Download University Physics AI No. 4 The Gravitational Force and the

University Physics AI
No. 4 The Gravitational Force and the Gravitational Field
Class
Number
Name
I．Choose the Correct Answer
1. The magnitude of the force of gravity between two identical objects is given by F0. If the mass of
each object is doubled but the distance between them is halved, then the new force of gravity
between the objects will be
（ A ）
(B) 4 F0
(C) F0
(D) F0/2
(A) 16 F0
Solution:
The magnitude of the gravitational force is Fgrav =
′ =
Fgrav
GMm
, according to the problem, we get
r2
4Gm 2
Gm 2
=
16
= 16 F0
(r / 2) 2
r2
2. A spherical symmetric nonrotating body has a density that varies appreciably with the radial
distance from the center. At the center of the body the acceleration of free fall is
( C )
(A) Definitely larger than zero. (B) Possibly larger than zero. (C) Definitely equal to zero.
Solution:
At the center, the force acted on the body is zero, so the acceleration of the body is zero.
3. The acceleration due to gravity in a hole dug into a nonuniform spherically symmetric body
( C )
(A) will increase as you go deeper, reaching a maximum at the center.
(B) will increase as you go deeper, but eventually reach a maximum, and then decrease until you
reach the center.
(C) can increase or decrease as you go deeper.
(D) must decrease as you go deeper.
Solution:
For a nonuniform spherically symmetric body, the force of gravity depends on the distance
from the center which is related to how the density of the body changed with respect to the distance
from the center, for instance: according to the Gauss’s law of gravity
r
r
∫ g ⋅ ds = 4πG ∑ M
s
i
= 4πG ∫ ρdV
r
when ρ = A r , the acceleration g will increase as you go deeper, when
2
r
acceleration g will decrease as you go deeper.
ρ = Ar the
II. Filling the Blanks
1. Two masses m1 and m2 exert gravitational forces of equal magnitude on each other. Show that if
the total mass M = m1+m2 is fixed, the magnitude of the mutual gravitational force on each of the
two masses is a maximum when m1 = m2 ( Fill < or = or > ).
Solution:
The magnitude of the gravitational force is Fgrav =
Gm1m2 Gm1 ( M − m1 )
,
=
r2
r2
G ( M − 2m1 )
M
dF
= 0 ⇒ m1 =
=0 ⇒
2
r
dm1
2
We get
m1 = m2 .
Then
2. A mass m is inside a uniform spherical shell of mass
M′ and a mass M is outside the shell as shown in Figure
1. The magnitude of the total gravitational force on m is
Mm
F =G
.
( s + R) 2 − d 2
m
d
M
R
s
M′
Solution:
r
r
r
Ftotal = FM 'on m + FM on m
r
Q FM ' on m = 0
r
r
∴ Ftotal = FM on m
GMm
GMm
∴ Ftotal =
=
2
r
( s + R) 2 − d 2
Fig.1
3. A certain neutron star has a radius of 10.0 km and a mass of 4.00 × 1030 kg, about twice the mass
of the Sun. The magnitude of the acceleration of an 80.0 kg student foolish enough to be 100 km
from the center of the neutron star is 2.67×1010 m/s2 . The ratio of the magnitude of this
acceleration and g is 2. 72×109 . If the student is in a circular orbit of radius 100 km about the
.
neutron star, the orbital period is 3.84×10-5 s
Solution:
(a) The magnitude of the force is F =
So a =
GMm
= ma
r2
GM 6.67 × 10−11 × 4 × 1030
=
= 2.67 × 1010 m/s 2
2
5 2
r
(1 × 10 )
(b) The ratio of the magnitude of this acceleration and g is
a 2.67 × 1010
=
= 2.72 × 109
g
9.81
4π 2 r
⎛ 2π ⎞
=
r
⎟
T2
⎝ T ⎠
2
(c) Since the acceleration is a = ω r = ⎜
2
The orbital period is
T=
4π 2 r
=
a
4π 2 r 3
4 × 3.142 × (1 × 105 ) 2
=
= 3.84 × 10− 5 ( s )
GM
6.67 × 10−11 × 4 × 1030
4. The magnitude of the total gravitational field at the point
P in Figure 2 is
2.37×10-3 m/s2 ,the magnitude of the
acceleration experienced by a 4.00 kg salt lick at point P is
2.37×10-3 m/s2 , the magnitude of the total gravitational
force on the salt lick if it is placed at P is 9.48 N .
Solution:
1.60×108m
P
90°
Moon
7.36×1022kg
ĵ
8
4.16×10 m
iˆ
(a) The total gravitational field at the point P is
r
r
r
GM
GM
gtotal = g M + g E = 2 M iˆ + 2 E (cosθ iˆ + sin θ ˆj )
r PM
r PE
Earth
5.98×1024kg
Fig.2
6.67 × 10−11 × 7.36 × 1022 ˆ 6.67 × 10−11 × 5.98 × 1024 1.6 × 108 ˆ 3.84 × 108 ˆ
i+
i+
j)
(
(1.6 × 108 ) 2
(4.16 × 108 ) 2
4.16 × 108
4.16 × 108
= 1.92 × 10− 4 iˆ + 2.3 × 10− 3 (0.38iˆ + 0.92 ˆj )
=
= (1.07 × 10− 3 )iˆ + (2.12 × 10− 3 ) ˆj
The magnitude of the total gravitational field at the point P is
gtotal = 2.37 × 10−3 m/s2
(b) The acceleration has no relation with anything at some point. So the magnitude of the
acceleration experienced by a 4.00 kg salt lick at point P is also equal to (a).
a = gtotal = 2.37 × 10−3 m/s 2
(c)The magnitude of the total gravitational force on the salt lick if it is placed at P is
F = ma = 4 × 2.37 = 9.48 N
III. Give the Solutions of the Following Problems
1. Several planets (the gas giants Jupiter, Saturn, Uranus,
y
and Neptune) possess nearly circular surrounding rings,
perhaps composed of material that failed to form a satellite. dM
m
x
In addition, many galaxies contain ring-like structures.
θ
R
Consider a homogeneous ring of mass M and radius R. (a)
x
Find an expression for the gravitational force exerted by
the ring on a particle of mass m located a distance x from
M
the center of the ring along its axis. See Fig.3. (b) Suppose
that the particle falls from rest as a result of the attraction
Fig.3
of the ring of matter. Find an expression for the speed with
which it passes through the center of the ring.
Solution:
(a) Set up the coordinate system shown in figure. Choose a pointlike differential masses dM shown
as in figure 3, the field produced by dM is
dg =
GdM
R2 + x2
GdM
⎧
⎪⎪dg x = −dg cosθ = − R 2 + x 2 cosθ
⇒ ⎨
⎪dg = dg sin θ = GdM sin θ
⎪⎩ y
R2 + x2
According to the symmetry of the ring-like structures,
∫ dg
y
= 0.
v
So the gravitational force is g = dg x iˆ .
∫
According to the graph, we get cos θ =
x
GxdM
, so dg x = − 2
2
( R + x 2 )3 / 2
R +x
2
Thus the gravitational force is
v
g = ∫ dg xiˆ = − ∫
Gx
GxM
dM iˆ = − 2
iˆ
2 3/ 2
(R + x )
( R + x 2 )3 / 2
2
(b) Apply Newton’s second law of motion, we have
v
v
F = mg = −
g=
Due to
GMmx ˆ
i
(R + x 2 )3 / 2
2
dv dv dx
dv
GMx
=
=
=v ,
2 3/ 2
dt dx dt
dx
(R + x )
2
We can get the speed
v
∫ vdv = ∫
0
0
x
GMx
1
1
dx ⇒ v = 2GM ( −
)
2 3/ 2
R
(R + x )
R2 + x2
2
2. A mass M is in the shape of a thin uniform
disk of radius R. Let the z-axis represent the
symmetry axis of the disk as indicated in Figure
4. The Milky Way Galaxy is modeled as such a
mass disk to a first approximation.
(a) Find the gravitational field of the disk at a
coordinate z along the symmetry axis of the
disk?
z
k̂
dθ
M
r
R
0
dr
Fig.4
r
(b) What is the expression for g for z >> R in
part (a)?
(c) Let σ be the surface mass density of the disk (the number of kilograms per square meter of the
disk), so that
σ=
M
. What is the gravitational field in part (a) as z → 0 along the positive
πR 2
z-axis？
Solution:
(a) The mass M is a uniform disk, so the surface mass density of the disk is σ =
M
.
πR 2
Choose a circular differential mass dM shown in figure, dM = 2πrdr ⋅ σ
The gravitational field established by the ring at a coordinate z is
v
dg = −
GzdM
Gzσ ⋅ 2πrdr ˆ
kˆ = − 2
k
2 3/ 2
(z + r )
(z + r 2 )3/ 2
2
Thus the gravitational field of the disk at a coordinate z along the symmetry axis of the disk is
M
z
⎧
]kˆ z > 0
− 2G 2 [1 − 2
2 1/ 2
⎪
R
R
(
z
R
)
+
Gzσ ⋅ 2πrdr ˆ ⎪
v
v
g = ∫ dg = − ∫
k =⎨
0 (z 2 + r 2 )3/ 2
z
⎪− 2G M [−1 −
]kˆ z < 0
2
2
⎪⎩
R
( z + R 2 )1 / 2
r
(b) For z >> R , the disk can be treated as a point mass, so g = −
(c) In part (a)
⎡
r
2GM ⎢
z
g = − 2 1−
R ⎢
R2 + z2
⎣
⎤
⎥ kˆ ,
1
⎥
2
⎦
(
)
r
2GM ˆ
k
R2
When z → 0 , the gravitational field is g = −
GM ˆ
k
z2
The surface mass density of the disk
σ=
M
, so the gravitational field is
πR 2
r
g = −2πGσ kˆ = constant
3. The Earth is not, in fact, a sphere of uniform density. A high-density core is surrounded by a shell
or mantle of lower-density material. Suppose we model a planet of radius R as indicated in Figure 5.
A core of density 2ρ and radius 3R/4 is surrounded by a mantle of density ρ and thickness R/4. Let
M be the mass of the core and m be the mass of the mantle. (This is not an accurate model of the
interior structure of the Earth but makes for an interesting and
Mantle density ρ
tractable problem.)
(a) Find the mass of the core .
R/4
3R/4
(b) Find the mass of the mantle
(c) What is the total mass of the planet (core + mantle) ?
(d) Find the magnitude of the gravitational field at the surface of
the planet.
(e) What is the magnitude of the gravitational field at the
interface between the core and the mantle ?.
Solution:
(a) The mass of the core is
Core density 2ρ
Fig.5
4 3
9
M = ρVcore ⇒ 2 ρ ⋅ π ( R) 3 = πR 3 ρ
3 4
8
(b) The mass of the mantle is m = ρ ⋅ Vmantle = ρ ⋅ [
(c) The total mass of the planet is M + m =
4 3 4 3 3 37 3
πR − π ( R) ] = πR ρ
3
3 4
48
9 3
37
91
πR ρ + πR 3 ρ = πR 3 ρ
8
48
48
(d) The magnitude of the gravitational field at the surface of the planet is
g surface =
G ( M + m) 91
= π G R ρ = 1.896πGR ρ
R2
48
(e) The magnitude of the gravitational field at the interface between the core and the mantle is
g int erface =
GM
9
16
= G ⋅ πR 3 ρ ⋅ 2 = 2π GRρ
3
8
9R
( R) 2
4
4. What is the flux of the total gravitational field of the
Earth and Moon through a closed surface that encloses the
Moon? If another surface encloses both the earth and
Moon, by how much is the flux of the gravitational field
changed?
Solution:
The surfaces are shown in figure.
(a) The flux of the total gravitational field of the Earth and
Moon through a closed surface S1 that encloses the Moon is
S2
S1
ME
MM
v v
Φ1 = ∫ g ⋅ ds = −4πGM M
S1
(b) The flux of the total gravitational field of the Earth and Moon through a closed surface S2
encloses both the earth and Moon is
v v
Φ2 = ∫ g ⋅ ds = −4πG ( M M + M E )
S2
The flux of the gravitational field changed is
∆Φ = Φ2 − Φ1 = −4πGM E