Actividad emprendedora general

−

=

× ⋅ ×

= ×

= ×

=

The theoretical Schwartzschild radius of the black hole is 8.86 mm.

(b) The very small radius of the black hole implies that it is extremely dense, according to 1₃

ρ 

 ∝

r  ^{. Thus,}

3 E

3 E

6 3

3 3

26 E

(6.38 10 m) (8.86 10 m) 3.73 10

r  r 

ρ  ρ 

ρ  ρ 

−

=

= ×

×

= ×

Therefore, the black hole is 3.73

×

 ¹⁰²⁶ times more dense than Earth.

Making Connections

12. The escape energy of an object from the Moon is much less than the escape energy of the same object from Earth. (In fact, calculations show that the ratio of the escape energy from Earth is more than 22 times as great as that from the Moon.) Thus, less fuel is required to send the spacecraft from the Moon to Earth than from Earth to the Moon.

CHAPTER 6 LAB ACTIVITIES

Lab Exercise 6.3.1: Graphical Analysis of Energies

(Page 295)

Procedure

1. Changing the numerical values in the given data to megametres and gigajoules is suggested just to make data entry in a software program easier. The results of the graphing and the analysis are not affected by this suggestion. The data are:

r  (Mm)

E g (GJ)

− − − − − − − − − −

2 – 5. The four lines required on the graph are shown below.

Analysis

(a) To analyze the energy data of a spacecraft-moon system, graphs of gravitational potential energy, binding energy, and kinetic energy are useful. All graphs should be as a function of the distance from the centre of the moon, r .

Evaluation

(b) Answers will vary. An advantage of a spreadsheet program is that it performs calculations and plots relationships on graphs very quickly. However, a disadvantage is that learning how to use the program for a specific application can be more time consuming.

Synthesis

(c) Energy considerations are important in planning space missions in order to determine the amount of fuel required to safely launch and transport the vehicles to and from the destination. The amount of fuel affects the total cost of any mission.

CHAPTER 6 SUMMARY Make a Summary

(Page 297)

Details on the diagram will vary. The figure below shows the start of the diagram. Urge students to add enough detail to help them understand and remember as much as possible about the chapter.

CHAPTER 6 SELF QUIZ

(Pages 298–299)

True/False

1. T 2. T 3. T

4. T

5. F“Evidence” is to the “Analysis” as Tycho Brahe’s work was to Kepler’s work.

6. F In Figure 1 where the path distances d 1 and d 2 are equal, the speeds along those path segments are not equal because the area swept out by these paths, and thus the time intervals, are different according to Kepler’s second law.

7. F Kepler’s third law constant for Earth is the same for the Moon as for all satellites in orbit around Earth; the constant depends on Earth’s mass.

8. F The gravitational potential energy of the Earth-Moon system is inversely proportional to the distance between the centres of the two bodies.

9. T

10. F Your escape energy and binding energy are equal because you are at rest, and therefore you have no kinetic energy.

Multiple Choice

20. (c) Kepler’s third law allows us to determine Earth’s mass according to the equation _{E 2}^E 4

23. (c) Let the subscript 1 represent the mass of the Sun at its current value and the subscript 2 represent the mass of the Sun at half its current value. Since M S2 = 1

2 M S1, C S2 = 1

24. (c) A satellite in geosynchronous orbit has a period of revolution of 24 h.

25. (a) The speed of the comet increases as it comes closer to the Sun. Position A is closet to the Sun, and therefore has the greatest speed. Position C is furthest from the Sun, and so has the slowest speed. Positions B and D are equidistant from the Sun and are between positions A and C. Thus, v_A

>

^vB = v_D

>

^vC.

CHAPTER 6 REVIEW

(Pages 300–301)

Understanding Concepts

1. The escape energy (and thus the escape speed) from the Sun is much greater than that from Earth, so the rocket given the speed needed to escape from Earth would not have enough speed to escape from the solar system. Space vehicles sent to explore distant planets have a much lower binding energy by the time they reach those distant locations, and could acquire enough energy to escape from the solar system by taking advantage of the force of gravity of the distant planet.

2. Since Earth rotates eastward, an eastward orientation of the rocket as it is being launched means that the rocket already has a component of the required velocity before blasting off. This means that less energy will be needed to launch the rocket eastward than would be required to launch it westward in order to achieve the same speed.

3.  g _U= 1.0 N/kg  M _U = 8.80

×

 ^1025 kg

r U = 2.56

×

 10⁷ m

U

Uranus has a gravitational field strength of 1.0 N/kg at an elevation of 5.1

×

 ¹⁰⁴ km above the surface.

4.  M  = 1.48

×

 10²³ kg

The magnitude of Ganymede’s gravitational field strength at a point in space 5.55

×

 10³ km from its centre is 0.318 N/kg.

5. Use the spacecraft-to-Earth line as the reference for the coordinate system.

 M _E = 5.98

×

 ^1024 kg

T,

The total gravitational field strength (magnitude and direction) of the Earth-Moon-spacecraft system is 4.23

×

 ^10−3 N/kg

[1.26

°

 from the spacecraft-to-Earth line].

6. r _E = 6.38

×

 ^106 m

d _M = 0.38d _E  g _E = 9.8 N/kg  g M = 0.38 g E

Since Mercury’s diameter is 0.38 times that of Earth’s, Mercury’s radius is also 0.38 times that of Earth. Therefore, r M = 0.38 r E.

Substituting g M = 0.38 g E:

( )

0.38 (9.80 N/kg) (0.38)(6.38 10 m) 6.67 10 N m /kg

Therefore, Mercury’s mass is 3.3

×

 ^1023 kg.

7. v = 7.15

×

 ^103 m/s

In terms of Earth’s radius, the satellite’s distance from Earth’s centre is

6

Thus, the satellite is 1.22 r E from Earth’s centre.

(b) altitude

= −

r r _E

E E

E

1.22 altit ude 0.22

r r  r 

= −

=

The satellite has an altitude of 0.22 r _E. 8. vTethys = 1.1

×

 10⁴ m/s

The orbital radius of Tethys is 3.1

×

 ^105 km.

(b) 2 r 

The orbital period of Tethys is 2.1 d.

9.  M _S = 1.99

×

 ^1030 kg

The average Sun-Venus distance is 1.08

×

 10¹¹ m.

10.  M _E = 5.98

×

 ^1024 kg

r E = 6.38

×

 10⁶ m

v = 9.00 km/s = 9.00

×

 ^103 m/s

m_R  = 4.60 kg

(a) r 

′

 ^{= ?}

Applying the law of conservation of energy:

( )( )

Let the altitude be A.

E

(b) At the altitude found in (a), the gravitational potential energy is negative, the kinetic energy is zero (because the speed is zero), and the binding energy, E _B

′

, is the extra energy needed to give the rocket a total energy of zero.

( )( ) ^{( )}

The escape speed from Titan is 2.64

×

 10³ m/s.

(b)  E esc = ? (escape speed)

At the surface of Titan, the rocket is at rest, so its kinetic energy is zero. Thus, its total energy is  E g and the escape energy is the extra energy needed to give the rocket a total energy of zero.

( )( )( )

The escape energy is 8.17

×

 ¹⁰⁹ J. This value can also be found by using the escape speed of the rocket in the equation

( )

²

The gravitational potential energy is

−

^3.99

×

 ^108 J.

(b) Since the total energy, E K  + E g, must be at least zero, the kinetic energy needed to escape is +3.99

×

 10⁸ J.

The escape speed from this position is 2.82

×

 10² m/s. The escape speed can also be found by applying the e scape energy found in (b) to the equation involving the kinetic energy, _esc 2 E ^K 

v

=

m ^{, where E} K  = E esc.

14. (a) M M = 3.28

×

 ^1023 kg

The escape speed from Mercury is 4.23 km/s.

(b)  M Moon = 7.35

×

 10²² kg

The escape speed from Earth’s Moon is 2.37 km/s.

15. (a) M star  = 3.4

×

 10³⁰ kg

The escape speed from a neutron star is 2.3

×

 ^108 m/s.

(b) c = 3.00

×

 ^108 m/s

Thus, the percentage equals v 100% 77%

  ×

c

=

  

  

^.

The escape speed from a neutron star is 77% the speed of light.

16. (a) r  =

(b) According to the data in Appendix C, the planet is Neptune. (There is a 6% difference in mass from Appendix C.)

Applying the law of conservation of energy:

( )

The escape speed is 3.1

×

 ¹⁰⁵ m/s at the location indicated; this is greater than the speed found in (a), so the proton will not escape.

18. When light strikes a piece of black paper, a small portion of the light is reflected. However, when light strikes a black hole, the light is absorbed, making the black hole even blacker than black paper.

19. m = 1.1

×

 10¹¹ M S

The Schwartzschild radius of the black hole is 3.2

×

 ^1014 m.

Applying Inquiry Skills

20. ^{Table 1} provides the missing answers concerning some of the moons of Uranus.

(a) Kepler’s third law constant for Uranus (C _U) can be calculated using t he ratio

3 2

r  T  . (b) The average of the C U values of the calculations in (a) is 1.48

×

 10¹⁴ m³/s².

(c)  M U = 8.80

×

 ^1025 kg

U

U 2

11 2 2 25

2 14 3 2 U

4

(6.67 10 N m / kg )(8.80 10 kg) 4

1.49 10 m /s C  GM 

C 

π 

π 

−

=

× ⋅ ×

=

= ×

The values agree (0.7% difference).

(d) Table completed using equations

3

U

T  r 

=

C  ^{and r}

=

^{3C T} _U ^{2 .}

(e) Students who speculate that only the larger moons can be observed using Earth-based telescopes are right. Thus, only the larger moons were discovered hundreds of years ago. Students who research the physical data of the moons that orbit Uranus will find that Titania and Oberon have diameters greater than 1500 km, whereas all the other moons listed are less than 100 km in diameter.

Table 1 Data of Several Moons of the Planet Uranus for Question 20

Moon Discovery ^r average (km) ^T  (Earth days) ^C U (m³ /s²)

Ophelia Voyager 2 (1986) 5.38

×

 10⁴ 0.375 1.48

×

 10¹⁴

Desdemona Voyager 2 (1986) 6.27

×

 10⁴ 0.475 1.46

×

 10¹⁴

Juliet Voyager 2 (1986) 6.44

×

 10⁴ 0.492 1.48

×

 10¹⁴

Portia Voyager 2 (1986) 6.61

×

 10⁴ 0.512 1.48

×

 10¹⁴

Rosalind Voyager 2 (1986) 6.99

×

 ^{104 0.556 1.48}

×

 ¹⁰¹⁴

Belinda Voyager 2 (1986) 7.52

×

 10⁴ 0.621 1.48

×

 10¹⁴

Titania Herschel (1787) 4.36

×

 10⁵ 8.66 1.48

×

 10¹⁴

Oberon Herschel (1787) 5.85

×

 10⁵ 13.46 1.48

×

 10¹⁴

21. (a) Some students may think the problem makes sense. However, many students will realize that the (theoretical) radius of an orbit that has a period of 65 min would be less than Earth’s radius. (Students may recall that the typical orbital  period of a satellite in lo w-altitude orbit is about 80 min. For example, see question 22 on page 168 of the text.) (b)  M E = 5.98

×

 10²⁴ kg

T  = 65 min = (65 min)(60 s/min) = 3.90

×

 ^103 s

r  = ?

( )( )( )

3

E

2 2

3 E 2

2

2 3 E

2

11 2 2 24 3 2

3

2

6

4

4

4

6.67 10 N m /s 5.98 10 kg 3.90 10 s 4

5.36 10 m GM 

r  T 

r GM  T 

GM T  r 

r 

π 

π 

π 

π 

−

=

  

=      

=

× ⋅ × ×

=

= ×

The theoretical radius of the orbit is 5.36

×

 ^106 m.

(c) Earth’s radius (6.38

×

 10⁶ m) is larger than the theoretical radius found in (b), so the calculated orbit cannot exist.

(d) The skill of analyzing a situation is valuable in order to reduce the chances of wasting time on calculations that don’t make sense and to increase the chances of being able to estimate whether or not a solution to a problem is logical.

22. (a) The rocket’s mass can be calculated from the gravitational potential energy at rest (on Earth’s surface at r E). From the graph E _g = –10

×

 ^{1010J = –1.0}

×

 ^1011J.

( )( )

( )( )

E g

g

E

6 11

11 2 2 24

3

6.38 10 m 1.0 10 J 6.67 10 N m /kg 5.98 10 kg 1.6 10 kg

 E  GM m r  m rE 

GM 

m

−

= −

= −

− × − ×

= × ⋅ ×

= ×

The rocket’s mass is 1.6

×

 ^103 kg.

(b) The escape energy can be determined using the value of gravitational potential energy at rest (1.0

×

 ^1011J).

(c) The launch speed of the rocket can be calculated using the value of the initial kinetic energy E _K  (on Earth’s surface at r _E). From the graph E _K  = –12

×

 ^{1010J = –1.2}

×

 ^1011J.

2 K 

K 

11

3

4

1 2 2

2(1.2 10 J) 1.6 10 kg 1.2 10 m/s

 E mv

v  E  m

v

=

=

= ×

×

= ×

The launch speed is 1.2

×

 ^104 m/s.

(d) Extrapolating from the graph, the kinetic energy E K , approaches 2.0

×

 10¹⁰ J as the distance approaches infinity, where  E _g would approach zero. This can be approximated: at 5r _E, the kinetic energy is 4.0

×

 ^{1010 J and E} g is –2.0

×

 ^1010 J.

2 K 

K 

10

3

3

1 2 2

2(2.0 10 J) 1.6 10 kg 5.0 10 m/s

 E mv

v  E  m

v

=

=

= ×

×

= ×

The speed is 5.0

×

 ^103 m/s.

Making Connections

23. (a) Turning the high-speed craft around would require a fairly large amount of energy, so mission control decided to have the craft continue on toward the Moon. The idea was to take advantage of the Moon’s gravity to act as a sort of sling-shot to help the craft accelerate in turning around and begin its return journey at the highest speed possible.

(b) One major risk was the chance that there would not be enough electrical power available to guide the craft around the Moon at the most crucial times.

Extension

24. Let L represent the large planet and S represent the small planet.

r L = 2r S

 DL = DS (densities)

Thus, using V  for volume, the ratio of the masses is:

The centripetal acceleration of the satellite is caused by the force of gravity in each case. Thus, using magnitudes:

2 2

The shortest possible period is 40 min.

25. Since the radius of the path is 2.0

×

 10¹¹ m, the distance between the stars is 2(2.0

×

 10¹¹ m) = 4.0

×

 10¹¹ m,

 M _S = 3.0

×

 ¹⁰³⁰ kg (mass of each star). The only force acting on each star is the force of gravity of the other star, which causes the circular motion of one star around the other.

( )

26.  E  is the amount of energy per unit area, and that area is proportional to 1₂

r  , where r  is the distance from the Sun to the  planet, so:

From Kepler’s third law:

3

where C  is Kepler’s third law constant for the Sun

Substituting into the first equation:

2 4

Thus, E  is proportional to

4

T 3

−

. (Solving for the “constant” is unnecessary.)

UNIT 2 PERFORMANCE TASK

In document Análisis de la situación del emprendimiento en Extremadura (página 45-49)