西南交大 大学物理 英文 试题 答案No.A1-4

force on the salt lick if it is placed at P is 9.48 N .
Solution:
1.60×108m
P
Moon 90° 7.36×1022kg
4.16×108m
ˆj iˆ
(a) The total gravitational field at the point P is
grtotal
=
grM
+
grE
=
GM M r 2PM
iˆ +
GM E r 2PE
(cosθ iˆ + sinθ
ˆj)
Earth 5.98×1024kg Fig.2
=
6.67 ×10−11 × 7.36 ×1022 (1.6 ×108 )2
iˆ +
6.67 ×10−11 × 5.98 ×1024 (4.16 ×108 )2
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
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:
a = gtotal = 2.37 ×10−3 m/s2
(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
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
each object is doubled but the distance between them is halved, then the new force of gravity
between the objects will be
（A）
(A) 16 F0 Solution:
(B) 4 F0
(a) The magnitude of the force is
F
=
GMm r2
=
ma
So
a
=
GM r2
= 6.67 ×10−11 × 4 ×1030 (1×105 )2
= 2.67 ×1010 m/s2
(b) The ratio of the magnitude of this acceleration and g is a = 2.67 ×1010 = 2.72 ×109 g 9.81
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
Solution:
According to the shell theorem 1, at the center of the body, 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
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.
r
QFtoFtrraMl
= FM 'on m ' on m =r 0
+
FM
on百度文库
m
∴ Ftotal = FM on m
∴ Ftotal
=
GMm r2
=
GMm (s + R)2 − d 2
m d
R M′
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
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
m
θ
x
x
the ring on a particle of mass m located a distance x from
the center of the ring along its axis. See Fig.3. (b) Suppose
M
that the particle falls from rest as a result of the attraction of the ring of matter. Find an expression for the speed with
(
1.6 ×108 4.16 ×108
iˆ +
3.84 ×108 4.16 ×108
ˆj)
= 1.92 ×10−4iˆ + 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).
(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
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
In addition, many galaxies contain ring-like structures.
Consider a homogeneous ring of mass M and radius R. (a)
R
Find an expression for the gravitational force exerted by
∫sgr ⋅ dsr = 4πG∑ Mi = 4πG∫ ρdV
when ρ = A r 2 , the acceleration gr will increase as you go deeper, when ρ = Ar the
acceleration gr will decrease as you go deeper.
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:
(C) F0
(D) F0/2
The magnitude of the gravitational force is
Fgrav
=
GMm r2
,
according
to
the
problem,
we
get
Fg′rav
=
4Gm2 (r / 2)2
=
16
Gm2 r2
= 16F0
2. A spherical symmetric nonrotating body has a density that varies appreciably with the radial
Fig.3
which it passes through the center of the ring.
The magnitude of the gravitational force is
Fgrav
=
Gm1m2 r2
=
Gm1(M r2
−
m1)
,
We get
dF = 0 dm1
⇒
G(M − r2
2m1)
=
0
⇒
m1
=
M 2
Then
m1 = m2 .
2. A mass m is inside a uniform spherical shell of mass
(c) Since the acceleration is
a = ω2r
= ⎜⎛ 2π ⎝T
⎟⎞2 r ⎠
=
4π 2r T2
The orbital period is
T=
4π 2r = a
4π 2r3 = GM
4 × 3.142 × (1×105 )3 6.67 ×10−11 × 4 ×1030
= 1.12 ×10−2 (s)
1. Several planets (the gas giants Jupiter, Saturn, Uranus,
and Neptune) possess nearly circular surrounding rings,
y
perhaps composed of material that failed to form a satellite. dM
M′ and a mass M is outside the shell as shown in Figure 1.
The magnitude of the total gravitational force on m is
F = G Mm . (s + R)2 − d 2
M s
Solution:
rr