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HSC Physics Marathon 2013-2015 Archive (1 Viewer)
- Thread starter Magical Kebab
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Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
re: HSC Physics Marathon Archive
I will send a Freddy frog to the first person to get part 2 of my previous question correct *
* conditions apply
I will send a Freddy frog to the first person to get part 2 of my previous question correct *
* conditions apply
Menomaths
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re: HSC Physics Marathon Archive
2)2.3*10^11m?
1)1.03*10^-3N towards Jupiter's centre
2)2.3*10^11m?
psychotropic
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re: HSC Physics Marathon Archive
2--- 2.87 x 10^11 m
2--- 2.87 x 10^11 m
psychotropic
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re: HSC Physics Marathon Archive
1. the earth under and on Mt Everest may be more dense than other parts of the earth, which would mean that there is a greater localised mass near Mt Everest. By the equation given above, that would increase the value of g at Mt Everest
2. Mt Everest is located between the equator and the poles. Because of the ellipsoid shape of the earth, the earth has a larger radius at the equator. This means that a mountain at the equator may have a lower value of g.
3. Further, the axial rotation of the earth creates a pseudo centrifugal force that acts outwards from the centre of the earth, reducing the value of g. Because the axial rotation is most prominent at the centre, so is this effect. Mt Everest is not very close to the equator, and hence its value of g is not greatly impacted by this effect. Rather, a mountain or high land feature at the equator would be.
Therefore, because of the above factors, Mt Everest understandably does not have the lowest acceleration due to g. Rather, it would be expected that a high land feature or mountain at the equator MAY have the lowest value of g.
Gravitational field strength is given by the formula g = GM/r^2. Although we may expect it to have the lowest value of g, since g decreases with the radius (and hence altitude) we are above the earth's surface, there are many reasons why Mt Everest may not have the lowest gravitational field strength.
1. the earth under and on Mt Everest may be more dense than other parts of the earth, which would mean that there is a greater localised mass near Mt Everest. By the equation given above, that would increase the value of g at Mt Everest
2. Mt Everest is located between the equator and the poles. Because of the ellipsoid shape of the earth, the earth has a larger radius at the equator. This means that a mountain at the equator may have a lower value of g.
3. Further, the axial rotation of the earth creates a pseudo centrifugal force that acts outwards from the centre of the earth, reducing the value of g. Because the axial rotation is most prominent at the centre, so is this effect. Mt Everest is not very close to the equator, and hence its value of g is not greatly impacted by this effect. Rather, a mountain or high land feature at the equator would be.
Therefore, because of the above factors, Mt Everest understandably does not have the lowest acceleration due to g. Rather, it would be expected that a high land feature or mountain at the equator MAY have the lowest value of g.
superSAIyan2
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re: HSC Physics Marathon Archive
redid part 2, got 2.87 x 10^11 m (same as psychotropic). I realised i didnt expand the perfect square properly
redid part 2, got 2.87 x 10^11 m (same as psychotropic). I realised i didnt expand the perfect square properly
omgiloverice
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re: HSC Physics Marathon Archive
2) 5.2x10^(11) m away from Jupiter ? o.o I doubt I'm right..
2) 5.2x10^(11) m away from Jupiter ? o.o I doubt I'm right..
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Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
nerdasdasd
Dont.msg.me.about.english
re: HSC Physics Marathon Archive
psychotropic
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nerdasdasd
Dont.msg.me.about.english
superSAIyan2
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re: HSC Physics Marathon Archive
You're correct since 5.2x10^11m from jupiter is the same as 2.8x10^11 m from saturn
Fizzy_Cyst
Owner @ Sigma Science + Phys Goat
re: HSC Physics Marathon Archive
Must. Come. Up. With. Something. Harder. For. BOS. Nerds.
Must. Come. Up. With. Something. Harder. For. BOS. Nerds.
hayabusaboston
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re: HSC Physics Marathon Archive
since one planet's mass is different to anothers, the question asks at what distance from one planet will the gravitational force of attraction to that planet be equal to the gravitational force of attraction to the other planet so giving no net force?
So its just modifying r in the formula right?
Am I interpreting the question right?
since one planet's mass is different to anothers, the question asks at what distance from one planet will the gravitational force of attraction to that planet be equal to the gravitational force of attraction to the other planet so giving no net force?
So its just modifying r in the formula right?
Menomaths
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re: HSC Physics Marathon Archive
ps i thought the answer was infinite because i never knew he meant the probe being in between the two planets
Yeah, when both of their 'F' is equal on the probe, it'll experience 0 net force but I don't know how to work it out :angry:
ps i thought the answer was infinite because i never knew he meant the probe being in between the two planets
psychotropic
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re: HSC Physics Marathon Archive
look at the field lines from the north and south poles
look at the field lines from the north and south poles
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