HSC Physics Marathon 2013-2015 Archive (6 Viewers)

Chris100 · Dec 31, 2013

Re: HSC Physics Marathon 2014

Length of day on Jupiter:9 hrs and 56 mins (35760s)
Mass of Jupiter: 317.8x Mass of Earth (317.8x6.0x10²⁴)
Equatorial radius: 69911km (69911000m)
Jupiters orbital radius :778 million km (778x10⁹m)

The NET acceleration due to gravity along Jupiter's equator may be determined by subtracting the acceleration due to gravity on Jupiter's surface from Jupiter's outward acceleration caused by the centripetal force from Jupiter's orbit around the Sun

Centripetal acceleration of Jupiter relative to the Sun:

Using the equation $F=\frac{mv^{2}}{r}$ , sub in $F=ma$ and $v_{av}=\frac{\Delta r}{\Delta t}$ ;

$ma=\frac{m(\frac{\Delta r}{\Delta t})^{2}}{r}$

$a=\frac{(\frac{\Delta r}{\Delta t})^{2}}{r}$

$a=\frac{(\frac{778e9}{35760})}{778e9}$

$=608.3935158ms^{-2} Outwards$ from the sun

Acceleration due to gravity on Jupiter's surface:
This can be calculated by equating the formula for the law of Universal gravitation and F=mg;

$m_{1}g=\frac{Gm_{1}m_{J}}{r^{2}}$
$g=\frac{Gm_{J}}{r^{2}}$
$g=\frac{(6.67e $negative$ 11)(317.8)(6e24)}{69911000^{2}}$
=26.02195662ms^-2 inwards to the sun

Therefore if we take the outward from the Sun direction as positive;
Net acceleration due to gravity on Jupiter's surface

=608.3935158-26.02195662
=582.3715592
=600ms^-2 Outwards from the Sun (1 s.f)

Fizzy_Cyst · Dec 31, 2013

Re: HSC Physics Marathon 2014

Correct idea! Something is incorrect in your working for centripetal acceleration though. The period you have used is the length of a day on Jupiter (the time it takes to rotate once), but the distance you have used is the orbital radius (distance between Sun and Jupiter), what you need to find the the tangential velocity at the equator on the surface of Jupiter.

At the moment, your answer is saying that an object on the surface of Jupiter would accelerate 'upwards' (away from surface of Jupiter) at 600ms^-2

You should expect that the centripetal acceleration is much smaller than 'g'

Chris100 · Dec 31, 2013

Re: HSC Physics Marathon 2014

Oh so "r" in the centripetal acceleration calculation needs to be the equatorial radius of Jupiter?

EDIT: Yes, r needs to be the equatorial radius of Jupiter

So that means the centripetal acceleration

= [(69911000/35760)^2]/778x10^9
= 0.00000491265ms^-2 Outwards from the Sun

Therefore, the net acceleration (if we take outwards from the Sun as the positive direction)

= 0.00000491265-26.02195171
= -26.02195171
=-30ms^-2 inwards towards the Sun (1 s.f)

Fizzy_Cyst · Dec 31, 2013

Re: HSC Physics Marathon 2014

To determine v, you need to consider how far a point on the surface of the Earth would move as Jupiter rotates once, it is not just the equatorial radius that it moves. Think about the path that it would take as it spins, how would you describe its path and hence how would you determine how far it moves?
When you then solve for v and use v^2/r, the r in this equation refers to the distance between the point that an object is rotating around and the object itself. Is the object itself rotating around the Sun? Or something else?

Chris100 · Dec 31, 2013

Re: HSC Physics Marathon 2014

ffs im so bad
pls share answers 2014ers

Chris100 · Dec 31, 2013

Re: HSC Physics Marathon 2014

YAY
finally got it, scroll up for the rest of the working out
Centripetal acceleration outwards:
a=[(2pi(69911000)/35760)^2]/(69911000)
=2.158292148ms^-2 Outwards

Therefore net acceleration on the surface of Jupiter, if we take the outward direction positive is:
Net acceleration=2.158292148-26.02195662
= -23.86366447ms^-2
=23.86ms^-2 Inwards, to the centre of Jupiter (4s.f)

Fizzy_Cyst · Dec 31, 2013

Re: HSC Physics Marathon 2014

Great stuff!

Chris100 · Dec 31, 2013

Re: HSC Physics Marathon 2014

Chris100 said:
The distance between the Moon and the Earth from their centres is 3.84x10^8 m. The moon circles the Earth once every 27.3 days and the mass of the Moon is 7.475x10^22kg.
Calculate the Moon's linear velocity

Worth 2 marks

Squar3root · Dec 31, 2013

Re: HSC Physics Marathon 2014

Lets keep the questions going:

"describe the significance of the MM experiment. In you answer; discuss the role of the MM experiment in making determinations about competing theories" [8 Marks]

edit: OP was a tricky question

QZP · Dec 31, 2013

Re: HSC Physics Marathon 2014

^ Not really a "marathon" question. I thought these marathon threads were for higher level applications of knowledge rather than extended responses

someth1ng · Dec 31, 2013

re: HSC Physics Marathon Archive

The new marathon can be continued here.

iloveschool1996 · Dec 31, 2013

Re: HSC Physics Marathon 2014

can i post a question???

Squar3root · Dec 31, 2013

Re: HSC Physics Marathon 2014

QZP said:
^ Not really a "marathon" question. I thought these marathon threads were for higher level applications of knowledge rather than extended responses

I was told conversely

Fizzy_Cyst · Dec 31, 2013

Re: HSC Physics Marathon 2014

Guys, please answer a question before posting another one!

Chris100 has a question about the Moons tangential velocity!

Tbh, the high mark qs are much more subjective and depending on whether you get a person who knows what they are on about regarding feedback, you may get some incorrect advice. I rkn best thing to do would be to stick to objective skills/calc questions until pre-half yearlies, particularly seeing as there were no real 'tricky' skills/calc qs this yr and perhaps the 3-4 mark explain type qs.

iStudent · Dec 31, 2013

Re: HSC Physics Marathon 2014

The film Apollo 13 contained many filmed sequences of astronauts when they were weightless, floating around in their space capsule. These sequences were filmed in a special NASA training plane that real astronauts use to prepare for space missions. This plane flies high up in the atmosphere and then plunges towards the Earth to produce weightlessness for the occupants inside.

Comment on the use of the term weightlessness in this context (3 marks)

Fizzy_Cyst · Jan 4, 2014

Re: HSC Physics Marathon 2014

Current Questions:

1. The distance between the Moon and the Earth from their centres is 3.84x10^8 m. The moon circles the Earth once every 27.3 days and the mass of the Moon is 7.475x10^22kg.
Calculate the Moon's linear velocity

2. The film Apollo 13 contained many filmed sequences of astronauts when they were weightless, floating around in their space capsule. These sequences were filmed in a special NASA training plane that real astronauts use to prepare for space missions. This plane flies high up in the atmosphere and then plunges towards the Earth to produce weightlessness for the occupants inside.

Comment on the use of the term weightlessness in this context (3 marks)

Been awfully quiet in here so far! I will give it another day or so before I answer and post up a new, hard question!!!

anthony_wheels · Jan 4, 2014

Re: HSC Physics Marathon 2014

Fizzy_Cyst said:
Current Questions:

1. The distance between the Moon and the Earth from their centres is 3.84x10^8 m. The moon circles the Earth once every 27.3 days and the mass of the Moon is 7.475x10^22kg.
Calculate the Moon's linear velocity

2. The film Apollo 13 contained many filmed sequences of astronauts when they were weightless, floating around in their space capsule. These sequences were filmed in a special NASA training plane that real astronauts use to prepare for space missions. This plane flies high up in the atmosphere and then plunges towards the Earth to produce weightlessness for the occupants inside.

Comment on the use of the term weightlessness in this context (3 marks)

Been awfully quiet in here so far! I will give it another day or so before I answer and post up a new, hard question!!!

Q.1. I haven't got a calculator around me atm.

Q.2. By definition, the crew are NOT weightless as they are accelerating due Earth's gravitational field, therefore, a force is acting upon them (which is a weight). However, they would feel to be weightless as they are not experiencing any reaction force, that is, they are experiencing 'zero g'. [ from the equation: g (force) = (a + g)/g-earth]

someth1ng · Jan 5, 2014

@Fizzy_Cyst this is normally the quietest time of year (end of HSC to start of T1). Not many students do much during their holidays so it's no surprise, really.

QZP · Jan 5, 2014

Re: HSC Physics Marathon 2014

I have yet to do space

anomalousdecay · Jan 5, 2014

Re: HSC Physics Marathon 2014

someth1ng said:
@Fizzy_Cyst this is normally the quietest time of year (end of HSC to start of T1). Not many students do much during their holidays so it's no surprise, really.

Generally this is true.
I was doing heaps of work though last year this time, but I didn't make an account yet.

QZP said:
I have yet to do space

Did you do Motors and Generators?

I have a few on that module to post up, but once all the questions on space have been answered.

HSC Physics Marathon 2013-2015 Archive (6 Viewers)

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