ORBITS AND KEPLER’S LAWS PRACTICE

=

The gravitational field strength is 3.20 N/kg at a distance of 0.75 r _E above Earth’s surface.

Applying Inquiry Skills

9. The diagram below shows examples of the required FBDs.

Making Connections

10. Based on the data in Table 1, some astronomers might argue that Pluto should not be considered to be a planet because the gravitational field strength of Pluto is so much smaller than the other planets. On average, Pluto’s gravitational field strength is less than one-tenth the gravitational field strength of the other planets.

6.2 ORBITS AND KEPLER’S LAWS PRACTICE

(Page 279)

Understanding Concepts

1. The Moon does not fall into Earth because it travels at a specific speed around Earth. This keeps the Moon at an approximately constant distance from Earth’s centre called the orbital radius.

2. Since the space probe is in a circular orbit, the direction of the gravitational force is perpendicular to the direction of the instantaneous velocity. Thus, the force of gravity does not do any work on the probe, and there is no change in the kinetic energy (or speed) of the probe.

3. Let the subscript s represent the satellite.

 M _E = 5.98

×

 ^1024 kg

r E = 6.38

×

 10⁶ m

r s = 525 km = 5.25

×

 ^105 m

r  = r E + r s

(a) _s GM ^E

v

=

r 

11 2 2 24

6 5

3 s

(6.67 10 N m / kg )(5.98 10 kg) (6.38 10 m) (5.25 10 m) 7.60 10 m/s

v

×

−

⋅ ×

= × + ×

= ×

The speed of the satellite is 7.60

×

 10³ m/s.

(b) 2 ^r 

v T 

=

π 

(

^{6 5}

)

3

3

2

2 6.38 10 m 5.25 10 m 7.60 10 m/s

5.71 10 s

T  r  v

T 

π 

π 

=

× + ×

= ×

= ×

The period of revolution of the satellite is 5.71

×

 ¹⁰³ s or (5.71

×

 ^{103 s) 1 h}

3600 s

  

  

  

 ^{= 1.59 h.}

4. Let the subscript s represent the satellite.

 M _Moon = 7.35

×

 ^1022 kg

r _Moon = 1.74

×

 ^106 m

Moon s

11 2 2 22

6

3 s

(6.67 10 N m / kg )(7.35 10 kg) 1.74 10 m

1.68 10 m/s

v GM 

r 

v

−

=

× ⋅ ×

= ×

= ×

The speed of the satellite is 1.68

×

 ^103 m/s.

Applying Inquiry Skills

5. (a) The speed of a satellite around a central body is inversely proportional to the square root of the radius of the satellite’s orbit. Thus, _s

s

v 1

∝

r  . (b)

Making Connections

6. Recommended web sites are:

www.space.com/spacewatch/space_junk.html

http://earthwatch.unep.net/solidwaste/spacejunk.html

Students are likely to encounter the following points about the problem of space junk:

•

a study done in 1999 estimated 4 million pounds of space junk in low-Earth orbit

•

objects that are baseball-size and bigger may threaten the safety of astronauts in space; collisions involving even the smallest of objects may be damaging due to the high speeds of the objects

•

the U.S. Space Command agency counted the number of objects in space as of June 21, 2000 and found 2671 satellites, 90 space probes, and 6096 chunks of debr is

•

some objects re-enter Earth’s atmosphere, but most burn up on re-entry, or land in water or uninhabited land

•

 NASA calculates that if the amount of debri s equal to or larger than 1 centimeter exceeds 150 000, it could make space flight impossible

•

as technology improves, more and more satellites and space objects are being launched into space, not only by space agencies, but other industries including telecommunications and organizations interested in astronomy

•

 possible solutions include better tracking of space junk, i mproved methods of bringing down satellites to Earth, and launching satellites to sweep up the space junk that is already in orbit

PRACTICE

(Page 283)

Understanding Concepts

7. Relative to the rest of the solar system, Earth’s frame of reference is accelerating, so the geocentric model is the

noninertial frame of reference. The heliocentric model is an inertial frame of reference if the solar system is considered to  be isolated. However, it is a noninertial frame with respect to the Milky W ay Galaxy and the rest of the universe.

8. Tycho Brahe (1546–1601) made precise, comprehensive astronomical measurements of the solar system and more than 700 stars. For 20 years, he made countless naked-eye observations using l arge instruments he made hi mself. He was able to collect data for Mercury, Venus, Earth, Mars, and Saturn, because those were planets that he could see. The other planets were beyond his scope of vision and the telescope was not invented until the early 17th century.

9. Using Kepler’s second law, Earth sweeps out equal areas in equal time intervals. Therefore, when Earth is closest to the Sun, it is moving fastest. Conversely, if the orbit is divided into 180

°

 “halves,” the portion closest to the Sun will have a smaller area, and therefore a shorter time. Since there are three fewer days between September 21 and March 21, Earth must be closest to the Sun at that time.

10. The ratio

3

2

r 

T  is calculated for each planet in Table 1.

Table 1

Object Mean

Radius of Orbit (m)

Period of Revolution of Orbit (s)

r ³

∝ ∝ ∝ ∝

T ²

3 2

C  r 

=

T  ^(m3 /s2)

Mercury 5.79

×

 10¹⁰  7.60

×

 10⁶  3.36

×

 10¹⁸ Venus 1.08

×

 10¹¹  1.94

×

 10⁷  3.35

×

 10¹⁸ Earth 1.49

×

 ^{1011  3.16}

×

 ^{107  3.31}

×

 ¹⁰¹⁸

Mars 2.28

×

 10¹¹  5.94

×

 10⁷  3.36

×

 10¹⁸ Jupiter 7.78

×

 10¹¹  3.75

×

 10⁸  3.35

×

 10¹⁸ Saturn 1.43

×

 10¹²  9.30

×

 10⁸  3.38

×

 10¹⁸ Uranus 2.87

×

 10¹²  2.65

×

 10⁹  3.37

×

 10¹⁸ Neptune 4.50

×

 ^{1012  5.20}

×

 ^{109  3.37}

×

 ¹⁰¹⁸

Pluto 5.91

×

 10¹²  7.82

×

 10⁹  3.38

×

 10¹⁸

All proportionality constants calculated in Table 1 are within 1.5% of the average value. This verifies Kepler’s third law.

11. (a) The average value (in SI base units) of the constant of proportionality in ^r 3

∝

^T 2 is 3.36

×

 10¹⁸ m³/s². (b) ^C _S = 3.36

×

 ^1018 m3/s2 (from (a))

S

S 2

2 S S

18 3 2 2

11 2 2

30 S

4 4

(3.36 10 m /s )4 6.67 10 N m / kg 1.99 10 kg

C  GM 

 M  C 

G

 M 

π  π 

π 

−

=

=

= ×

× ⋅

= ×

The mass of the Sun is 1.99

×

 ^1030kg.

12. (a)^r Moon = 3.84

×

 ^108 m

Thus, Kepler’s third law constant, ^C _E, is 1.02

×

 10¹³ m³/s² for objects orbiting Earth.

(b) Let the subscript s represent the satellite.

 M _E = 5.98

×

 10²⁴ kg

The satellite must be 1.3

×

 ¹⁰⁴ km above the centre of Earth.

( )

Applying Inquiry Skills

13.

In the sample diagram above, the shape of each figure is approximately that of a triangle, so the approximate areas are:

1 1 1

14. (a) Let the subscript C represent the central body around which another body (subscript B) revolves in an orbit of known

3

(b) According to the equation derived in (a) above,^G is known and if the mass of the star can be estimated, the period of revolution of the planet must be found in order to solve for the radius of the orbit. Astronomers have discovered that the mass of a star can be estimated by determining its luminosity. (This applies to “main-sequence” stars, by far the majority of stars.) The period of revolution is determined by measuring the period of the wobble of the star as the  planet tugs on the star. The radius of the orbit is found using the equation

2

Section 6.2 Questions

(Page 284)

Understanding Concepts

1. According to Kepler’s first law, a comet travels in an elongated elliptical orbit. Kepler’s second law implies that the  portion of the comet’s orbit cl oset to the Sun will have a much smaller area than the rest of its orbit. Thus, the time spent

in this region (where the comet may be visible to observers on Earth) will be far less compared to the total orbital period.

2. Kepler’s second law states that Earth sweeps out equal areas in equal time intervals. Therefore, on January 4 when Earth is closest to the Sun, it is moving most rapidly because the distance travelled is greatest for equal time intervals. The Earth is moving least rapidly on July 5 when it is farthest from the Sun.

3. Although the nonrotating frame of reference is placed at the centre of the Sun, the Sun is in orbit around the centre of the Milky Way Galaxy, so it is constantly accelerating. This means it is a noninertial frame of reference within the galaxy.

4. ^r  = 4.8

×

 10¹¹ m

The orbital period of the asteroid is 1.8

×

 ^108 s.

5. Let the subscript P represent the unknown planet.

3

The small planet would be 1.6 times farther from the Sun than Earth.

6.  ^M E = 5.98

×

 ^1024 kg

The period of revolution is 4.0 h.

7. Let the subscript P represent Phobos, the subscript D represent Deimos, and the subscript M represent Mars.

T _D = 30 h 18 min = 30 h 3600 s

Substitute the value of ^C M into the equation for Phobos:

3

Applying Inquiry Skills

8.

Therefore, the SI base units are metres per second.

9. Since^r 3 =^CT 2 (from Kepler’s third law), the line on the graph is straight and the slope is Kepler’s third law constant for the Sun.

Making Connections

10. (a) Galileo was born near Pisa, on February 15, 1564. In 1609, after learning that a telescope had been invented in

Holland, he built his own telescope of 20 times magnification. The strength of this magnification allowed Galileo to see mountains and craters on the Moon, and to discover the four largest satellites of Jupiter.

Tycho’s work was done between 1581 and 1601, during which he made numerous naked-eye observations with large instruments. Kepler began analyzing Tycho’s data in 1601. In 1609, Kepler published his first two laws. He  published his third law in 1619.

(b) Relating Kepler’s third law and the law of universal gravitation, we find that the mass of Jupiter is given by

2 2 M

J 2

M

4 ^r 

 M 

GT 

=

π  ^{, wherer} M and^T _M are the radius of the orbit and period of revolution, respectively, of any moon around Jupiter. Thus, Galileo would need to know the values of^r _M and^T _M for at least one moon, as well as the universal gravitation constant,^G.

(c) Calculating the mass of Jupiter was not possible until the value of^G was determined, which was not possible until Kepler’s third law was formed. (As mentioned in Section 3.3, it was Cavendish who first determined that value in 1798.)

6.3 GRAVITATIONAL POTENTIAL ENERGY IN GENERAL

In document SPH4U SOLUTIONS (UNIT 2) (Page 124-130)