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# SPACE Section 2 - PowerPoint PPT Presentation

SPACE Section 2. Projectile Motion. A projectile is an object that is projected (that means thrown, dropped or launched) into the air, but not propelled as is a rocket.

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### SPACE Section 2

• A projectile is an object that is projected (that means thrown, dropped or launched) into the air, but not propelled as is a rocket.

• This includes a ball being thrown, a football being kicked, a golf ball being struck, a bullet being fired, or cargo being dropped from a plane.

• Basically Projectile = Unpowered

• Aristotle:

• Everyobject has a certain amount of impetus and thenwhenthisruns out, the objectfalls to Earth.

• Galileo:

• Studied projectile motion and came to realiseit has a parabolic nature.

What is the nature of this curved path?

How to test this out? Things fall way too fast!

• Galileo’sinclined plane:

Problem 1: Things fall too fast.

Soln: Inclined planes make things fall a lot slower.

Problem 2: How to accurately time?

Soln: Being a musician GG realised that the ear is very good at keeping track of time. Thus he designed his inclined plane to have movable bells that rung everytime an object went down the plane. He adjusted these backwards and forwards until he got a perfectly even beat as the object travelled down. This allowed him to work out the distance of the object in equal time intervals.

1

4

9

16

Galileo's genius:

Distance proportional to time squared

in a sense: y = x2 which is...A PARABOLA!

Galileo was able to use MATHS to prove that the nature of any projectile is a parabola.

Galileo showed that horizontal motion is completely independent of vertical motion.

Looking at the horizontal motion we can see the

ux = vx

However vertical motion has a square relationship

Galileo showed that horizontal motion is completely independent of vertical motion.

Looking at the horizontal motion we can see the

ux = vx

However vertical motion has a square relationship

Galileo's analysis of Projectile Motion Summary

· Inclined plane

· Distance ∝to time2

· Horizontal motion independent of vertical motion

· Projectiles follow a parabolic path

A launched projectile, in the absence of any air resistance, will follow a parabolic path until it strikes the Earth. This path is called the trajectory of a projectile.

• To analyse the trajectory of a projectile separate it into horizontal and vertical motion (NOTE: We ignore air resistance).

• The horizontal motion, is not subject to any forces, and therefore experiences no acceleration.

• The vertical motion, is subject to the vertical weight force, and therefore experiences the downwards acceleration due to gravity. (Recall that there is only one force acting – the vertical weight.)

1. A rifle with a muzzle velocity (the speed the bullet comes out of the barrel) of 450 ms-1 is fired level at the horizon. Determine:

how fast the bullet is travelling 0.3 seconds after firing

how far it has travelled horizontally in that time.

ANS a) Our first formula tells us that vx = ux, that is, the final velocity equals the initial velocity over any time period. In other words, the horizontal velocity is the same all the way through the motion. Therefore, the velocity after 0.3 s is still 450 ms-1 (horizontally).

ANS b)

x = uxt

= 450 x 0.3

= 135 m

That is, after 0.3 s the bullet has travelled 135 m.

2. An air hockey puck is pushed so that it glides along its table at 0.15 ms-1. If the table is 1.2 m long, determine:

how long the puck takes to travel the length of the table

its velocity when it gets there

ANS a)

ux= 0.15 ms-1, x = 1.2 m, t = ?

x = u x t

1.2 = 0.15 t

t = 8.0 s

ANS b)

vx= ux = 0.15 ms-1

3. A stone is thrown horizontally at 8.0 ms-1. If it takes 0.5 s to fall to the ground, how far horizontally will it have travelled in this time?

ANS

ux= 8.0 ms-1, t = 0.5 s, x = ?

x = uxt

= 8.0 x 0.5

= 4.0 ms-1 That is, the trajectory of the stone will have a range of 4.0 m.

• In this regard it is much like any object thrown straight up. If thrown up from ground level, an object will rise up to a peak height, stop momentarily in the air, then return to the ground, speeding up as it does so.

• The second half of the motion is symmetrical with the first part, so that the time taken for the fall will equal the time taken for the rise. Also, the speed with which the object strikes the ground will equal the speed with which it was launched (only the direction will be down instead of up).

An arrow is fired directly upwards with a velocity of 55 ms-1. Assume that it is fired from ground level and that there is no air resistance.

a) How fast is the arrow moving when it returns to the ground?

b) What is the time of flight of the arrow?

Solution:

• By the symmetry of the motion, you can say that the arrow will have a velocity of 55 ms-1 down.

• A useful strategy to solve this problem is to focus on the arrow’s rise up to its peak height. Assume that ‘up’ is the positive direction. You can now say that:

uy = 55 ms-1, vy = 0 ms-1, ay = -9.8 ms-2, t = ?

Theequationto use is:

v = u + at

0 = 55 + (–9.8)t

t = 5.6 s

That is, the arrow will take 5.6 s to rise to its peak height. By symmetry, it must take just as long to fall back, so:

trip time = 2 x5.6 = 11.2 s.

The mathematical expressions you need to use therefore are:

• the horizontal component, ux = u cos 

• the vertical, uy= u sin , where is measured from horizontal.

Putting it all together

• In order to find the velocity of the projectile at other times during its flight you need to separately calculate the velocity in the x direction and the y direction, and then add them together using two-dimensional vector addition.

• Resolve initial velocity u into components ux and uy.

• Note that the horizontal velocity is uniform and doesn’t change (vx = ux).

• Consider the vertical motion and calculate the velocity vy after the specified time.

• You must now add vx and vy together as shown in the following diagram:

• A tennis ball is struck, this time at 5.0 ms-1, 55° above horizontal. What is the velocity of the tennis ball 1.2 s after being struck?

a) The components of the initial velocity are:

The horizontal component, ux = u cos = 5.0 cos 55 = 2.87 ms-1

Theverticalcomponent, uy = u sin = 5.0 sin 55 = 4.10 ms-1

b) vx = ux = 2.87 ms-1

c) In analysing the vertical motion weshalltake ‘up’ to be the positive direction.

vy = uy + at=4.10 + (-9.8 x 1.2) = -7.66 ms-1= 7.66 ms-1down

d) The final step is toaddvxandvytogether as shown in thediagram.

That is, after 1.2 s the velocity of the tennis ball is 8.2 ms-1 at 69° below horizontal.

Maximum height and trip time

• The strategy for working out these quantities is to consider just the vertical motion up to the peak. At the peak, the projectile has stopped moving in the vertical direction so that you can say that vy = 0 in this portion of the motion. The method then follows these steps.

a) Resolve the vertical component uy of the initial velocity u.

b) Consider the vertical motion up to the peak.

c) Note that vy = 0 for this portion of the motion.

d) Select an acceleration equation that enables calculation of unknowns from the data. then either calculate y, which is the maximum height.

or

calculate t, which is the time for the projectile to rise up to the peak.

e) Double this time to find the trip time. (Making use of the symmetry of the motion here, because it takes as long to fall as it does to rise up to the peak.)

• A tennis player plays a half volley off the ground, so that the ball leaves the racquet with a velocity of 7.2 ms-1 at 36° above horizontal. Calculate the maximum height achieved by the ball, and the time it takes to bounce for the first time (that is, the trip time).

The first step is to calculate uy:

uy = 7.2 sin 36° = 4.2 ms-1

uy = 4.2 ms-1, vy = 0 ms-1, ay = -9.8 ms-2, y = ?

Thecalculationneededtofindthemaximumheightis:

vy2 = uy2 + 2ayy

0 = 4.22 + (-9.8)y

y = 0.9 m

Tocalculatethetrip time youneedtofindthe time toreachthepeak.

uy = 4.2 ms-1, vy = 0 ms-1, ay = -9.8 ms-1, y = 0.9 m, t = ?

A suitablecalculationtofindthe time tothepeakis:

v = u + at

0 = 4.2 + (-9.8)t

t = 0.43s

and hence, trip time = 2t = 2 0.43 = 0.86 s

• Your final strategy is concerned with calculating the maximum horizontal displacement of a projectile, that is, the range of the trajectory. In order to make this calculation you will need to follow these steps:

a) Resolve initial velocity u into components uy and ux.

b) Analyse the vertical motion to find the trip time as shown above.

c) Consider only the horizontal motion and calculate the range using,

x = uxt

• Back to the tennis shot played in the earlier sample problem. The ball was struck from ground level at 7.2 ms-1 at 36° above horizontal. What will be the range of its trajectory?

The first step, as usual, is to resolve the initial velocity into components.

ux = 7.2 cos 36 = 5.8 ms-1

uy = 7.2 sin 36= 4.2 ms-1

Normallyyouwouldhavetocalculatethetrip time byanalysingthe vertical motion, butweknowthatthetrip time is 0.86s.

The final stepthenistoanalysethe horizontal motiontofindthemaximumdisplacement.

ux = 5.8 ms-1, trip time t = 0.86s, x = ?

The required calculation is:

x = uxt

= 5.8 X 0.86

= 5.0m

Newton’s concept of escape velocity

Isaac Newton was the first to write about the possibility of an artificial satellite of the Earth.

He imagined a cannon ball fired from a very tall mountain.

Such a projectile could follow a trajectory as shown, eventually hitting the Earth.

!!

boom

trajectory

Newton’s concept of escape velocity

• Fired at a greater velocity, the projectile travels further…

• The projectile increases its speed as it falls due to the conversion of gravitational potential energy to kinetic energy

• The ball reaches its greatest speed as it hits the ground.

Newton’s concept of escape velocity

• Fired at a still greater velocity, the projectile follows a trajectory closely matching the curvature of the Earth, so that it never hits the ground.

• If the orbit is elliptical, the projectile increases its speed as it gets closer to the Earth.

• The force of gravity results in the the projectile being in a continuous state of free fall…

• There is a net force, gravity, towards the centre of the Earth.

This force is the centripetal force

Newton’s concept of escape velocity

Escape velocity

Fired with a sufficiently large velocity, the projectile would never return to Earth.

The minimum velocity required to achieve this result is called the escape velocity.

The escape velocity from the Earth’s surface is . . .~ 11.2 km s-1

Newton’s concept of escape velocity

Escape velocity

A satellite given a velocity equal to or greater than the escape velocity never returns because its initial kinetic energy exceeds the change in potential energy as it increases its altitude (distance from the central body)

Escape velocity is the minimum velocity that must be imparted to an object at a specific location to cause it to completely escape from the gravitational pull of the planet

A projectile with a large velocity may travel so fast that it escapes the Earth’s gravitational influence.

Newton’s concept of escape velocity

Not on sheet

Learn it!

Newton’s concept of escape velocity

For a rocket at rest

• The downward force due to gravity…

is equal to

• the upward reaction force of the Earth against the rocket.

• The two forces are in equilibrium.

• The net force on the rocket is zero.

• As the rocket lifts off

the weight of the rocket (W)

is less than

the force produced by the thrust of the engines (FT)

• The two forces are not in equilibrium.

• Because the net force on the rocket is upward…

the rocket accelerates in upward direction (Newton’s second law in action)

aR

W = mg

FT> mg

Fweight = mg

Fthrust

Forces on an Astronaut During Launch

• As a rocket takes off, the thrust produced by the rocket engines produces a force that exceeds the weight of the vehicle

• The resulting net upward force causes the rocket to accelerate away from the Earth’s surface

• The acceleration increases as the mass decreases as fuel is burned [F=ma] because the mass of the rocket is rapidly decreasing.

Atlantis blasts off - September 10th 2006

During Launch

Freaction

• The astronaut experiences two forces

• A gravitational force downward

• A reaction force upward

• These are equal in magnitude and opposite in direction - there is zero net force on the astronaut

• The astronaut is said to be experiencing a force of “1G”

Rocket at Rest

• The astronaut experiences two forces

• A downward gravitational force

• - which remains constant

• An upward reaction force

• - which exceeds that of gravity

• The sum of these two forces (the resultant) produces a net upward force.

• If the rocket is accelerating upward at 9.8 m s–2, the astronaut experiences a reaction force of “2G”

Freaction

W=mg

W=mg

• If the thrust produced by the engines remains constant…

• As the mass of the rocket decreases due to the fuel being expelled…

• - the acceleration of the rocket, and hence the astronaut in the rocket, increases

• Hence the upward reaction force on the astronaut increases…

• - reaching a typical maximum during a launch of 3G (e.g. the space shuttle)

• As the rocket mass decreases, the engines may be throttled back to avoid excessive accelerations which could damage the rocket

• The reaction force that the astronaut experiences is often called a “g-force”.

• Once the spacecraft is orbit, there is no reaction force - gravity is the only force acting on the astronaut - a condition sometimes called “zero g”

Freaction

As the rocket accelerates upwards…

W=mg

• The term ‘g force’ is used to express a person’s apparent weight as a multiple of their normal true weight (that is, weight when standing on the surface of the Earth).

A person with mass, m, who is located at or near the surface of the Earth will always have some weight W=mg.

When a person stands on a scale, the reading (the number of kilograms or newtons) on the scale is actually the Normal Force that the scale exerts back towards the person to support the person's mass/weight.

This Normal Force is what we call Apparent Weight. However when the person and the scale experience acceleration, things get complicated.

Consider a person travelling in an elevator whilst standing on measuring scales.

If the acceleration of the elevator is zero, then there are two possible scenarios; the elevator can be at rest (stationary, zero velocity) or moving with a constant speed (no acceleration if velocity does not change).

In this case, the action and reaction force pair between the person and the scale is just the weight.

The person pushes down on the scale with a force of -W=-mg (negative direction) and the scale pushes back up against the man with a Normal Force in the positive direction.

Because the reading on the scale is the magnitude of the normal force, the scale will read the true weight when the elevator is NOT accelerating.

If the elevator is going up, there is an acceleration on the person and the scales.

The inertia of the person would prefer to stay stationary, so the elevator floor and scale must push up on the person to accelerate him upward along with the elevator. (The person doesn't sink into the floor when the elevator accelerates up. The elevator and the scale and the person all move together.)

The scale therefore has to push upward with extra force on the person to accelerate the person's mass upward. This results in a greater contact force between the scale and the person.

Therefore the Normal Force is larger, so the reading on the scale is a number that is GREATER than the true weight.

The inertia of the person would prefer to stay at rest, so the elevator floor and scale effectively drop out a little bit from underneath the person as the elevator accelerates down.

The person doesn't float upward here also, because again the elevator and the person move together, but the contact force between the person and the scale is reduced.

The scale therefore has to push upward with less force on the person to support the person's weight. Therefore the Normal Force is smaller, so the reading on the scale is a number that is LESS than the true weight.

If the elevator is going down, there is a deceleration on the person and the scales.

If the elevator cable were to break, the whole elevator-scale-person system would all begin to accelerate downward due to the force of gravity.

All objects in freefall accelerate downward with the same magnitude (acceleration due to gravity, g). The scale and the person are freefalling together, so there is NO contact force (Normal Force) between the scale and the person.

(When they are both falling together, there is no way that the scale can support any of the person's weight.)

As a rocket ascends from Earth’s surface

FT

The G-force experienced by the astronaut when the rocket is accelerating vertically is…

W=mg

a = Fnet/m

Fnet = FT – W

The Earth’s gravitational acceleration is almost constant over the distances involved in LEO.

LEO orbits range up to about 500 km altitude.

• Identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch. (4M)

The force applied to the astronaut by the seat is the sum of the reaction force against gravity plus the additional force needed to produce the acceleration. To the astronaut this force feels just like the reaction force of the seat when the rocket is stationary - only greater. The force experienced feels like that of gravity - the “g-force” - only larger, so the force is called the “g-force”. The astronaut feels him/herself being “pushed into the seat” however, rather than being aware of the net upward force. [Compare this to the effect felt by a person in a car when the car accelerates rapidly.] Therefore, “g-force” is used because it quantitatively relates the effect of acceleration on an astronaut during a rocket launch to the familiar effect, called 1-g, that is produced by the force of gravity when the person is at rest on Earth’s surface.

The seat supporting the astronaut provides an upward force on the astronaut that counteracts the force of gravity, even when the astronaut is at rest on the launch pad. During launch a net upward force must act on the astronaut to increase the astronaut’s upward speed. This force is also applied through the seat.

• G-forces encountered during the launch of a Saturn V rocket were significantly greater than those experienced during a space shuttle launch.

• When the rocket accelerates vertically upward at 9.8 m s–2, the astronaut experiences a reaction force of “2G”.

Variation in ‘g’ forces during launch

• The g-forces experienced by the astronauts during the second and third stage burns are less than those experienced during the first-stage burn because

• the trajectory of the rocket is curving over and becoming closer to being parallel to the Earth’s surface in preparation for the insertion into orbit and so the reaction force against gravity is reduced

• the engines in the 2nd and 3rd stages have less thrust than the 1st stage engines because the mass they have to act on is much less

Variation in ‘g’ forces during launch

• G-forces experienced by an astronaut during the third stage engine burn are less than 1G because at this region of the trajectory, the rocket is travelling close to parallel to the Earth’s surface.

• The G-forces experienced are almost entirely due to the increasing speed of the rocket.

• The Earth’s gravity provides a net centripetal force causing the rocket to travel in a near circular orbit - the rocket at this stage is in free-fall, but increasing the component of its speed parallel to the Earth’s surface.

As the rocket enters Earth orbit, the trajectory becomes parallel to the Earth’s surface…

The rocket engines shut down and the only force acting on the rocket, and the astronaut is gravity.

This is referred to as “weightlessness”

The rocket motion is in “free fall”

W=mg

Problem

Use the information in the following table to calculate the initial acceleration of the space shuttle from the launch pad and use this to predict the g-force experienced by the astronaut. The shuttle has three main engines and two solid rocket booster engines.

• Steps involved

• calculate the total thrust produced by the five engines – this is upwards so call it positive

• determine the weight force – this is downwards, so call it negative

• calculate the net force which is the sum of the forces above, taking direction into account

• use Newton’s second law to determine the acceleration

Total mass m = 2 x 106 kg

SRB thrust = 2 x 15 x 106 N

Main engine thrust = 3 x 1.75 x 106 N

Total engine thrust = 3.525 x 107 N (adding)

Weight of rocket: W = mg

mg = 2 x 106 x 9.8 = 1.96 x 107 N

Net force on rocket = (3.525 - 1.96) x 107 N

Fnet= 1.565 x 107 N

a = F/m

= 1.565 x 107 / 2 x 106

= 7.825 m s–2

G force = (9.8 + a) / 9.8 = 1.8 g

• Newton’s third law of motion

• When a force acts on an object, an equal and opposite force acts on the object producing that force

• Or, specifically for a rocket

• The force acting on the gases produced by the rocket engine, propelling those gases out of the rocket engine, results in an equal and opposite force on the rocket, propelling it forward

• Since the magnitude of the force propelling the gases backwards equals the magnitude of the force on the rocket in the other direction, and the duration of the force on each is the same, the momentum change of the gases must equal the momentum change of the rocket in the other direction.

• Momentum is conserved in the rocket propulsion process

• Hot gases, having mass, m2, are ejected with a high speed, v2, from a rocket engine, causing the rocket of mass, M1, to receive an impulse driving it in the opposite direction at a more moderate speed, v1.

• As m2, the propellant, leaves at speed v2 with respect the rocket, the remaining rocket mass m receives a boost in speed such that….

 fuel momentum =  rocket momentum

m2v2 = M1v1

Rocket Launch and Earth’s Motion

• To minimise the fuel required for a launch, rockets are launched from a point on the Earth’s surface that is close to the equator, and in the direction of the Earth’s rotation on its axis.

• The motion of the Earth imparts an additional velocity equal to 0.45 km/s (1700 km/h) at the equator.

• At the Kennedy Space Center, this drops to about 0.40 km/s, because it is not at the equator.

• Given that a satellite must reach an orbital velocity of about 7 km/s, the effect of the Earth’s rotation is significant.

Rocket Launch and Earth’s Motion

• The Earth travels around the Sun at a speed of 29 km s-1

• This motion can be used to advantage when choosing when to fire rocket engines to send a satellite from Earth orbit to other planets.

• To leave Earth orbit, a satellite must almost reach the escape velocity at the point from which it is leaving its orbit around the Earth (~11 km s-1).

• Rocket engines are fired when the satellite is in a position in the orbit such that it is travelling in the same direction as the Earth around the Sun.

satellite motion

Rocket Launch and Earth’s Motion

When interplanetary flights are being carried out, the satellite is fired out of its Earth orbit in the direction that that Earth is moving around the Sun.

This takes advantage of the Earth’s orbital speed around the Sun, which is about 30 km/s.

The final velocity of the satellite relative to the Sun is the sum of…

vo, the orbital velocity of the Earth

va, the velocity due to axial rotation

vs, the satellite’s acquired velocity

Such a manoeuvre was used in getting the Mars satellites to Mars in 2003/4

Interplanetary satellites take advantage of the Earth’s orbital motion

Rocket Launch and Earth’s Motion

If a spacecraft is launched from a site near Earth's equator, it can take optimum advantage of the Earth's rotation about its axis.

Sitting on the launch pad near the equator, it is already moving at a speed of over 1650 km per hour.

This can be applied to the speed required to orbit the Earth (approximately 28,000 km per hour). Compared to a launch far from the equator, the equator-launched vehicle would need less propellant, or a given vehicle can launch a more massive spacecraft.

See animation (JAVA)

http://www2.jpl.nasa.gov/basics/bsf14-1.html

Rocket Launch and Earth’s Motion

This animation shows how a rocket is used to send a satellite on an interplanetary mission from Earth.

The trajectory and its timing takes advantage of both the Earth’s rotational motion on its axis and its orbital motion around the Sun.

The view is from above the Earth’s north pole so the rotation of the Earth is anticlockwise on its axis and the orbital motion of the Earth from this point of view is anti-clockwise around the Sun.

Rocket Launch and Earth’s Motion – Mission to Pluto

• Launched January 19, 2006 at 2.00 p.m. (Eastern Standard Time, Florida)

• Velocity after 3rd stage rocket burnout 15.63 km/s, making it the fastest spacecraft ever launched from Earth

• Slingshot effect will occur at Jupiter in February 2007 to increase its velocity to enable it to reach Pluto

• Pluto-Charon encounter July 2015

• The primary scientific objectives are to investigate the global geology and morphology of Pluto and Charon and to determine the atmospheric and surface composition of these objects

• The mission will also investigate other objects in the Kuiper belt

• The net force on a vehicle travelling at a constant speed in a circle is towards the centre of the circle

• The acceleration of the vehicle is thus towards the centre of the circle

force

• What causes centripetal force?

• The distance of a satellite in a circular orbit from the centre of the Earth is called the orbital radius

• The time it takes for the satellite to complete one revolution around the Earth, or one orbit, is called the orbital period

• The speed at which the satellite is travelling is called the orbital velocity

The International Space Station has a mass of 188 tonnes. It orbits the Earth at an altitude of 350 km and takes 92 minutes to complete each circular orbit. The radius of the Earth is 6400 km.

Calculate the centripetal force acting on the International Space Station.

v = 2r/T

= 2 x 3.142 x 6 750 000 / (92 x 60)

= 7680 m s–1

F = mv2/r

= 188 000 x 76802 / 6 750 000

= 188 000 x 8.738 N

= 1.64 x 106 N

At an altitude of 350 km the strength of the gravitational field is 8.738 N kg–1

Compare the centripetal force on the International Space Station in orbit with its weight on the Earth’s surface .

F = mv2/r

= 188 000 x 76802 / 6 750 000

= 188 000 x 8.738 N

= 1.64 x 106 N

On Earth

F = mg

= 188 000 x 9.81

= 1.84 x 106 N

A useful way of making a quantitative comparison is find the ratio (or percentage of one quantity to the other.

Conclusion

Thus in its orbit, the centripetal force* is 89% of the weight of the ISS on Earth

*The force is provided by gravity - the only force on the space craft

Low Earth orbit

• Altitude above 250km (or too much drag from atmosphere)

• Altitude below 1000km (or interference from Van Allen radiation belt)

• Generally orbit from Pole to Pole (N, S, N, S, ...)

• Orbital period as little as 90 minutes (1&1/2 hours)

• Used for observation (mapping, weather, military...)

• Limited life (slowed by atmosphere, spiral in, crash & burn!)

Geostationary orbit

• Stays “stationary” above one point on the Earth’s equator.

• Radius 42 168km (to match Earth’s rotation)

• Must be above the equator (or it will crash)

• main use communications (TV, GPS... receiver dishes pointing at it)

Geostationary vs Geosynchronous

• Both geostationary and geosynchronous orbits have periods of 23 hours 56 minutes

• A satellite in a geostationary orbit has its orbit directly above the equator

• The orbit of a satellite that is geosynchronous is not necessarily above the equator - but may be at an angle to the equator

• All geostationary satellites are geosynchronous - but the reverse is not true

• Geostationary satellites appear in a fixed position in the sky

• Geosynchronous satellites that are not geostationary appear to move along a north-south line in the sky, completing one cycle every 24 h

Kepler’s Law of Periods

• Johannes Kepler (1571-1630) searched for an underlying pattern of the motion of planets.

• Studying these motions, and using measurements taken by Tycho Brahe, Kepler deduced three laws of planetary motion.

• Kepler was the first person to reject the assumption that planets moved in perfect circular motion.

Kepler’s Law of Periods

A little revision: Orbital velocity

This is the velocity of a satellite as it moves around another body under the influence of the force of gravity.

Orbital velocity is constant in magnitude if the satellite travels in a circular orbit.

The direction of the orbital velocity is at a tangent to the orbit.

How is magnitude of the orbital velocity is related to the period (T) and the orbital radius (r)?

V = 2r/T

Orbital velocity

direction

Kepler’s Law of Periods

Orbital velocity

If the orbit is elliptical, the speed is least when the distance between the objects is the greatest and it increases as they get closer together.

The speed change is a consequence of the law of conservation of energy - as the potential energy decreases the kinetic energy increases correspondingly.

Minimum speed

Maximum speed

Kepler’s Law of Periods

First two laws…

• Satellites move in elliptical orbits with the central body at one focus of the ellipse

• A planet moving in its orbit sweeps out equal areas in equal times

[not explicitly in HSC course]

But the third law (the law of periods) is…

Kepler’s Law of Periods

• The square of a planet’s orbital period (T) is proportional to the cube of the mean distance (r) of the planet from the Sun

T2 = k r3

• Where k is a constant dependant on the mass of the central body (e.g. the Sun) and the universal gravitational constant

Kepler’s Law of Periods

The square of a planet’s orbital period is proportional to the cube of the mean distance of the planet from the Sun

T - orbital period (s)

G - universal gravitational constant

M - mass of central body

Saturn

Kepler’s Law of Periods

Calculate the ratio of the [radius3/period2] for the Earth, and use this to calculate the orbital period of Saturn, given its mean orbital radius 9.54 au.

REarth orbit 1.50 x 1011 m

TEarth365.25 days

G 6.67 x 10–11

M (mass of Sun)1.99 x 1030kg

1AU = 1.5 x 1011

Orbital Decay

• More than 90% of the molecules in the Earth’s atmosphere are below the top of Mount Everest (pictured from ISS).

• There are still a few molecules of the Earth’s atmosphere extending to an altitude of about 500 km.

• Satellites in orbits between 150 km and 500 km altitude are described as being in low Earth orbit (LEO).

Orbital Decay

• Satellites in LEO encounter frictional drag as they orbit at these altitudes.

• The effect of this drag is to cause the satellite to lose energy resulting in the satellite’s moving closer to the Earth.

• If the orbit is to be maintained, booster engines must be fired periodically to increase the satellite’s altitude.

Re-Entry

• For a satellite in LEO, the kinetic energy is about ten times the potential energy and they are both very significant quantities of energy.

• To land safely, a spacecraft must reduce its speed by 90% as it approaches the Earth.

Re-Entry

• The speed reduction is accomplished through

• Retro-rocket firing (slows the vehicle by about 1%)

• Frictional drag in the atmosphere

Frictional drag through the Earth’s atmosphere converts the satellite’s energy to heat energy

Re-Entry

• For spacecraft intended to for return to Earth, dissipation of the heat energy generated by during re-entry is a major consideration in the spacecraft design and re-entry process.

• Key strategies employed to ensure the spacecraft does not burn up include the use of:

• heat resistant (high melting point) materials

• materials with very low thermal conductivity

• materials with a very low heat capacity

• ablation (burning off of material from the craft)

• heat radiation from the heated surface of the spacecraft

Re-Entry

• Retro-rockets slow the spacecraft slightly, causing its orbit to decay.

• The lower orbit results in much greater frictional drag, greatly slowing the spacecraft.

• The angle at which the spacecraft enters the atmosphere is critical

• too steep an angle results in a re-entry that is too fast and the craft would burn up

• Too shallow an angle causes the craft to bounce off the atmosphere and not make a successful re-entry

A weather satellite captured this image of the heating effect produced as a spacecraft reentered the Earth’s atmosphere

Re-Entry

ablative heat shield

Spacecraft - capsule

Re-Entry

• Protection of the shuttle during re-entry is achieved by insulating tiles made of silica and placed on the underside of the craft.

Re-Entry

• Too shallow an angle will cause the satellite to bounce off the atmosphere and re-enter space

• Too steep an angle will cause too great an increase in drag, causing the spacecraft to burn up in the atmosphere(Russian MIR re-entry pictured below)

Re-Entry

• There is thus an optimum angle at which a spacecraft returning to Earth must enter the atmosphere

• For safe re-entry, the angle at which the satellite enters the atmosphere must be between 5° and 7°

• Image - MIR re-entry (artist’s impression)