MuffinMan
Juno 15/4/08 :)
I've got one just finished...took me 5 hours
i'll post it tomorrow cause im too tired
tis 1am now
i'll post it tomorrow cause im too tired
tis 1am now
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Who needs Space Syllabus Summaries? (2 Viewers)
- Thread starter MuffinMan
- Start date
MuffinMan
Juno 15/4/08 :)
* define weight as the force on an object due to a gravitational field
* explain that a change in gravitational potential energy is related to work done
* define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field
Ep = -Gm1m2/r
Gravitational potential energy, Ep, is the energy of a mass due to its position within a gravitational field.
Gravitational potential energy = Ep = work done to move to the point = force required x distance moved (since work = Fr)
= (mg) x h = mgh
Hence in this case, Ep = mgh. We choose the ground as our defined starting point, Ep = 0. Note that since work must be done on the object to lift it, it acquires energy. Hence at this point x, Ep > 0.
On a larger, planetary scale, we need to rethink our approach. Due to the inverse square relationship in the Law of Universal Gravitation, the force of attraction between a planet and an object will drop to zero only at an infinite distance from the planet. For this reason, infinity is chosen as our level of zero potential energy.
Therefore Ep at infinity > Ep at point x
But Ep at infinity = 0
So that Ep at point x has negative value!
Using the same approach as earlier, the gravitational potential energy, Ep of an object at point, x, in a gravitational field is equal to the work done to move the object from zero energy level at infinity to point x.
It can be shown mathematically as:
Ep = - Gm1m2/r
where m1 = mass of planet (kg)
m2 = mass of object (kg)
r = distances separating masses (m)
* explain that a change in gravitational potential energy is related to work done
* define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field
Ep = -Gm1m2/r
Gravitational potential energy, Ep, is the energy of a mass due to its position within a gravitational field.
Gravitational potential energy = Ep = work done to move to the point = force required x distance moved (since work = Fr)
= (mg) x h = mgh
Hence in this case, Ep = mgh. We choose the ground as our defined starting point, Ep = 0. Note that since work must be done on the object to lift it, it acquires energy. Hence at this point x, Ep > 0.
On a larger, planetary scale, we need to rethink our approach. Due to the inverse square relationship in the Law of Universal Gravitation, the force of attraction between a planet and an object will drop to zero only at an infinite distance from the planet. For this reason, infinity is chosen as our level of zero potential energy.
Therefore Ep at infinity > Ep at point x
But Ep at infinity = 0
So that Ep at point x has negative value!
Using the same approach as earlier, the gravitational potential energy, Ep of an object at point, x, in a gravitational field is equal to the work done to move the object from zero energy level at infinity to point x.
It can be shown mathematically as:
Ep = - Gm1m2/r
where m1 = mass of planet (kg)
m2 = mass of object (kg)
r = distances separating masses (m)
MuffinMan
Juno 15/4/08 :)
Section 2
* describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components
* describe Galileo’s analysis of projectile motion
The trajectory of a projectile is the path that it follows during its flight. To understand and analyse this motion of a projectile can be regarded as two separate and independent motions superimposed upon each other. The first is a vertical motion, which is subject to acceleration due to gravity, and the second is horizontal motion, which experiences no acceleration. Because the two motions are perpendicular, and therefore independent, we can treat them separately and analyse them separately.
The Vertical Motion
When a ball is thrown directly up, it is accelerated due to gravity directly down. As a result it would rise up, slow to a halt in the air and then fall back to Earth. As it falls it would speed up until when back at its starting point, it is going as fast as when it was thrown. Furthermore, the time taken to fall from its peak height to the ground exactly equals the time taken to rise to the peak height.
The Horizontal Motion
If a ball is pushed horizontally, ideally, once it is under way, it experiences no acceleration at all in the direction of the motion. If no acceleration is experienced, the disk will travel with a uniform, unchanging velocity. Once free of the ground, there is no friction for a thrown ball, other than air resistance.
* describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components
* describe Galileo’s analysis of projectile motion
The trajectory of a projectile is the path that it follows during its flight. To understand and analyse this motion of a projectile can be regarded as two separate and independent motions superimposed upon each other. The first is a vertical motion, which is subject to acceleration due to gravity, and the second is horizontal motion, which experiences no acceleration. Because the two motions are perpendicular, and therefore independent, we can treat them separately and analyse them separately.
The Vertical Motion
When a ball is thrown directly up, it is accelerated due to gravity directly down. As a result it would rise up, slow to a halt in the air and then fall back to Earth. As it falls it would speed up until when back at its starting point, it is going as fast as when it was thrown. Furthermore, the time taken to fall from its peak height to the ground exactly equals the time taken to rise to the peak height.
The Horizontal Motion
If a ball is pushed horizontally, ideally, once it is under way, it experiences no acceleration at all in the direction of the motion. If no acceleration is experienced, the disk will travel with a uniform, unchanging velocity. Once free of the ground, there is no friction for a thrown ball, other than air resistance.
MuffinMan
Juno 15/4/08 :)
* explain the concept of escape velocity in terms of the:
– gravitational constant
– mass and radius of the planet
* outline Newton’s concept of escape velocity
Isaac Newton wrote that it should be possible to launch a projectile fast enough that it achieves an orbit around the Earth. His reasoning is as follows: if a stone was thrown from a tall tower, it would cover a considerable range before striking the ground. If it was thrown faster, it would travel faster before stopping. If it was thrown faster still, it would have an even greater range. If thrown fast enough then, as the stone falls, the Earth’s surface curves away so that the falling stone never actually lands on the ground and orbits the Earth.
If this specific velocity is exceeded slightly, then the orbit would follow an elliptical orbit around the Earth. If the specific velocity is exceeded further still, then the object will follow a parabolic or hyperbolic path. Escape velocity is the initial velocity required by a projectile to rise vertically and just escape the gravitational field of a planet is called escape velocity.
– gravitational constant
– mass and radius of the planet
* outline Newton’s concept of escape velocity
Isaac Newton wrote that it should be possible to launch a projectile fast enough that it achieves an orbit around the Earth. His reasoning is as follows: if a stone was thrown from a tall tower, it would cover a considerable range before striking the ground. If it was thrown faster, it would travel faster before stopping. If it was thrown faster still, it would have an even greater range. If thrown fast enough then, as the stone falls, the Earth’s surface curves away so that the falling stone never actually lands on the ground and orbits the Earth.
If this specific velocity is exceeded slightly, then the orbit would follow an elliptical orbit around the Earth. If the specific velocity is exceeded further still, then the object will follow a parabolic or hyperbolic path. Escape velocity is the initial velocity required by a projectile to rise vertically and just escape the gravitational field of a planet is called escape velocity.
MuffinMan
Juno 15/4/08 :)
* identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch
Your body is a mass lying somewhere within a gravitational field, and therefore experiences a true weight, W=mg. The sensation of weight that you feel, however, derives from your apparent weight, which is equal to the sum of the contact forces resisting your true weight.
The term ‘g force’ is used to express a person’s apparent weight as a multiple of his/her normal true weight.
Hence, g force = apparent weight/normal true weight.
Prior to lift off, a rocket has zero acceleration because of the balance that exists between the weight force and the reaction force plus thrust. The astronaut within is experiencing a one g load. The initial condition will not change until the building thrust exceeds the weight of the rocket, at which point the rocket will lift off.
Since the thrust now exceeds the weight, there is a net force upwards on the rocket, which begins to accelerate the rocket upwards. The g forces experienced by the rocket will now have a value greater than one. From this point, the mass of the rocket begins to decrease as fuel is consumed and, hence, the rate of acceleration and subsequent g force steadily climbs, reaching maximum values just before the rocket has exhausted its fuel.
A multi-stage rocket drops its spent stage away, momentarily experiencing zero g conditions as it coasts. The second stage rocket fires, and quickly exceeds the necessary thrust to exceed the effective weight at its altitude, and then starts to accelerate again. Thus the same process happens again.
Your body is a mass lying somewhere within a gravitational field, and therefore experiences a true weight, W=mg. The sensation of weight that you feel, however, derives from your apparent weight, which is equal to the sum of the contact forces resisting your true weight.
The term ‘g force’ is used to express a person’s apparent weight as a multiple of his/her normal true weight.
Hence, g force = apparent weight/normal true weight.
Prior to lift off, a rocket has zero acceleration because of the balance that exists between the weight force and the reaction force plus thrust. The astronaut within is experiencing a one g load. The initial condition will not change until the building thrust exceeds the weight of the rocket, at which point the rocket will lift off.
Since the thrust now exceeds the weight, there is a net force upwards on the rocket, which begins to accelerate the rocket upwards. The g forces experienced by the rocket will now have a value greater than one. From this point, the mass of the rocket begins to decrease as fuel is consumed and, hence, the rate of acceleration and subsequent g force steadily climbs, reaching maximum values just before the rocket has exhausted its fuel.
A multi-stage rocket drops its spent stage away, momentarily experiencing zero g conditions as it coasts. The second stage rocket fires, and quickly exceeds the necessary thrust to exceed the effective weight at its altitude, and then starts to accelerate again. Thus the same process happens again.
MuffinMan
Juno 15/4/08 :)
* discuss the effect of the Earth‘s orbital motion and its rotational motion on the launch of a rocket
Consider the Earth is revolving around the sun at approximately 107 000 kmh-1 relative to the sun. In addition, the Earth rotates once on its axis per day so that a point on the Equator has a rotational velocity of approximately 1700kmh-1 relative to the sun. Hence, the Earth itself can be exploited in a rocket launch to gain a boost in velocity.
Launching is done in the direction of the Earth’s motion, that is, launching towards the east. In this way, the rotational velocity of the launch site is relative to the sun will add to the orbital velocity achieved by the rocket relative to the sun.
In a similar way, engineers planning a rocket mission heading further into space can exploit the Earth’s revolution around the sun by planning a launch for a time of year when the direction of the Earth’s orbital velocity corresponds to the desired heading. The rocket is allowed to proceed around its orbit until the direction of its orbital velocity corresponds to Earth’s, and then its engines are fired to push it out of orbit and further into space. In this way, Earth’s orbital velocity relative to the sun adds to the rocket’s orbital velocity relative to the Earth, to produce a higher velocity achieved by the rocket relative to the Sun.
Consider the Earth is revolving around the sun at approximately 107 000 kmh-1 relative to the sun. In addition, the Earth rotates once on its axis per day so that a point on the Equator has a rotational velocity of approximately 1700kmh-1 relative to the sun. Hence, the Earth itself can be exploited in a rocket launch to gain a boost in velocity.
Launching is done in the direction of the Earth’s motion, that is, launching towards the east. In this way, the rotational velocity of the launch site is relative to the sun will add to the orbital velocity achieved by the rocket relative to the sun.
In a similar way, engineers planning a rocket mission heading further into space can exploit the Earth’s revolution around the sun by planning a launch for a time of year when the direction of the Earth’s orbital velocity corresponds to the desired heading. The rocket is allowed to proceed around its orbit until the direction of its orbital velocity corresponds to Earth’s, and then its engines are fired to push it out of orbit and further into space. In this way, Earth’s orbital velocity relative to the sun adds to the rocket’s orbital velocity relative to the Earth, to produce a higher velocity achieved by the rocket relative to the Sun.
MuffinMan
Juno 15/4/08 :)
nalyse the changing acceleration of a rocket during launch in terms of the:
– Law of Conservation of Momentum
– forces experienced by astronauts
The forward motion of the rocket can be understood by recalling the Law of Conservation of Momentum. The law states that during any interaction in a closed system the total momentum of a system remains unchanged. This means that the backward momentum of gases is equal in magnitude to the forward momentum of the rocket endowing the rocket with forward velocity. This is Newton’s third law of motion. The law states that for every force, there is an equal but opposite force, and this is the case here.
Second section answered in d.p. 2.2.5 (identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch)
*analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth
Uniform Circular Motion is circular motion with an uniform orbital speed. For example, a rock tied onto a string is whirled around in a horizontal plane. The object will continue in uniform motion in a straight line unless acted upon by a force. In the case of the rock, the force keeping it within a circular path is the tension in the string, and is always directed back towards the centre of the circle. Without this force, the rock will travel in a straight line.
The same is true of a spacecraft orbiting the Earth, or any object in circular motion – some force is needed to keep it there, and the force is directed back towards the centre of the circle. In the case of the spacecraft, it is the gravitational attraction between the Earth and the spacecraft that acts to maintain in that orbit.
The force required to maintain circular motion, known as centripetal force can be determined using the following equation
Fc = (mv2) / r
Fc = centripetal force (N)
m = mass of object in motion (kg)
v = instantaneous or orbital velocity of the mass (ms-1)
r = radius of circular motion (m)
* compare qualitatively low Earth and geo-stationary orbits
Geostationary satellites orbit the Earth over the equator with a period of 23 hrs, 56 min and 4 sec – one sidereal day. This is the time for Earth to rotate once on its axis. A geostationary satellite will occupy the same position above the Earth
A low Earth orbit is generally an orbit higher than approximately 250km in order to avoid atmospheric drag, and lower than approximately 1000km. The space shuttle utilises a low Earth orbit somewhere between 250km and 400km.
* define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods
Orbital velocity is the instantaneous speed (magnitude) in the direction indicated by an arrow (directional) drawn as a tangent to the point of interest on the orbital path.
Quantitative relationship: (where d = average distance between centres of the two masses; v = orbital velocity; M = larger mass; ms = smaller mass; G = gravitational constant; T = period of orbit)
– Law of Conservation of Momentum
– forces experienced by astronauts
The forward motion of the rocket can be understood by recalling the Law of Conservation of Momentum. The law states that during any interaction in a closed system the total momentum of a system remains unchanged. This means that the backward momentum of gases is equal in magnitude to the forward momentum of the rocket endowing the rocket with forward velocity. This is Newton’s third law of motion. The law states that for every force, there is an equal but opposite force, and this is the case here.
Second section answered in d.p. 2.2.5 (identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch)
*analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth
Uniform Circular Motion is circular motion with an uniform orbital speed. For example, a rock tied onto a string is whirled around in a horizontal plane. The object will continue in uniform motion in a straight line unless acted upon by a force. In the case of the rock, the force keeping it within a circular path is the tension in the string, and is always directed back towards the centre of the circle. Without this force, the rock will travel in a straight line.
The same is true of a spacecraft orbiting the Earth, or any object in circular motion – some force is needed to keep it there, and the force is directed back towards the centre of the circle. In the case of the spacecraft, it is the gravitational attraction between the Earth and the spacecraft that acts to maintain in that orbit.
The force required to maintain circular motion, known as centripetal force can be determined using the following equation
Fc = (mv2) / r
Fc = centripetal force (N)
m = mass of object in motion (kg)
v = instantaneous or orbital velocity of the mass (ms-1)
r = radius of circular motion (m)
* compare qualitatively low Earth and geo-stationary orbits
Geostationary satellites orbit the Earth over the equator with a period of 23 hrs, 56 min and 4 sec – one sidereal day. This is the time for Earth to rotate once on its axis. A geostationary satellite will occupy the same position above the Earth
A low Earth orbit is generally an orbit higher than approximately 250km in order to avoid atmospheric drag, and lower than approximately 1000km. The space shuttle utilises a low Earth orbit somewhere between 250km and 400km.
* define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods
Orbital velocity is the instantaneous speed (magnitude) in the direction indicated by an arrow (directional) drawn as a tangent to the point of interest on the orbital path.
Quantitative relationship: (where d = average distance between centres of the two masses; v = orbital velocity; M = larger mass; ms = smaller mass; G = gravitational constant; T = period of orbit)
MuffinMan
Juno 15/4/08 :)
yay! it is now small enough to post as an attatchment
yeh sorry about multiposting it was just too long and got me confuzzed
free to correct it tho...i did it late at night
yeh sorry about multiposting it was just too long and got me confuzzed
free to correct it tho...i did it late at night
can someone pls explain what galileo's analysis of projectile motion was... different books say diff things. some say that he proved different mass objects fall at the same rate --> what does that have to do with projectile motion as such?HSC_sUcKsSsS said:
physician
Some things never change.
Abtari said:
HSC sUcKsSsS... covered the whole concept quite well although he decided to integrate the 2 dot pts together...
anyways i'll give a decsription for that dot pt...
describe Galileo's analysis of projectile motion
It's always nice to give a general statement about a scinetist.. and how terrific their contribution has been...
"Our understanding of projectile motion owes a great debt to Galileo, who in his work entitled 'Dialogues Concerning Two New Scineces', presnted his classic analysis of such motion. " ok enough chit chat lets get to the good stuff....
basically what Galieloe did is that he argued that projectile motion was a compound motion made up of both horizontal and vertical components... 'which our friend Mr HSC has allready explained'. The horizontal motion had a steady speed in a fixed direction, while the vertical motion was one of downwards acceleartion. Using a geometric arguemnt, Galileo went on to show that the path of a projectile undergoing such motion was a parabola.
He aslo admitted to the fact that his assumptions and results were only approximations to the real world, and that due to resistence of the medium-for instance- a projectiles horizontal motion cannot be truly constant in speed.
more chit chat...
" Galileo became perhaps the first scinetist to demonstarte this modern scientific attitude. His approach was certainly very different from that of the ancinet Greek geometers, who were only interested in exact results"...
Hope that helps...
so his analysis was basically that projectile motion was parabolic, and composed of two components?
so do i ignore what it says in one book where Galileo's analysis is represented as:
"He stated his theory that all objects fall at the same rate, neglecting air resistance. he did this by experimenting with highly polished inclined planes..."
this is what it says in Jacaranda. in that book, it doesn't mention anything about galileo's analysis being one of components etc...
what do i do?
so do i ignore what it says in one book where Galileo's analysis is represented as:
"He stated his theory that all objects fall at the same rate, neglecting air resistance. he did this by experimenting with highly polished inclined planes..."
this is what it says in Jacaranda. in that book, it doesn't mention anything about galileo's analysis being one of components etc...
what do i do?
physician
Some things never change.
well that may have contributed to his conclusion and analysis that the motion of projectiles was parabolic... notice where it says "neglecting air resistance".. because if air resistance was to be considered.. then the motion would not be completely parabolic... thus the difference between his new outlook compared to those of the ancinet Greek geometers...Abtari said:
also... when we calculate projectile motion questions.. we don't take into concideration the mass of objects... "all objects fall at the same rate"..
s=ut+(1/2)at2
v2=u2 + 2as
v=u+at
so i'm assuming that is why the Jacranda book explains it as such...
anyone feel free to correct me
fair enough. i see what u mean. btw, is there any way of finding out what the board of studies answer is for this particular idea of galileo's analysis.
i would feel much more secure with something authentic (rather than different opinions from different sources: in jacaranda, all objects fall at same rate; in a worksheet i have, horizontal and vertical components of velocity).
i would feel much more secure with something authentic (rather than different opinions from different sources: in jacaranda, all objects fall at same rate; in a worksheet i have, horizontal and vertical components of velocity).
slingshot effect
sorry but can i bring up another dotpoint in the same module?
identify that a slingshot effect can be provided by planets for space probes.
despite the keyword identify, do we still have to know more than just a statement. e.g. how it provides a slingshot effect, and other finer details about it?
dont know to what depth i should understand this concept. jacaranda goes into monstrous detail...
sorry but can i bring up another dotpoint in the same module?
identify that a slingshot effect can be provided by planets for space probes.
despite the keyword identify, do we still have to know more than just a statement. e.g. how it provides a slingshot effect, and other finer details about it?
dont know to what depth i should understand this concept. jacaranda goes into monstrous detail...
physician
Some things never change.
Sure u can!!!Abtari said:
U don't need to know too much.. but its always good to have a little more knoweldege about the concept..
In my half yearly i was asked the following questions:
When sending space probes to distant planets, scientists make use of the so called 'garvity-assist' (or 'slingshot') trajectories,
(a) Explain the maning of this term (3 marks)
(b) Clearly explain how this assists in reducing the fuel required to reach the distant planet. (2marks)
Ok i'll give u info whch i think is more than suffcent for this dot pt.
Sometimes the garvitational fields of planets can be used to increase the speed of spacecraft relative to the Sun and ths reduce travel times and minimise fuel and energy demans. Such spacecraft are said to be on "garvity-assist trajectories" and the whole process is often referred to as the "slingshot" effect. Essentially, the spacecraft moves behind the planet as viewed from the Sun, and is accelerated by this moving garvity field, much as a surfer is pushed forward by a wave. The energy gained by the spacecraft does not actually come from the garvitational field but from the kinetic energy of the moving planet, which is slowed by a tiny amount in its orbit, causing it to drop veryslighlty closer to the Sun.
I think that should be enough info...