# HSC Physics Marathon 2016 (1 Viewer)

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#### anomalousdecay

The HSC Physics Marathon is an open chain of questions between students. It works by answering a question then posting another question and allowing the cycle to repeat itself.

Rules:
- After answering a question, always provide a new one - this is what keeps the thread alive.
- Allocate a number of marks for any question that you post.
- Do not cheat, if you cannot answer a question, do not search how to answer the question but rather, allow other students to answer the question.
- No copyrighted questions (eg CSSA and Independent) should be posted.

Tips:
- You may post more than one question.
- When possible, after questions have been answered, you can peer mark using the marking scheme.

#### anomalousdecay

This would present a nice start to the thread:

Fizzy_Cyst said:
Lets start off with a question about acceleration due to gravity!

Question:
Determine the NET (effective) acceleration due to gravity along Jupiters equator: (4 marks)
Length of day on Jupiter: 9 hours 56mins
Mass of Jupiter: 317.8 x Mass of Earth
Jupiters orbital radius: 778 million km

#### leehuan

##### Well-Known Member
Eyy good luck 16ers

Also, why is this a 4 marker? Am I dumb?

Use two formulae, and there's quite a fair bit of extraneous information.
_______________________
NEXT QUESTION:

Distinguish between the terms 'mass' and 'weight'. (2)

#### RachelGreen

##### Member
Mass is the absolute measurement of the amount of matter the object posses (scalar quantity). Whereas weight is the force acting on a mass in a gravitational field and is a vector quantity.

#### Drsoccerball

##### Well-Known Member
This would present a nice start to the thread:
HINT: ||Gravitational acceleration| - |centrifugal acceleration| |

#### RachelGreen

##### Member
V = 2.pi.r / T
= (2.pi.x (69911 x 10^3)) / (35760 seconds)
V = 12,283.6 m/s

Centripetal acceleration = v^2 / r
= (12,283.6)^2 / (69911 x 10^3)
a = 2.15829 m/s^2

Gravitational acceleration = GM/r^2
= (6.67x10^-11 x 317.8 x 6.0 x 10^24) / (69911 x 10^3)^2
= 26.0219 m/s^2

Net acceleration due to gravity = 26.0219 - 2.12829 = 23.89 m/s^2

I hope this is correct, if I'm not, someone please correct me

#### Ambility

##### Active Member
I guess I'll provide the next question. State the period of a geostationary satellite and hence calculate the altitude of a geostationary satellite's orbit.

#### leehuan

##### Well-Known Member
HINT: ||Gravitational acceleration| - |centrifugal acceleration| |
In the HSC course, don't we assume Fg=Fc?

(I will also do one answer every 2-3 days if this one is inactive as well.)

Mass is the absolute measurement of the amount of matter the object posses (scalar quantity). Whereas weight is the force acting on a mass in a gravitational field and is a vector quantity.
Yep, 2/2

#### Ambility

##### Active Member
In the HSC course, don't we assume Fg=Fc?
We only assume Fg=Fc if the object is in orbit. When it is on the equator of Jupiter, it will have a gravitational force and a centripetal force of different magnitudes.

#### Fizzy_Cyst

##### Well-Known Member
I guess I'll provide the next question. State the period of a geostationary satellite and hence calculate the altitude of a geostationary satellite's orbit.
Period = 24 hours.

Thus using Keplers 3rd Law (r^3/T^2 = GM/4(Pi)^2)

Solving for r we get

r = 42,298km (this is radius of the earth + altitude)

Rather than assume radius of Earth, wedetermine radius of the Earth, we use the formula for 'g' --> g=GM/r^2 (should derive this first), then sub in the values on the data sheet and solve for 'r' and we get r =6390km

Therefore altitude = 42,298 - 6390 = 35,908km

New Question:

Using the values given in the first question, determine the position of a geostationary orbit above Jupiter (perhaps a jovostationary orbit?)

#### leehuan

##### Well-Known Member
We only assume Fg=Fc if the object is in orbit. When it is on the equator of Jupiter, it will have a gravitational force and a centripetal force of different magnitudes.
That completely went over my head.

#### RachelGreen

##### Member
New Question:

Using the values given in the first question, determine the position of a geostationary orbit above Jupiter (perhaps a jovostationary orbit?)[/QUOTE]

Rearranging Kepler's Third Law to find 'r' with values:
T = 9 hours, 56 minutes
= 35760 seconds
M = 317.8 x 6.0 x 10^24
Therefore r = 1.6 x 10^8 when solving.

g = GM/r^2
23.89 m/s^2 = (6.67 x 10^-11 x 317.8 x 6.0 x 10^24) / r^2 --------> ('g' from previous part)
Rearrange, to find 'r' (radius of Jupiter) = 7.29 x 10^7 m

Therefore, altitude of geostationary orbit = 1.6 x 10^8 - 7.29 x 10^7
= 8.71 x 10^7m

#### leehuan

##### Well-Known Member
NEXT QUESTION:
Explain why there exists an optimum re-entry angle that must be satisfied for a rocket to ensure they return to Earth, and safely. (4)

#### Glyde

##### Member
NEXT QUESTION:
Explain why there exists an optimum re-entry angle that must be satisfied for a rocket to ensure they return to Earth, and safely. (4)
For a rocket to return to earth is must obviously re-enter the earths atmosphere. During the rockets re-entry a lot can go wrong if it's angle of entry is not between 2 and 7 degrees ( probably a bit off ). This angle is defined as the optimum re-entry angle. If the rockets angle of re entry is too large it will undergo significant amounts of atmospheric drag coding the shuttle to burn up upon re-entry and in the process kill the passenger due to the change in g-force. If the angle of re-entry is too small it will 'bounce' off the earths atmosphere and continue our into space. This is because of the differences in the mediums density. An analogy would be a rock skimming against the surface of a pond where the pond is earths atmosphere and the rock is the rocket. The rocket will then have to perform a fuel expensive manoeuvre to place it back on track for a 2nd attempt.

#### leehuan

##### Well-Known Member
For a rocket to return to earth is must obviously re-enter the earths atmosphere. During the rockets re-entry a lot can go wrong if it's angle of entry is not between 2 and 7 degrees ( probably a bit off ). This angle is defined as the optimum re-entry angle. If the rockets angle of re entry is too large it will undergo significant amounts of atmospheric drag coding the shuttle to burn up upon re-entry and in the process kill the passenger due to the change in g-force. If the angle of re-entry is too small it will 'bounce' off the earths atmosphere and continue our into space. This is because of the differences in the mediums density. An analogy would be a rock skimming against the surface of a pond where the pond is earths atmosphere and the rock is the rocket. The rocket will then have to perform a fuel expensive manoeuvre to place it back on track for a 2nd attempt.
First bold - Just the wording is awkward. No marks deducted for it.
Second bold - 5.2-7.2 deg
Third bold - Yes, but you might want to note that g-forces are a measure of acceleration. So the high g-forces reached are also a valid reason. (Note that humans can withstand any velocity but never extremely high acceleration)

So the info on the angle being too low was done well. In extreme cases, one could argue that when it's too low it gets bounced so far out of orbit... and won't even have a chance at returning. Info on when it's too high was correct, but maybe a bit rushed. Atmospheric drag causing heat build up and burning of the shuttle should be seperate from extreme g-forces as one causes the shuttle harm and one causes the astronaut harm (black-out could be a good example). 3/4

#### Drsoccerball

##### Well-Known Member
New question - Explain how multi-staged rockets allow for an increase in ability to explore space.

#### Glyde

##### Member
Multi-staged rockets are a necessity when it comes to space exploration. In order to launch a rocket off the ground, it will need an enormous velocity to escape the pull of gravity. This 'escape velocity' is hard to achieve due to the mass of liquid rocket fuel (approx 0.01% of the volume of its gas state) as well as the container it rests in. In order to maintain an acceleration the rocket will either need stronger engines and in the long run more fuel to allow the rocket to escape the earths gravitational field. This is where multi staged rockets come in. When a fuel cell has been depleted, it is ejected back down to the surface. When we observe the F=MA equation we notice that as our weight has been minimised the gravitational force acting down on the rocket becomes smaller, consequently giving the rocket a boost in acceleration. Hence the final rocket after all ejections will be much smaller allowing for better manoeuvres with more efficiency and give the rocket extra speed in space to cover larger distances

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#### Drsoccerball

##### Well-Known Member
Multi-staged rockets are a necessity when it comes to space exploration. In order to launch a rocket off the ground, it will need an enormous velocity to escape the pull of gravity. This 'escape velocity' is hard to achieve due to the mass of liquid rocket fuel (approx 0.01% of the volume of its gas state) as well as the container it rests in. In order to maintain an acceleration the rocket will either need stronger engines and in the long run more fuel to allow the rocket to escape the earths gravitational field. This is where multi staged rockets come in. When a fuel cell has been depleted, it is ejected back down to the surface. When we observe the F=MA equation we notice that as our weight has been minimised the gravitational force acting down on the rocket becomes smaller, consequently giving the rocket a boost in acceleration. Hence the final rocket after all ejections will be much smaller allowing for better manoeuvres with more efficiency and give the rocket extra speed in space to cover larger distances

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You have the right idea but the purpose behind multi-rockets isn't what you said it was. For example why can't we just use a bigger fuel tank to reach the same place? It actually isn't to increase acceleration but rather to reduce it! This is especially useful in manned rockets where there needs to be a limit on g force to maximise the survival potential. $\bg_white G Force = 1+ \frac{a}{g}$ Therefore by using multi-staged rockets we can stop the acceleration or reduce it to reasonable measures for the on board passengers. * Add some bs at beginning and end about space travel*

#### RachelGreen

##### Member
I saw this question from Syd Grammar's old trial paper and thought was a great question:
"The Earth exerts the exact same gravitational force on the Sun as the Sun does on the Earth due to Newtons 3rd Law". Assess the above statement (4 marks)

#### Glyde

##### Member
You have the right idea but the purpose behind multi-rockets isn't what you said it was. For example why can't we just use a bigger fuel tank to reach the same place? It actually isn't to increase acceleration but rather to reduce it! This is especially useful in manned rockets where there needs to be a limit on g force to maximise the survival potential. $\bg_white G Force = 1+ \frac{a}{g}$ Therefore by using multi-staged rockets we can stop the acceleration or reduce it to reasonable measures for the on board passengers. * Add some bs at beginning and end about space travel*
Ohh so do you mean I should talk about the re-entry ? Because stoping acceleration upon launching would seem counterintuitive :/ Was my answer complete rubbish or did I get some marks

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