# Thread: HSC Physics Marathon 2013-2015 Archive

1. ## Re: 2013 HSC physics marathon

Originally Posted by Immortalp00n
new question:
Discuss the issues associated with safe re entry into the Earth's Atmosphere and the precautions and safety measures taken to overcome said issues.
Issue: There is an optimum angle for safe re-entry for a manned spacecraft. Failure to meet such an angle can result in fatality and destruction. Other factors that need to be taken into consideration and appropriately treated in order for a safe re-entry are acceleration of the spacecraft, g-forces experienced by the spacecraft and lastly, the heat of the spacecraft.

Heat: The considerable amount of kinetic and potential energy possessed by an orbiting spacecraft must be lost during re-entry. As the atmosphere decelerates the spacecraft, the energy is converted into a large amount of heat, which must be tolerated and/or minimized. Spacecrafts with a 'blunt nose' produce a shockwave of air in front of them, which allows for a large amount of the frictional heat to be absorbed. Heat can also be tolerated through use of heat shields that use ablating tiles. The ablation layer vaporises or ablates under extreme heats and therefore, dissipates heat during re-entry. Porous Silica tiles are an alternative that can be placed on the exterior of the spacecraft. They dissipate (instead of ablating) large amounts of heat and are extremely good insulators. They are beneficial is that they don't require any excess of fuel in order to be carried, since they are relatively very light (composed of 90% air). Lastly, heat can be minimized by taking a longer period of time to re-enter, thereby lengthening the time over which energy is converted into heat. This essentially allows for an avoidance of any substantial accumulation of heat that can damage the spacecraft and the occupants within it.

Ionisation Blackout: Radio blackouts occur during a period of re-entry. They are caused by overheated air particles ionizing as they collide with the spacecraft. Radio signals are unable to penetrate these air particles and therefore, raises concern if contact is required between the spacecraft and Earth. The issue can only really be avoided or minimized, through planning the mission more carefully such that when they go into that period, the astronauts are self-sufficient and therefore, don't require communication through radio contact.

Decelerating G-Forces: The deceleration of a re-entering spacecraft produces g-forces, which are typically greater than those experienced at launch (~20g). High g-forces can potentially kill occupants and certain procedures are vital for safe re-entry. Such high g-forces are able to be tolerated by reclining the astronaut, such that blood in the body is not forced away from the brain (it also fully supports the body). Such reclining of the body is achieved through the use of a moulded fibreglass couch. G-forces can further be minimised by extending the re-entry, slowing the rate of descent.

2. ## Re: 2013 HSC physics marathon

Here's a somewhat tricky question for you guys to answer.

HSC 2006 question. (3 marks)

An object is stationary in space and located at a distance 10000km from the centre of a certain planet. It is found that 1.0MJ of work needs to be done to move the object to a stationary point 20000km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80000km from the centre of the planet.

4. ## Re: 2013 HSC physics marathon

Originally Posted by iSplicer
+1

5. ## Re: 2013 HSC physics marathon

Originally Posted by RivalryofTroll
Here's a somewhat tricky question for you guys to answer.

HSC 2006 question. (3 marks)

An object is stationary in space and located at a distance 10000km from the centre of a certain planet. It is found that 1.0MJ of work needs to be done to move the object to a stationary point 20000km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80000km from the centre of the planet.
work done is the change in potential energy right
ep=-g x m1m2/r
from 10000 to 20000 it took 1.0MJ ( 1x10^6 J)
1x10^6=- G X M1M2/20000 X 10^3 - (-G x m1m2/10000 x 10^3)
20000x 10^3 x1x10^6=-Gm1m2 + 2Gm1m2
Gm1m2=2 x 10^13

to move from 20000--> 80 000km
-G x m1m2/80000 x 10^3 + G x m1m2/r
= - 2x 10^13/80000 x 10^3 + 2x10^13/20000x 10^3
Change in potential energy= -250000 + 1000 000 = 750000
divide by 10^6
work done= 0.75 MJ

6. ## Re: 2013 HSC physics marathon

fuark wish i knew how to latex

7. ## Re: 2013 HSC physics marathon

Originally Posted by Immortalp00n
work done is the negative change in potential energy right
ep=-g x m1m2/r
from 10000 to 20000 it took 1.0MJ ( 1x10^6 J)
1x10^6=- G X M1M2/20000 X 10^3 - (-G x m1m2/10000 x 10^3)
20000x 10^3 x1x10^6=-Gm1m2 + 2Gm1m2
Gm1m2=2 x 10^13

to move from 20000--> 80 000km
-G x m1m2/80000 x 10^3 + G x m1m2/r
= - 2x 10^13/80000 x 10^3 + 2x10^13/20000x 10^3
Change in potential energy= -250000 + 1000 000 = 750000
divide by 10^6
work done= 0.75 MJ
3/3.

Excellent work champ.

8. ## Re: 2013 HSC physics marathon

Originally Posted by premskies
Issue: There is an optimum angle for safe re-entry for a manned spacecraft. Failure to meet such an angle can result in fatality and destruction.What happens if its coming in too shallow/steep? Other factors that need to be taken into consideration and appropriately treated in order for a safe re-entry are acceleration of the spacecraft, g-forces experienced by the spacecraft and lastly, the heat of the spacecraft.

Heat: The considerable amount of kinetic and potential energy possessed by an orbiting spacecraft must be lost during re-entry. As the atmosphere decelerates the spacecraft, the energy is converted into a large amount of heat, which must be tolerated and/or minimized. Spacecrafts with a 'blunt nose' produce a shockwave of air in front of them, which allows for a large amount of the frictional heat to be absorbed. Heat can also be tolerated through use of heat shields that use ablating tiles. The ablation layer vaporises or ablates under extreme heats and therefore, dissipates heat during re-entry. Porous Silica tiles are an alternative that can be placed on the exterior of the spacecraft. They dissipate (instead of ablating) large amounts of heat and are extremely good insulators. They are beneficial is that they don't require any excess of fuel in order to be carried, since they are relatively very light (composed of 90% air). Lastly, heat can be minimized by taking a longer period of time to re-enter How do you do this?, thereby lengthening the time over which energy is converted into heat. This essentially allows for an avoidance of any substantial accumulation of heat that can damage the spacecraft and the occupants within it.

Ionisation Blackout: Radio blackouts occur during a period of re-entry. They are caused by overheated air particles ionizing as they collide with the spacecraft. Radio signals are unable to penetrate these air particles and therefore, raises concern if contact is required between the spacecraft and Earth. The issue can only really be avoided or minimized, through planning the mission more carefully such that when they go into that period, the astronauts are self-sufficient and therefore, don't require communication through radio contact.

Decelerating G-Forces: The deceleration of a re-entering spacecraft produces g-forces, which are typically greater than those experienced at launch (~20g). High g-forces can potentially kill occupants and certain procedures are vital for safe re-entry. Such high g-forces are able to be tolerated by reclining the astronaut, such that blood in the body is not forced away from the brain (it also fully supports the body). Such reclining of the body is achieved through the use of a moulded fibreglass couch. G-forces can further be minimised by extending the re-entry, slowing the rate of descent.

9. ## Re: 2013 HSC physics marathon

Originally Posted by nightweaver066

The spacecraft must touch down softly onto the surface of the earth.
- use of russian capsules ( parachutes to slow the capsule down until astronauts could eject and fall using a parachute down to earth.
- Early American craft ( slowed down with parachutes and resulted in a soft landing on the ocean).
Since the space shuttle is unpowered, it has one chance at landing on the landing strip, so many measures must be taken to ensure that it is successful.
-The shuttle must be accurately guided, against drag and lift forces causing spacecraft to deviate from the original path of motion.
- The Spacecraft is designed with " lift" like an aeroplane so it can be manouvered and controlled during descent and landing.

10. ## Re: 2013 HSC physics marathon

Originally Posted by RivalryofTroll
3/3.

Excellent work champ.
thanks man

11. ## Re: 2013 HSC physics marathon

Originally Posted by RivalryofTroll
Here's a somewhat tricky question for you guys to answer.

HSC 2006 question. (3 marks)

An object is stationary in space and located at a distance 10000km from the centre of a certain planet. It is found that 1.0MJ of work needs to be done to move the object to a stationary point 20000km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80000km from the centre of the planet.
\begin{align*}&\Delta E_{p}=GmM\left ( \frac{1}{r_{2}}-\frac{1}{r_{1}} \right )\\&6.67\times 10^{-11}\times mM \left ( \frac{1}{1 \times 10^{7}}-\frac{1}{2 \times 10^{7}} \right )=1 \times 10^{6}\\&\therefore mM=2.999 \times 10^{23}\\\\&\Delta E_{p}=6.67\times 10^{-11}\times 2.999 \times 10^{23}\left ( \frac{1}{2 \times 10^{7}}-\frac{1}{8 \times 10^{7}} \right )=750~000 ~J\end{align*}

EDIT: Too slow D: Not latexing next time....

12. ## Re: 2013 HSC physics marathon

Originally Posted by iBibah
\begin{align*}&\Delta E_{p}=GmM\left ( \frac{1}{r_{2}}-\frac{1}{r_{1}} \right )\\&6.67\times 10^{-11}\times mM \left ( \frac{1}{1 \times 10^{7}}-\frac{1}{2 \times 10^{7}} \right )=1 \times 10^{6}\\&\therefore mM=2.999 \times 10^{23}\\\\&\Delta E_{p}=6.67\times 10^{-11}\times 2.999 \times 10^{23}\left ( \frac{1}{2 \times 10^{7}}-\frac{1}{8 \times 10^{7}} \right )=750~000 ~J\end{align*}

EDIT: Too slow D: Not latexing next time....
much quicker method haha

13. ## Re: 2013 HSC physics marathon

Originally Posted by nightweaver066

Yeah I wasn't sure whether to mention landing, based on the wording of the question. I also probably went too in depth in some areas and not enough in others.

If the spacecraft re-enters at an angle too shallow (typically greater than 10degrees), the spacecraft may "skip" off the atmosphere and go back deep into space as a result of the compression of the atmosphere (high pressured) atmosphere below it.

With the time taken to re-enter, the most efficient way to lengthen the time of re-entry is to maximise the angle of re-entry, such that it takes a longer amount of time to re-enter (so less heat is produced as a result of friction, because the spacecraft is travelling slower, or decelerating at a greater rate), however is still at a great enough angle to not "skip off the atmosphere."

Thanks nightweaver066.

14. ## Re: 2013 HSC physics marathon

Q. Explain how the slingshot effect works.

15. ## Re: 2013 HSC physics marathon

Originally Posted by premskies
Q. Explain how the slingshot effect works.
The slingshot effect, defined as the increase in velocity given to a spacecraft because it enters the gravitational field of a planet as it passes by, is a technque used by Space agencies in accordance with physicists in order to take full advantage of a planets rotational motion in order to increase a particle's ( spacecraft) final speed.
The slingshot effect is used by many spacecraft which travel within and beyond the solar system.
A spacecraft passes close to a a planet such that its gravity pulls the spacecraft into an arc or circular motion.It leaves with the same speed relative to the planet, but its speed is increased when viewed from an alternate frame of reference such as the Sun.
The acquired speed is large enough so that the spacecraft can travel away from the planet.
Comparatively, a ball thrown from a person who is stationary would have speeds less than a ball thrown from a person who is running at an increasing or stationary speed. Thus the slingshot effect, also known as gravity assist trajectory is used in order to attain a significant change in speed and direction despite the very little expenditure of fuel, improving flight efficiency.

16. ## Re: 2013 HSC physics marathon

Originally Posted by Immortalp00n
The slingshot effect is used by many spacecraft which travel within and beyond the solar system.
A spacecraft passes close to a a planet such that its gravity pulls the spacecraft into an arc or circular motion.It leaves with the same speed relative to the planet, but its speed is increased when viewed from an alternate frame of reference such as the Sun.
The acquired speed is large enough so that the spacecraft can travel away from the planet.
Good stuff, maybe instead of the first sentence though, I would have defined the slingshot effect.

17. ## Re: 2013 HSC physics marathon

Originally Posted by premskies
Good stuff, maybe instead of the first sentence though, I would have defined the slingshot effect.
improved version^^

18. ## Re: 2013 HSC physics marathon

Originally Posted by Immortalp00n
improved version^^
It's probably better defined as being the increase in velocity given to a spacecraft because it enters the gravitational field of a planet as it passes by.

19. ## Re: 2013 HSC physics marathon

okay haha that's sorted, thanks prem!
new question:
At present, space travel is very expensive. NASA estimates that:
· the cost of sending 1 kg of cargo into Earth orbit is about $30 thousand · the cost of sending one astronaut on a Space Shuttle mission for 2 weeks is about$200
million
· the cost of an unmanned probe to the surface of Mars is about \$300 million
Use this information to account for the high cost of space travel and identify some measures that can be taken to improve flight efficiency.

~!!

21. ## Re: 2013 HSC physics marathon

Time to revive this bitch.

Explain why the speed of light is the limiting speed of any object in the universe.

22. ## Re: 2013 HSC physics marathon

If an object travelled at relavistic speeds it would experience mass dilation and its mass would increase. As a result of its mass increasing, its acceleration would decrease, assuming net force is constant. If this trend continues, the final outcome would be that as the object approaches the speed of light, its mass would approach infinity whilst its acceleration would approach zero, meaning that the object would not be able to further accelerate and hence the object would never be able to actually obtain the speed of light and thus it is referred to as being the limiting speed of an object in the universe.

Not too sure on the answer, havent actually covered space at school

23. ## Re: 2013 HSC physics marathon

Originally Posted by kunal96
If an object travelled at relavistic speeds it would experience mass dilation and its mass would increase. As a result of its mass increasing, its acceleration would decrease, assuming net force is constant. If this trend continues, the final outcome would be that as the object approaches the speed of light, its mass would approach infinity whilst its acceleration would approach zero, meaning that the object would not be able to further accelerate and hence the object would never be able to actually obtain the speed of light and thus it is referred to as being the limiting speed of an object in the universe.

Not too sure on the answer, havent actually covered space at school
Yep, the general idea is right. I recommend using equations in your future responses.

24. ## Re: 2013 HSC physics marathon

Originally Posted by bleakarcher
Yep, the general idea is right. I recommend using equations in your future responses.
haha thank you for the advice so which equations other than f=ma?

25. ## Re: 2013 HSC physics marathon

Originally Posted by kunal96
haha thank you for the advice so which equations other than f=ma?
Yeah, I just meant applying the mass dilation formula to F=ma:
a=F/(m/sqrt(1-v^2/c^2))=F*sqrt(1-v^2/c^2)/m
As v->c, a->0.

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