere.
gravitational acceleration of a particle at points (a) R and (b) 3R from the center of the planet?
某一行星模型，其核心半徑為 R，質量為 M，包圍的外殼（包含半徑為 R 內部）
，半徑 2R，
質量 4M 。如果 M = 4.1× 1024 kg 和 R = 6 × 106 m ，問一個粒子在距離行星中心（a）R 和（b）
3R 時的重力加速度？
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一均勻固體球質量 1×104 kg ，半徑 1m。球對粒子 m 的重力大小為何？（a）粒子固定在 1.5m
的地方（b）粒子固定在距離球心 0.5m 的地方（c）粒子距離球心小於 1m，寫出重力的一
般表達式。
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4
<解>：球體積： V = π R 3
3
球密度： ρ =
M total 1.0 × 104 kg
=
= 2.4 × 103 kg/m3 .
π R3 43 π (1.0 m )3
When the particle of mass m (upon which the sphere, or parts of it, are exerting a
gravitational force) is at radius r (measured from the center of the sphere), then whatever
mass M is at a radius less than r must contribute to the magnitude of that force (GMm/r2).
(a) At r = 1.5 m, all of Mtotal is at a smaller radius and thus all contributes to the force:
(b) At r = 0.50 m, the portion of the sphere at radius smaller than that is
⎛4
⎞
M = ρ ⎜ π r 3 ⎟ = 1.3 × 103 kg.
⎝3
⎠
Thus, the force on m has magnitude GMm/r2 = m (3.3 × 10−7 N/kg).
(c) Pursuing the calculation of part (b) algebraically, we find
Fon m =
Gm ρ ( 43 π r
r
2
) = mr ⎛ 6.7 ×10
⎜
⎝
−7
MJ
KJ
R
M R
(b)
= J = J e = 28.5
K e M e M e RJ
Re
8th Ed【Problem 13-35】：9th Ed【Problem 13-37】
The three spheres in Fig. 13-45, with masses mA = 80 g , mB = 10 g , and mC = 20 g , have their
centers on a common line, with L = 12cm and d = 4cm . You move sphere B along the line until
its center-to-center separation from C is d = 4cm . How much work is done on sphere B (a) by you
and (b) by the net gravitational force on B due to spheres A and C?
GmM total
= m ( 3.0 × 10−7 N/kg ) .
r2
3
東海大學物理系
Mm
Km
R
M R
(a)
= m = m e = 0.0451
M e M e Rm
Ke
Re
4
3
Fon m =
Fundamentals of Physics, 8th Ed
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如圖13-45。三顆球質量分別為 mA = 80 g ， mB = 10 g 和 mC = 20 g 。球中心在同一直線上，其
中 L = 12cm ， d = 4cm 。沿著直線移動B球，直到B球與C球距離 d = 4cm ，（a）問你對B球
作功多少？（b）A球和C球對B球淨重力作功多少？
N ⎞
⎟.
kg ⋅ m ⎠
8th Ed【Problem 13-31】：9th Ed【Problem 13-33】
What multiple of the energy needed to escape from Earth gives the energy needed to escape from (a)
the Moon and (b) Jupiter?
一物體要由（a）月球（b）木星上脫離，所需能量為由地球脫離能量的幾倍？
potential energy of the three-sphere system.
<解>： M e = 5.98 ×1024 kg 。 Re = 6.37 ×106 m
The initial potential energy is U i = −
M m = 7.35 ×10 kg 。 Rm = 1.74 ×10 m
22
6
GmA mB GmA mC GmB mC
−
−
d
L
L−d
and the final potential energy is U f = −
M J = 1.89 ×10 kg 。 RJ = 7.18 ×10 m
27
（圖 13-45）
<解>：(a) The work done by you in moving the sphere of mass mB equals the change in the
7
GmA mB GmA mC GmB mC
.
−
−
L−d
L
d
The work done is
GMm
K=
R
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⎛1
⎛
1⎞
1
1 ⎞
− 10 ⎟
v = GM ⎜ − ⎟ = (6.67 × 10−11 m3 / s 2 ⋅ kg) (1030 kg) ⎜
5
⎜r
⎟
ri ⎠
⎝ 2 × 10 m 10 m ⎠
⎝ f
⎡ ⎛1
1 ⎞
1 ⎞⎤
⎛ 1
W = U f − U i = GmB ⎢ mA ⎜ −
− ⎟⎥
⎟ + mC ⎜
d
L
d
L
d
d
−
−
⎠
⎝
⎠⎦
⎣ ⎝
⎡
L − 2d
L − 2d
2d − L ⎤
= GmB ⎢ mA
+ mC
= GmB (mA − mC )
d ( L − d ) ⎥⎦
d (L − d )
⎣ d (L − d )
0.12 m − 2(0.040 m)
= (6.67 × 10−11 m3 / s 2 ⋅ kg) (0.010 kg)(0.080 kg − 0.020 kg)
(0.040 m)(0.12 − 0.040 m)
= + 5.0 × 10−13 J.
(b) The work done by the force of gravity is −(Uf − Ui) = −5.0 × 10−13 J.
東海大學物理系
= 1.8 × 107 m/s.
8th Ed【Problem 13-45】：9th Ed【Problem 13-47】
The Sun, which is 2.2 ×1020 m from the center of the Milky Way galaxy, revolves around that
center once every 2.5 ×108 years. Assuming each star in the Galaxy has a mass equal to the Sun’s
mass of 2 × 1030 kg , the stars are distributed uniformly in a sphere about the galactic center, and the
Sun is at the edge of that sphere, estimate the number of stars in the Galaxy.
th
th
8 Ed【Problem 13-39】：9 Ed【Problem 13-41】
Two neutron stars are separated by a distance of 1×1010 m . They each have a mass of 1× 1030 kg
and a radius of 1×10 m . They are initially at rest with respect to each other. As measured from that
rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial
5
太陽離銀河中心 2.2 × 1020 m ，繞銀河運行，運轉週期為 2.5 × 108 年。假設銀河每個星球質量
都和太陽一樣是 2 × 1030 kg 。全部星球均勻分佈在以銀河中心為球心的球型裡，太陽位在球
型的邊緣。請估計銀河系中的星球數量。
value and (b) they are about to collide?
兩個中子星相距 1×1010 m 。質量均為 1×1030 kg ，半徑為 1× 105 m 。一開始靜止，兩中子星因
重力互相靠近時（a）當兩者距離為最初的一半時，兩個星球速率為何？（b）碰撞瞬間，兩
個星球速率為何？
<解>：(a) The momentum of the two-star system is conserved, and since the stars have the same
mass, their speeds and kinetic energies are the same. We use the principle of
conservation of energy. The initial potential energy is Ui = −GM2/ri, where M is the mass
of either star and ri is their initial center-to-center separation. The initial kinetic energy is
zero since the stars are at rest. The final potential energy is Uf = −2GM /ri since the final
separation is ri/2. We write Mv2 for the final kinetic energy of the system. This is the
2
2
sum of two terms, each of which is ½Mv . Conservation of energy yields
−
GM 2
2GM 2
=−
+ Mv 2 .
ri
ri
GM
=
ri
(6.67 × 10−11 m3 / s 2 ⋅ kg) (1030 kg)
= 8.2 × 104 m/s.
1010 m
G ( NM ) M
r2
太陽向心加速度： a =
v=
2π R
T
a=
4π 2 R
T2
F=
GNM 2
4π 2 R
=M
2
R
T2
v2
R
4π 2 R 3
.
GT 2 M
8
T = 2.5 × 10 y = 7.88 ×1015 s
N =
N =
The solution for v is
v=
<解>： F =
4π 2 (2.2 × 1020 m)3
= 5.1 × 1010.
(6.67 × 10−11 m3 / s 2 ⋅ kg) (7.88 × 1015 s) 2 (2.0 × 1030 kg)
8th Ed【Problem 13-49】：9th Ed【Problem 13-43】
(b) Now the final separation of the centers is rf = 2R = 2 × 105 m, where R is the radius of
(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 160 km
above Earth’s surface? (b) What is the period of revolution?
either of the stars. The final potential energy is given by Uf = −GM2/rf and the energy
equation becomes −GM2/ri = −GM2/rf + Mv2. The solution for v is
（a）距離地球表面 160 公里，以圓軌跡運行的衛星，問線速度為何？（b）運轉週期？
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<解>：(a)
東海大學物理系
GMm mv 2
=
r2
r
v=
GM
r
⎛M ⎞ ⎛ a ⎞
(T/TE ) 2 = ⎜ o ⎟ ⎜ ⎟
⎝ M ⎠ ⎝ rE ⎠
r = 6.37 × 106 + 160 × 103 = 6.53 × 106 m
(b) T =
3
where TE = 365.25 days is Earth’s orbital period and rE = 1.50 × 1011 m is its mean
distance from the Sun. In this case, it is perfectly legitimate to take logarithms and
obtain
Re = 6.37 × 106 m
GM
=
r
東海大學物理系
If we divide Kepler’s law (applied to the Jupiter-moon system, where M is mass of
Jupiter) by the law applied to Earth orbiting the Sun (of mass Mo), we obtain
M e = 5.98 × 1024 kg
v=
Fundamentals of Physics, 8th Ed
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Halliday & Resnic
(6.67 × 10−11 m3 / s 2 ⋅ kg) (5.98 × 1024 kg)
= 7.82 × 103 m/s.
6.53 × 106 m
⎛r ⎞ 2
⎛T ⎞ 1
⎛M ⎞
log ⎜ E ⎟ = log ⎜ E ⎟ + log ⎜ o ⎟
a
3
T
3
⎝ ⎠
⎝ ⎠
⎝ M ⎠
(written to make each term positive) which is the way we plot the data (log (rE/a) on the
vertical axis and log (TE/T) on the horizontal axis).
2π r 2π (6.53 × 106 m)
=
= 5.25 × 103 s = 87.5 min
7.82 × 103 m/s
v
8th Ed【Problem 13-53】：9th Ed【Problem 13-55】
In 1610, Galileo used his telescope to discover four prominent moons around Jupiter. Their mean
orbital radii a and periods T are as follows: (a) Plot log a (y axis) against log T (x axis) and show
that you get a straight line. (b) Measure the slope of the line and compare it with the value that you
expect from Kepler’s third law. (c) Find the mass of Jupiter from the intercept of this line with the y
axis.
Name
Io
Europa
Ganymede
Callisto
a ( 108 m)
4.22
6.71
10.7
18.8
(b) When we perform a least-squares fit to the data, we obtain
log (rE/a) = 0.666 log (TE/T) + 1.01,
which confirms the expectation of slope = 2/3 based on the above equation.
(c) And the 1.01 intercept corresponds to the term 1/3 log (Mo/M) which implies
Mo
Mo
= 103.03 ⇒ M =
.
M
1.07 × 103
T (days)
1.77
3.55
7.16
16.7
1610 年伽利略利用望遠鏡發現木星的四個衛星，其軌道半徑 a 與週期 T 如表。（a）以 log a
為 y 軸， logT 為 x 軸，證明你會得到一條直線。（b）由圖得到斜率，並與刻卜勒第三定律
得到的理論值比較（c）由直線與 y 軸的交叉點，求出木星質量。
<解>：(a) If we take the logarithm of Kepler’s law of periods, we obtain
2 log (T ) = log (4π 2 /GM ) + 3 log (a ) ⇒ log (a ) =
2
1
log (T ) −
log (4π 2 /GM )
3
3
where we are ignoring an important subtlety about units (the arguments of logarithms
cannot have units, since they are transcendental functions). Although the problem can be
continued in this way, we prefer to set it up without units, which requires taking a ratio.
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Plugging in Mo = 1.99 × 1030 kg (see Appendix C), we obtain M = 1.86 × 1027 kg for
Jupiter’s mass. This is reasonably consistent with the value 1.90 × 1027 kg found in
Appendix C.
8th Ed【Problem 13-56】：9th Ed【Problem 13-54】
Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary
(two-star) system. This visible star has orbital speed v = 270km / s , orbital period T = 1.7days ,
and approximate mass m1 = 6M S , where M S is the Sun’s mass, 1.99 ×1030 kg . Assume that the
visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. 13-49).
What multiple of M S gives the approximate mass m2 of the dark star?
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Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
狩獵黑洞。觀測雙 星系統的 光線。看得到的那顆星軌道 速度 v = 270km / s ，軌道週期
T = 1.7days ，和近似質量 m1 = 6M S ，在此 M S 是太陽的質量 1.99 ×1030 kg 。假設可見星和它
的伴星（黑暗，看不見的），都在圓形軌道（圖 13-49）。問伴星質量 m2 為太陽的幾倍？
Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
東海大學物理系
The equation has one real solution: α = 9.3 , which implies m2 / M s ≈ 9 .
8th Ed【Problem 13-61】：9th Ed【Problem 13-63】
An asteroid, whose mass is 2 ×10−4 times the mass of Earth, revolves in a circular orbit around the
Sun at a distance that is twice Earth’s distance from the Sun. (a) Calculate the period of revolution
of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic
energy of Earth?
一個小行星，其質量是地球的 2 × 10−4 倍，繞太陽圓形軌道運轉，離太陽的距離是地球離太
陽距離的的兩倍。
（a）以年為單位計算運轉週期。
（b）小行星的動能與地球動能比例？
<解>：(a) 由 Kepler’s law of periods： T 2 =
（圖 13-49）
<解>：The two stars are in circular orbits, not about each other, but about the two-star system’s
center of mass (denoted as O), which lies along the line connecting the centers of the two
stars. The gravitational force between the stars provides the centripetal force necessary to
keep their orbits circular. Thus, for the visible, Newton’s second law gives
F=
Gm1m2 m1v 2
=
r2
r1
r1 =
M = 1.99 ×1030 kg
r = 2re = 2 ×1.5 ×1011 m = 3 × 1011 m
T =
where r is the distance between the centers of the stars. To find the relation between r and
r1 , we locate the center of mass relative to m1 . Using Equation 9-1, we obtain
m1 (0) + m2 r
m2 r
m + m2
=
⇒ r= 1
r1 .
m1 + m2
m1 + m2
m2
4π 2 3
r
GM
4π 2 r 3
4π 2 (3 × 1011 m)3
=
= 8.96 × 107 s = 2.8 y
−11
GM
(6.67 × 10 m 3 / s 2 ⋅ kg) (1.99 × 1030 kg)
(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given by K
= GMm/2r, where m is the mass of the asteroid or planet. We note that it is proportional
to m and inversely proportional to r. The ratio of the kinetic energy of the asteroid to
On the other hand, since the orbital speed of m1 is v = 2π r1 / T , then r1 = vT / 2π and the
the kinetic energy of Earth is K/Ke = (m/me) (re/r). We substitute m = 2.0 × 10−4me and r
= 2re to obtain K/Ke = 1.0 × 10−4.
m + m2 vT
.
expression for r can be rewritten as r = 1
m2 2π
8th Ed【Problem 13-65】：9th Ed【Problem 13-65】
Substituting r and r1 into the force equation, we obtain
A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r 2
4π 2Gm1m23
2π m1v
F=
=
(m1 + m2 ) 2 v 2T 2
T
m23
v3T
(2.7 ×105 m/s)3 (1.70 days)(86400 s/day)
=
=
= 6.90 × 1030 kg
2
or (m1 + m2 )
2π G
2π (6.67 × 10−11 m3 /kg ⋅ s 2 )
= 3.467 M s ,
because A = π r 2 . Determine how the following properties of the satellite depend on r: (a) period,
(b) kinetic energy, (c) angular momentum, and (d) speed.
一衛星是在一個圓形地球軌道，半徑 r。軌道包含的面積跟 r 2 有關，因為 A = π r 2 。確定衛
星以下的特性跟 r 有關：（a）週期，（b）動能，（c）角動量，和（d）速度。
where M s = 1.99 × 1030 kg is the mass of the sun. With m1 = 6M s , we write m2 = α M s
and solve the following cubic equation for α :
<解>：(a) 由 Kepler’s law of periods： T 2 =
α3
− 3.467 = 0 .
(6 + α ) 2
4π 2 3
r
GM
T ∝ r 3/ 2
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Fundamentals of Physics, 8th Ed
Principle of Physics, 9th Ed
Halliday & Resnic
(b) K =
東海大學物理系
GMm 1
∝
r
2r
(c) L = r 2 mω
1
1
(d) K = mv 2 ∝
r
2
v∝
1
r
(c) and (d) From the previous part, knowing that K is proportional to v2, we find that v is
proportional to 1/ r . Thus, by Eq. 13-31, the angular momentum (which depends on
the product rv) is proportional to r/ r = r .
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