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AP Physics C Textbook Problems
Chapter 13
Pages 412 – 416
HW-16:
03.
A 200-kg object and a 500-kg object are separated by 0.400 m.
(a)
Find the net gravitational force exerted by these objects on a 50.0-kg object placed midway between them.
F = FL + FR = - G mL m0 / rL0² + G mR m0 / rL0² = G ( - mL / rL0² + mR / rR0²) m0
-11
-5
-5
= (6.673 x 10 ) [ - (200) / (0.2)² + (500) / (0.2)² ] (50) = 2.502375 x 10 = 2.50 x 10 N, toward the 500-kg object
(b)
At what position (other than an infinitely remote one) can the 50.0-kg object be placed so as to experience a
net force of zero?
Anywhere that:
500 / x² = 200 / (0.4 – x)²
2 x² = 5 (x² - 0.8 x + 0.16) = 5x² - 4 x + 0.8
3x² - 4 x + 0.8 = 0 
x1 = 0.245 m = distance from 500-kg object
x2 = 0.155 m = distance from 300-kg object
05.
Three uniform spheres of mass 2.00 kg, 4.00 kg,
and 6.00 kg are placed at the corners of a right
triangle as shown in the figure at the right. Calculate
the resultant gravitational force on the 4.00-kg
object, assuming the spheres are isolated from the
rest of the universe.
-11
F24 = G m2 m4 / r24² = (6.673 x 10 ) (2) (4) / (3)²
-11
= 5.93156 x 10 , +y direction
-11
F64 = G m6 m4 / r64² = (6.673 x 10 ) (6) (4) / (4)²
-10
= 1.00095 x 10 , -x direction
F = (-100 i + 59.3 j) pN
07.
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses
lead spheres with masses of 1.50 kg and 15.0 g whose centers are separated by 4.50 cm. Calculate the
gravitational force between these spheres, treating each as a particle located at the center of the sphere.
-11
F12 = G m1 m2 / r12² = (6.673 x 10 ) (1.5) (0.015) / (0.045)²
-10
= 7.41 x 10 N
09.
When a falling meteoroid is at a distance above the Earth’s surface of 3.00 times the Earth’s radius, what is its
acceleration due to the Earth’s gravitation?
-11
24
6
aE = G mE / (4 rE)² = (6.673 x 10 ) (5.98 x 10 ) / (4 x 6.37 x 10 )²
= 0.614644 = 0.615 m/s², toward the Earth
11.
On the way to the Moon the Apollo astronauts reached a point where the Moon’s gravitational pull became
stronger than the Earth’s.
(a)
Determine the distance of this point from the center of the Earth.
ME / RES² = MM / RMS²
RMS² = RES² (MM / ME) = (RES)² (7.36) / (598) = RES² / 81.25

RMS = RES / 9.01388
8
RMS + RES = 3.84 x 10 = RES / 9.01388 + RES = RES (1 / 9.01388 + 1) = 1.11094 RES
8
8
8
RES = 3.84 x 10 / 1.11094 = 3.45653 x 10 = 3.46 x 10 m
(b)
What is the acceleration due to the Earth’s gravitation at this point?
8
-11
24
8
-3
aE = G mE / (3.45653 x 10 )² = (6.673 x 10 ) (5.98 x 10 ) / (3.45653 x 10 )² = 3.33996 x 10 m/s²
HW-17:
13.
Plaskett’s binary system consists of two stars
revolving in a circular orbit about a center of mass
midway between them. This means that the masses
of the two stars are equal. Assume the orbital
speed of each star is 220 km/s and the orbital period
of each is 14.4 days. Find the mass M of each star.
(For comparison purposes, the mass of our Sun is
1.99 x 1030 kg.)
Centripetal Force:
Gravitational Force:
Equating the two forces:
M = 4 v² r / G
M v² / r
G M M / (2 r)²
Mv² / r = G M² / (4r²)
Now, the distance traveled in orbit is:
2 π r = v T  r = v T / (2 π)
22
-11
M = (4 v² / G) (v T / 2 π) = (2 v³ T) / (π G) = (2) (220,000)³ (14.4 x 24 x 3600) / π / G = 2.64956 x 10 / 6.673 x 10
32
= 1.26 x 10 kg
17.
Comet Halley approaches the Sun to within 0.570
AU, and its orbital period is 75.6 years. (AU is the
symbol for astronomical unit, where 1 AU =
1.50 x 1011 m is the mean Earth-Sun distance.)
How far from the Sun will Halley’s comet travel
before it starts it return journey?
semi-major axis of orbit:
a = (0.570 AU + y) / 2
Note that since T² = k a³: when we measure T in terms of Earth
years a in terms of AU we have k = 1
T² = k a³
(75.6)² = (1) (0.570 + y)³ / 8
(0.570 + y)³ = 8 (75.6)² = 45722.88
0.570 + y = 35.758381
y = 35.188 AU = 35.2 AU
/π
19.
A synchronous satellite, which always remains above the same point on a planet’s equator, is put in orbit around
Jupiter to study the famous red spot. Jupiter rotates about its axis once every 9.84 h. Use the data in the table
below to find the altitude of the satellite.
Centripetal Force:
Gravitational Force:
Equating the two forces:
MS v² / r = MS [ 2π (RJ + y) / T ]² / (RJ + y) = 4π² MS (RJ + y) / T²
G MS MJ / r² = G MS MJ / (RJ + y)²
4π² MS (RJ + y) / T² = G MS MJ / (RJ + y)²
4π² / T² = G MJ / (RJ + y)³
-11
27
24
(RJ + y)³ = G MJ T² / (4 π²) = (6.673 x 10 ) (1.90 x 10 ) (9.84 x 3600)² / (4π²) = 4.030047709 x 10
8
(RJ + y) = 1.59136595 x 10
7
7
(6.99 x 10 ) + y = 8.9236595 x 10 m
21.
Suppose the Sun’s gravity were switched off. The planets would leave their nearly circular orbits and fly away in
straight lines, as described by Newton’s first law. Would Mercury ever be further from the Sun than Pluto? If so,
find how long it would take for Mercury to achieve this passage. If not, give a convincing argument that Pluto is
always further from the Sun.
Centripetal Force:
Gravitational Force:
Equating the two forces:
MP v² / r
G MS MP / r²
G MS MP / r² = MP v² / r
 v² = G MS / r
-11
30
10
9
-11
30
12
7
For Mercury:
v² = G MS / r = (6.673 x 10 ) (1.991 x 10 ) / (5.79 x 10 ) = 2.29463 x 10
v = 47,902 m/s
For Pluto:
v² = G MS / r = (6.673 x 10 ) (1.991 x 10 ) / (5.91 x 10 ) = 2.24804 x 10
v = 4,741 m/s
With greater speed, Mercury will ultimately move further from the Sun than Pluto.
(Mercury distance from sun)² = dM0² + (vM t)²
(Pluto distance from sun)²
= dP0² + (vP t)²
 dM0² + (vM t)² = dP0² + (vP t)²
(dp0² - dM0²) = (vM² - vP²) t²
12
10
7
t² = (dp0² - dM0²) / (vM² - vP²) = [(5.91 x 10 )² - (5.79 x 10 )²] / [(229.463 – 2.24804) x 10 ]
20
7
13
= (349247 x 10 ) / (227.21496 x 10 ) = 1537.077 x 10
8
t = 1.239789 x 10 s = 34,439 hr = 1435 days = 3.93 years
Answer book indicates 393 years
HW-18:
23.
Three objects of equal mass are located at three corners of a
square of edge length l as shown in the figure at the right. Find the
gravitational field at the fourth corner due to these objects.
g = (G m / l²) j + (G m / 2 l²) (cos 45 i + cos 45 j) + (G m / l²) i
__
= (G m / l²) (1 + ½ cos 45) (i + j) = (G m / l²) (1 + √(2) / 4) (i + j)
__
___
| g |² = (G m / l²)² (2) (1 + √(2) / 4 )² = (G m / l²)² (2) (√(2) + 2 / 4 )² =
___
| g | = (G m / l²) (√(2) + ½), toward the opposite corner
25.
Compute the magnitude and direction of the gravitational field at a point P
on the perpendicular bisector of the line joining two objects of equal mass
separated by a distance 2a as shown in the figure at the right.
Let l² = a² + r²
g = (G M / l²) ( - cos ϴ i + sin ϴ j) + (G m / l²) ( - cos ϴ i - sin ϴ j)
= - 2 (G M / l²) cos ϴ i = - 2 (G M / l²) (r / l) i
| g | = 2 G M r / l³, in the –x direction
27.
How much energy is required to move a 1000 kg object from the Earth’s surface to an altitude twice the Earth’s
radius?
For object far removed from the Earth’s surface:
U =-GMm/r
∆U = U(3 RE) - U( RE) = - G M m [ (1 / 3RE) - ( 1 / RE) ] = - (G M m / RE) (1/3 – 1) = (2/3) (G M m / RE)
-11
24
6
10
∆U = (2/3) (6.673 x 10 ) (5.98 x 10 ) (10³) / (6.37 x 10 ) = 4.1763 x 10 J
29.
After our Sun exhausts its nuclear fuel, its ultimate fate may be to collapse to a white dwarf state, in which it has
approximately the same mass as it has now, but a radius equal to the radius of the Earth. Calculate
(a)
the average density of the white dwarf,
30
6
12
ρ = MS / V = MS / [(4/3) π RE³] = (1.991 x 10 ) / (4/3) / π / (6.37 x 10 )³ = 0.001838926 x 10
(b)
the free-fall acceleration, and
-11
30
6
7
9
= 1.84 x 10 kg/m³
6
g = G MS / RE² = (6.673 x 10 ) (1.991 x 10 ) / (6.37 x 10 )² = 0.327426 x 10 = 3.27 x 10 m/s²
(c)
the gravitational potential energy of a 1.00-kg object at its surface.
-11
30
6
13
U = - G MS m / RE = - (6.673 x 10 ) (1.991 x 10 ) (1) / (6.37 x 10 ) = - 2.09 x 10 J
31.
A system consists of three particles, each of mass 5.00 g, located at the corners of an equilateral triangle with
sides of 30.0 cm.
(a)
Calculate the potential energy of the system.
-11
Utotal = U12 + U13 + U23 = - 3 (G m1 m2 / r12) = - (3) (6.673 x 10 ) (0.005) (0.005) / (0.3) = - 1.66825 x 10
(b)
If the particles are released simultaneously, where will they collide?
At the center of the triangle.
-14
J
HW-19:
33. A space probe is fired as a projectile from the Earth’s surface with an initial speed of 2.00 x 104 m/s. What will its
speed be when it is very far from the Earth? Ignore friction and the rotation of the Earth.
Einitial = ½ m vI² - G ME m / rI
Efinal = ½ m vF² - G ME m / rF = Einitial = ½ m vI² - G ME m / rE
½ m vF² = G ME m / rF + ½ m vI² - G ME m / rE
4
-11
24
6
8
vF² = 2 G ME / rF + vI² - 2 G ME / rE = 0 + (2 x 10 )² - (2) (6.673 x 10 ) (5.98 x 10 ) / (6.37 x 10 ) = 2.74711 x 10
vF = 16,600 m/s
37.
A satellite of mass 200 kg is placed in Earth orbit at a height of 200 km above the surface.
(a)
With a circular orbit, how long does the satellite take to complete one orbit?
mv² / r = G m ME / r²
T = ∆t = 2 π r / v


v² = G ME / r
T² = 4 π² r² / v = 4 π² r³ / G ME
6
6
6
r = RE + 200,000 = 6.37 x 10 + 0.2 x 10 = 6.57 x 10 m
6
-11
24
22
14
T² = (4 π²) (6.57 x 10 )³ / (6.673 x 10 ) / (5.98 x 10 ) = (1.11958184 x 10 ) / (3.990454 x 10 ) = 28056503
T = 5297 s = 88.3 min = 1.47 hr
(b)
What is the satellite’s speed?
6
v = 2 π r / T = (2π) (6.57 x 10 ) / (5297) = 7793 m/s
(c)
What is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance, but
include the effect of the planet’s daily rotation.
Einitial = ½ m vI² - G ME m / rI
Efinal = ½ m vF² - G ME m / rF = Einitial + ∆E
vI = (2π RE) / (3600 x 24) = 463.239 m/s
6
rI = 6.37 x 10 m
∆E =
=
=
=
41.
Efinal – Einitial = ½ m vF² - G ME m / rF - ½ m vI² + G ME m / rI = ½ m (vF² - vI²) + G ME m (1 / rI - 1 / rF)
-11
24
6
6
(0.5) (200) [ (7793)² - (463.239)² ] + (6.673 x 10 ) (5.98 x 10 ) (200) [ 1 / (6.37 x 10 ) - 1 / (6.57 x 10 ) ]
7
16
-9
(100) (6.051626 x 10 ) + (7.98091 x 10 ) (4.77887 x 10 )
9
8
9
9
(6.051626 x 10 ) + (3.813973 x 10 ) = 6.4330233 x 10 = 6.43 x 10 J
Determine the escape speed for a rocket on the far side
of Ganymede, the largest of Jupiter’s moons. The
radius of Ganymede is 2.64 x 106 m, and its mass is
1.495 x 1023 kg. The mass of Jupiter is 1.90 x 1027 kg
and the distance between Jupiter and Ganymede is
1.071 x 109 m. Be sure to include the gravitational effect
due to Jupiter, but you may ignore the motion of Jupiter
and Ganymede as they revolve about their center of
mass.
-11
23
6
6
8
UGanamede = - G M1 M2 / r = - (6.673 x 10 ) (1.495 x 10 ) (M2) / (2.64 x 10 ) = - (3.7783 x 10 ) M2 J = - (0.037783 x 10 ) M2 J
-11
27
9
8
= - G M1 M2 / r = - (6.673 x 10 ) (1.90 x 10 ) (M2) / (1.071 x 10 ) = - (1.1838189 x 10 ) M2 J
UJupiter
8
Utotal = - 1.2216019 x 10 M2 J
8
Required Kinetic Energy = 1.2216019 x 10 M2 J = ½ M2 v²
8
v² = 2.4432038 x 10 J

4
v = 1.56307 x 10 = 15.6 km/s
45.
A comet of mass 1.20 x 1010 kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges
between 0.500 AU and 50.0 AU.
Note: 1 AU = one astronomical unit = the average distance from Sun to Earth = 1.496 x 1011 m.
(a)
What is the eccentricity of its orbit?
The semi-major axis is:
(50.0 + 0.5) / 2 AU 25.25 AU = a
a + c = 50 AU  c = 50 – 25.25 = 24.75 AU
e = c / a = 24.75 / 25.25 = 0.980198 = 0.980
(b)
What is its period?
T² = k a³ = (1) (25.25)³ = 16098.45
T = 126.880 yr = 127 yr
(c)
At aphelion what is the potential energy of the comet-Sun system?
-11
30
10
11
U = - G M m / r = - (6.673 x 10 ) (1.991 x 10 ) (1.20 x 10 ) / (50 x 1.496 x 10 )
17
17
= - 2.13 x 10 J
= - 2.131434 x 10

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