Module 1 Space

HSC Physics Module 1: Space Summary

Space: 1. The Earth’s gravitational field x

Define weight as the force on an object due to a gravitational field

Weight is defined as the force of an object due to a gravitational field. It is a vector quantity with the unit Newton (N). Mathematically, weight can be expressed as W=mg, where m is the mass of the object and g is the acceleration due to gravity. x

Explain that a change in gravitational potential energy is related to work done

Consider the work done to move an object from the Earth’s surface to a height, h: Work done = force x displacement Work done = forcegravity x displacement Work done = [mass x gravity] x displacement Work done = [mass x gravity] x height

(where forcegravity is the weight) (as the displacement over which the work is done is equal to the height of the lift)

Mathematically, W = mgh x

Perform an investigation and gather information to determine a value for acceleration due to gravity using pendulum motion or computer assisted technology and identify reason for possible variations from the value 9.8ms-2

Investigation: Determining a value for gravity Aim: To determine a value for g, by observing the motion of a swinging pendulum Equipment: -50g mass or large nut to act as pendulum -about 1.2m length of string -a support at least one metre above the ground (e.g. hook on ceiling) -stopwatch -another person to assist Method: 1/ Adjust the length of the pendulum, l, to one metre. 2/

Set up equipment as shown:

Robert Lee Chin

1

HSC Physics Module 1: Space Summary

l

Time taken for one complete swing is called the period, T

3/

Set pendulum gently by pulling it back 10˚ from vertical. Starting at the extreme of the motion, time 10 full periods, finishing back at the same side.

4/

Divide time by 10 and record in table. Calculate a value for g, using g

5/

Repeat steps 3-4 four more times, shortening string by 5cm each time.

6/

Calculate an average value for g, using all 5 trials

4S 2 T2

Results: Trial 1 2 3 4 5

Pendulum length (m) 1.01 0.97 0.91 0.85 0.81

Period (s) 2.00 1.93 1.97 1.89 1.88 Average

g (ms-2) 9.9680 10.281 9.256 9.394 9.041 9.59

T2 4.00 3.72 3.88 3.57 3.53 n/A

Alternatively, plot T2 against l on an x-y graph and draw the line of best fit: 2

T squared (s ) 4 3.72 3.88 3.57 3.53 0.00

Robert Lee Chin

length (m) 1.01 0.97 0.91 0.85 0.81 0.00

2

HSC Physics Module 1: Space Summary 4π 2 gradient

Calculate ‘g’ using the formula: g

T squared vs L 4.5 y = 4.0965x

T squared (s squared)

4 3.5 3 2.5 2 1.5 1 0.5 0 0

0.2

0.4

0.6

0.8

1

1.2

le ngth (m )

Gradient ≈ 4.10 (3 s.f.)

' g'

4S 2 4.10

9.62888... # 9.63 (3 s.f.)

Discussion: Variations in g were due to reaction time in timing the swing- the accuracy of the stopwatch is much more accurate than reaction time. Also, friction between the string and its attachment as well as air resistance, which would have slowed the swing of the pendulum. The reliability of the results could be improved by performing multiple trials for the same length. Also, any results that do not conform with the majority of the results e.g. a value of 7ms-2 for ‘g’, should be ignored and the trial repeated until a measurement that conforms to the majority is obtained. The validity of the experiment could be improved by using an attachment for the string that would produce as little friction as possible, without causing the pendulum to swing erratically. The experiment was conducted slightly above sea level (approximately 50 metres). While this altitude may not seem like much, it could have has some effect on the results. Finally, two people were involved with performing the experiment- one to time the pendulum and another to let go of the pendulum. It would be better to have only one person to perform both of these steps because there are differences in reaction times when two or more people are involved.. The first method gave a fairly reliable value for g, (9.59ms-2), only 0.21ms-2 less than the published value of 9.8ms-2. Using the second method gave a slightly more accurate value of 9.63ms-2. The second method is more valid because it relies on the line of best fit, which helps to rule out any discrepant results

Robert Lee Chin

3

HSC Physics Module 1: Space Summary x

Gather secondary information to predict the value of acceleration due to gravity on other planets

The value of g on a planet is given by: m g G Planet 2 where, ( rPlanet ) G Universal gravitational constant m Planet mass of planet in kg rPlanet

6.67 u 10 11 m 3 kg 1 s 2

radius of planet in metres

Planet

Value Calculations of g (ms-2) Mercury m Planet 3.30 u 10 23 kg 3.70 Venus

g

m Planet

Jupiter

g

G

m Planet

24.8 Saturn

g

G

m Planet

10. 4 8.83

g

0.623

G

m Planet g

Robert Lee Chin

G

m Planet g

Pluto

G

9.80 3.70

Uranus

G

m Planet

8.87 Earth Mars

g

G

m Planet ( rPlanet )

2

6.67 u 10 11 u 24

4.87 u 10 kg m Planet

( rPlanet ) 2

( rPlanet ) 2

( rPlanet )

2

5.69 u 10 kg ( rPlanet )

2

8.68 u 10 kg ( rPlanet ) 2

( rPlanet ) 2

rPlanet

6.67 u 10 11 u

1.30 u 10 22 kg m Planet

rPlanet

6.67 u 10 11 u 25

m Planet

rPlanet

6.67 u 10 11 u 26

m Planet

rPlanet

6.67 u 10 11 u

1.90 u 10 27 kg m Planet

rPlanet

6.67 u 10 11 u

6.42 u 10 23 kg m Planet

rPlanet

rPlanet

6.67 u 10 11 u

2.44 u 10 6 m (3.30 u 10 23 ) 6 2

( 2.44 u 10 )

3.69709... # 3.70ms 2 (3 s.f.)

6.05 u 10 6 m ( 4.87 u 10 24 ) (6.05 u 10 6 ) 2

8.87450.. # 8.87 ms 2 (3 s.f.)

3.40 u 10 6 m (6.42 u 10 23 ) (3.40 u 10 6 ) 2

3.70427.. # 3.70ms 2 (3 s.f.)

7.15 u 10 7 m (1.90 u 10 27 ) 7 2

(7.15 u 10 )

24.7894.. # 24.8ms 2 (3 s.f.)

6.03 u 10 7 m (5.69 u 10 26 ) 7 2

(6.03 u 10 )

10.43766.. # 10.4ms 2 (3 s.f.)

2.56 u 10 7 m (8.68 u 10 25 ) ( 2.56 u 10 7 ) 2

8.83416.. # 8.83ms 2 (3 s.f.)

1.18 u 10 6 m (1.30 u 10 22 ) (1.18 u 10 6 ) 2

0.62273.. # 0.632ms 2 (3 s.f.)

4

HSC Physics Module 1: Space Summary x

Analyse information using the expression F=mg to determine the weight force for a body on Earth and for the same body on other planets

The magnitude of the weight force on an object can be calculated using a slightly altered form of Newton’s second law: F = mg. The direction of the weight force is the same as the direction of the gravitational field at the location of the object i.e. towards the centre of the Earth. The acceleration due to gravity at the Earth’s surface is 9.8ms-2. The weight of a 465kg llama can be calculated as follows:

m W

465kg g 9.8ms -2 mg 465 u 9.8 4557 # 4600 N (2 s.f.)

Acceleration due to gravity for the same 465kg llama on different planets: Planet Value of g Weight Force Calculations -2 (ms ) Mercury

3.70

Venus

8.87

Earth

9. 8

Mars

3.70

Jupiter

24. 8

Saturn

10. 4

Uranus Pluto

1700

4100

W

mg

465 u 3.70 1720.5 # 1700 N (2 s.f.)

W

mg

465 u 8.87

4600 1700

12000

4124.55 # 4100 N (2 s.f.)

W

mg

465 u 3.70 1720.5 # 1700 N (2 s.f.)

W

mg

465 u 24.8 11532 # 12000 N (2 s.f.)

W

mg

465 u 10.4

4836 # 4800 N (2 s.f.)

8.83

4800 4100

W

mg

465 u 8.83

4105.95 # 4100 N (2 s.f.)

0.623

290

W

mg

465 u 10.4

293.88 # 290 N (2 s.f.)

Robert Lee Chin

5

HSC Physics Module 1: Space Summary Define gravitational potential energy as the work done to move an object from a mm very large distance away to a point in a gravitational field, E p G 1 2 r The gravitational potential energy of an object at some point within a gravitational field is defined as: the work done to moving the object from an infinite distance to a point in a gravitational field x

At an infinite distance away from a gravitational field, the gravitational potential energy is zero. As an object moves towards the source of the field i.e. the centre of a planet, gravitational potential energy becomes negative. Although this value is negative, it represents a change in Ep i.e. work done. Work is therefore required to push an object away from, as well as towards the Earth. Mathematically:

m1m2 where, r

Ep

G

Ep

grav. potential energy

G universal grav. constant 6.67 u 10 -11 m 3 kg 1s 2 m1 an object moving within the field m2 r

Earth

mass of planet radius of planet (m 2 )

r

+

Ep d

-

Robert Lee Chin

6

HSC Physics Module 1: Space Summary

Space: 2. A successful rocket launch x

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using:

vx

2

v vy

2

ux

u  at 2

'x

2

u y  2a y 'y ux t

1 a yt 2 2 A projectile is any object that is launched, and then moves only under the influence of gravity. Examples are a ball that is struck, a bullet shell once it is fired. A rocket or missile is NOT a projectile. 'y

uyt 

The trajectory of a projectile can be analysed as two separate motions: Horizontal (x-axis) motion, representing constant velocity Vertical (y-axis) motion, representing constant acceleration at “g”, downwards

Horizontal velocity, Vx

Δy Uy

U

Vertical velocity, Vy

θ Angle of launch Ux

Δx= range = total horizontal distance

Ux

Equations for Projectile motion: 1)

Resolve initial launch velocity into vertical and horizontal component

U uy

u . sin T & u x

Uy

u . cos T

θ Ux

Robert Lee Chin

7

HSC Physics Module 1: Space Summary 2)

Horizontal motion is constant, so all you need is

3)

Vertical motion is constant acceleration at “g”

To find vertical velocity, use

vy

u y  a yt

To find vertical displacement, use

'y

u yt 

(a y

vx

u x and 'x

u xt

9.8ms-2 )

1 2 a yt 2

(a y

9.8ms-2 )

Points to remember: -the motion is symmetrical, so the elapsed time is half the full flight -vertical displacement, 'y has a negative sign -at its peak height, v y

0

-Maximum range is achieved with a launch angle of 45˚, i.e. T -horizontal velocity is constant i.e. v x

45q

ux

-Use v y

u y  a y t to find “t” at max height (when vy=0) or find vy at a given time

-Use 'y

u yt 

1 a y t 2 to find Δy at a given time, or find the time to fall through a given 2 height (if uy=0)

Example Problems: 1) A projectile is fired with a velocity of 50m/s at an angle of 30° to the horizontal. Determine the range of the projectile. u y 50 sin 30q u x 50 cos 30q a y 9.8ms 2 Consider the half - flight, v y t1 2

vy  uy ay

0  50 sin 30q  9.8

?t

2(50 sin 30q) 9.8

'x

u xt

50 cos 30q u

0ms 1 50 sin 30q 9.8

2(50 sin 30q) 9.8

220.924847... # 220m (3 s.f.)

2) Military bombs must be dropped from an altitude of 15 000m when the plane is flying level at 300ms-1.

Robert Lee Chin

8

HSC Physics Module 1: Space Summary a) How far in front must the bombs be released? 'y

15000m u x

300ms 1 u y

0ms 1

1 ayt 2 2 1  15000 0t  ( 9.8)t 2 2 15000 t 55.238333... # 55.3s (3 s . f .) 4.9 'y

u yt 

'x

u xt

300(55.3) 16590m

b) How fast will they be going (in magnitude) when they hit the ground? t

ux

vy

u y  ayt

x

0ms 1 a y

9.8ms 2

300ms 1

vx v

300ms 1 u y

55.3s u x

2

vx  v y

0  9.8(55.3) 2

541.94ms 1

(300) 2  ( 541.94) 2

619.43439...ms 1 # 6.19 u 10 2 ms 1 (3 s.f.)

Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components.

The trajectory of a projectile (ignoring air resistance) is parabolic, with constant downward acceleration at “g”. The trajectory can be analysed by considering its horizontal and vertical components at particular instances during the flight. The horizontal motion of the projectile is a constant velocity. Its vertical motion is changing all the time due to gravity, which causes the projectile to accelerate at 9.8 m s-2 downwards. x

Describe Galileo’s analysis of projectile motion

Notice that none of the equations used for projectile motion ever use the mass of the projectile. All objects, regardless of their mass, accelerate with gravity at the same rate. Galileo (1564-1642) performed an experiment to verify this. He dropped objects of the same size and shape, but with different mass, from the leaning tower in Pisa. He found that regardless of mass, all objects hit the ground at the same time. He also rolled cannon balls down an incline (there was friction) and was able to see 2 motions: constant horizontal velocity and constant vertical acceleration.

Robert Lee Chin

9

HSC Physics Module 1: Space Summary x

Perform a first-hand investigation, gather secondary information and analyse data to calculate initial and final velocity, maximum height reached, range, time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis

Experiment: Projectile launcher Aim: To construct a working model of a projectile launcher and to launch it at a target to gather data for calculating various aspects of the projectile’s flight including force, velocity and range. Equipment: - projectile launcher - steel ball bearings (small enough to fit in projectile launcher barrel) - scientific scales - digital video camera - target (circle of diameter 5 % of range) Note: The experiment requires the assistance of another person during the launching of the projectile. Procedure: 1) Weigh and record the mass of the steel ball with a scientific scale.

2)

Set up equipment as shown below:

Range Target

3)

Place steel ball in barrel of projectile launcher. Pull back end of solid plunger rod at least 20cm until a strong tension force can be felt. Get a friend to help hold down the back end of the launcher base.

Robert Lee Chin

10

HSC Physics Module 1: Space Summary 4)

Ask an assistant to set up the video camera ready to record the projectile’s flight when launched.

5)

Aim the launcher at the target and adjust the angle of launch according to the range of the target.

6)

Release the end of the rod and watch the ball bearing go!

7)

Repeat steps 1-6 as many times as possible until accuracy is improved. Once the desired amount of stretch and angle of launch is determined, keep these variables controlled (constant) in each trial to provide reliable results.

8)

Measure the displacement of the projectile in the launcher (this will depend on how far back the plunger rod is pulled) and the angle of launch for each trial.

9)

Use gathered data to calculate/determine values for: -horizontal and vertical components of the initial velocity -the resultant initial velocity -average acceleration & net force of projectile while in launcher. -range if launched from 50m cliff -velocity one quarter through its flight

Robert Lee Chin

11

HSC Physics Module 1: Space Summary Results:

Gathered Data Trial No.

Time of flight t, (s)

Range 'x , (m)

Hit on target/distance off from target (m)

1

0.70

4.0

Off target- 0.09

2

0.67

4.0

Off target - 0.11

3

0.78

4.0

Off target - 0.15

Average

0.72

4.0

n/a

Displacement of projectile in launcher, s = 0.284 m Angle of launch, T = 31˚ Mass of projectile, m = 0.015 kg Discussion: Initial horizontal velocity, ux = 5.6 ms-1 Initial vertical velocity, uy = 3.4 ms-1 Initial resultant velocity, v = 6.6 ms-1, 031˚ above horizontal Average acceleration of projectile while in launcher, aav = 76.7 ms-2 Net Force on projectile while in launcher, Fnet = 1.15 N Range if launched from 50 m cliff = 20.1 m Velocity one quarter through time of flight = 5.8 ms-1, 016˚ above horizontal

Robert Lee Chin

12

HSC Physics Module 1: Space Summary Quantity

Data

Initial horizontal velocity (ux)

'x t av

4.0ms 0.72 s

ux

'x t

Initial Vertical Velocity (uy)

ux

5.6ms 1

uy

u x tan T

Resultant initial velocity (v)

T

Calculations (all answers to 2 s.f.)

5.6ms 1

3.4ms 1

5.6 u tan 31q

31q

uy

3.4ms 1

v

ux  u y

ux

5.6ms 1

?v

6.6ms 1 ,031q above horizontal

a av

v2 2s

Fnet

mv 2 2s

T

31q

Average acceleration, (aav)

v s

6.6ms 1 0 . 284 m

Net Force, (Fnet)

m

0.015kg

v s

6.6ms 1 0.284m

Range if launched from 50m cliff

4.0 0.72

'y

50m 1

uy

3.4ms

ay

9.8ms  2

ux

5.6ms 1

2

2

5.6 2  3.4 2

6.6 2 2 u 0.284

6.6ms 1

77 ms 2

0.015 u 6.6 2 2 u 0.284

1.2 N

1 u yt  a yt 2 2 1  50 3.4t  u 9.8t 2 ? 4.9t 2  3.4t  50 0 2 Equation is in the form ax2+bx+c=0. Using the quad. formula, t is given by +ve value of x: 'y

 b r b 2  4ac   3.4 r 2a ? t 3.6 s 'x u x t 5.6 u 3.6 20.1m x

Velocity one quarter of time through flight

vx

5.6ms 1

t av 4 0.72 0.18s 4 u y 3.4ms 1

vy

Robert Lee Chin

T

2

vx  v y § vy tan 1 ¨¨ © vx

2 u 4.9

3.6  2.9

3.4  9.8 u 0.18 1.6ms 1

u y  ayt

t

v

 3.4 2  4 u 4.9 u 50

2

· ¸¸ ¹

5.6 2  1.6 § 1.6 · tan 1 ¨ ¸ © 5.6 ¹

2

5.82ms 1

016q ? v

5.8ms 1 ,016q above horizontal

13

HSC Physics Module 1: Space Summary x

Explain the concept of escape velocity in terms of the: -gravitational constant -mass and radius of the planet

Escape velocity is defined as: the launch velocity required for a projectile to escape from the Earth’s gravitational field. Mathematically, Escape velocity, where,

Ve

G

2GmE rE universal grav. constant

6.67 u10 -11 m 3 kg 1s 2

The escape velocity is therefore proportional to the gravitational constant, G and mass of the planet i.e. the greater the mass of the planet, the higher the escape velocity .It is inversely proportional to the radius of the planet i.e. the larger the planet, the lower the escape velocity. Note: The escape velocity of a planet is the same for all objects, independent of their mass. The escape velocity on earth is approximately 11.2km/s. x

Outline Newton’s concept of escape velocity

Sir Isaac Newton, the scientist who developed the mathematics behind projectile motion, created the following thought experiment: a person climbed a very high mountain and launched a cannon ball from the peak. The cannon ball follows a parabolic trajectory before hitting Earth. If the cannon ball were to be launched with increasing velocities, it would travel around the Earth because, as it falls, the curvature of the Earth curves away from it. Thus, the projectile would move in a circular orbit at a fixed height above the Earth’s surface. If the cannon ball were to be launched even faster than this, its orbit would changes from being a circle to an ellipse. Faster still and it would follow a parabolic or hyperbolic path away from the Earth, escaping it entirely.

Earth

Robert Lee Chin

14

HSC Physics Module 1: Space Summary x

Identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch

The term “g-force” refers to a mass’s apparent rate as a ratio of its true weight. True weight refers to the weight as mass experiences when stationary (or in relative motion) on the ground e.g. sitting in a vehicle moving horizontally across a flat surface at constant velocity, standing still at ground level. True weight cannot be felt because it is the force due to gravity; gravity being a force-at-a-distance. Apparent weight is the weight a mass experiences due to contact forces or an acceleration (other than gravity). These forces are also known as inertial forces because they arise from the body’s inertia or resistance to having its momentum changed. g - force

apparent weight true weight

Humans and g-force: -the maximum g-force a properly supported human can withstand is 20g (using fiberglass seating in spacecraft) -most people experience tunnel vision and lose colour perception at 4g G-force experienced by astronaut during a launch into low orbit:

4

g-force

1st stage jettisoned

3

2nd stage jettisoned

2 1

(decreasing mass) 1st stage (decreasing mass)

Ignition of 2nd stage

time

Robert Lee Chin

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HSC Physics Module 1: Space Summary Comparison of forces of a launch to a roller coaster ride: g-force Period during rocket launch Period during roller coaster ride >1g Rockets burning Accelerating upwards 1g At point of lift off on horizontal surfaces and when it reaches peak