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
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.
Fig.3
which it passes through the center of the ring.
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
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
each object is doubled but the distance between them is halved, then the new force of gravity
between the objecttion:
(B) 4 F0
r
QFtoFtrraMl
= FM 'on m ' on m =r 0
+
FM
on
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
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
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
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
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
(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
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).
(
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
(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
(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)
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
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:
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
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:
from the center, which is related to how the density of the body changed with respect to the distance