question 16 ? (2 Viewers)

[ianc]

what a gay question

oh well you'd still get most of the marks if you put 0.75MJ, because you used the formula right and everything

 

[RingerINC]

3 marker, i did it the wrong lame 6 MJ way, i think the other way makes sense... Damnit i want the question paper.

 

[dartm2]

I got exactly 750,000J or 750KJ.

I have attached my solution, I hope it makes sense (and that its right)...

I got the same thing,

THe way i remember the question, it gave you 1Mj from centre of planet to 20 000 then asked you to work out how much more to 80 000

Does anyone have a copy of the question?

 

[followme]

1.75mj?? yay thats what i got!

 

[nittyc]

yeah i got that answer too! woohoo! my brain was struggling with that one=D

 

[drewgcn]

The reason why you can't use W=fs or W=mgh is because when you move far away from the earth's surface, the value of g changes, especially 10 000 km up.
w=mgh can be used but only when you're close to the surface.

I think they'd probably accept 1.75 and 0.75 though. They've got to be more interested in your method and working than interpreting their difficultly worded question.

 

[Sober]

what i remeber was that the object is in a staionary position 10000 km away from the centre of a planet, to move the object from the centre of the planet to a stationary position 20000km from the centre of the planet requires 1 Mj. how much more work would be required to move the object from the centre of the planet to a stationary position 80000km away.

Not from the centre of the planet but from 10,000 as well. The equation we use for E_p treats the planet as a point gravity source (much like a black hole), so if you try to work out the enery required to move from the exact centre you will get a value of infinite (r is on the denonminator).

 

[sja]

i just used the graviational potential energy formulae assuming tht work done to move an object a certain distance is stored in the objects potential energy.

hence i just assumed M1 x m2 is a konstant, worked it out. and had 250,000 joules to move the object 80,000 km. .. which DOESN"T make sense!!!!

wah.

hey shinji i think you are right; the reason why you think you are wrong is because you are forgetting that as you get further and further away from a planet, then the gravitational pull becomes lower and lower. hence for this planet, going beyond the 20,000km region obviously exceeds most of its gravitational pull due to the distance being covered.

 

[dmac]

The reason why you can't use W=fs or W=mgh is because when you move far away from the earth's surface, the value of g changes, especially 10 000 km up.
w=mgh can be used but only when you're close to the surface.

I think they'd probably accept 1.75 and 0.75 though. They've got to be more interested in your method and working than interpreting their difficultly worded question.

if it askes how much more than it should be 0.75 ... if it doesnt then its 1.75 ... meh who cares

 

[suchet_i]

i used E = mc^2 and so i found the mass and then i sub that into the equation
F = mv^2 / r with v being the speed of light and r being 80000 and then i converted this force into energy by dividing by pi.. my teacher said i m 100% rite

 

[typical_azn]

hmmm, im not quite sure about you answers but i got 6MJ, it kinda seems dumb but it makes sense.

oh yeah, if you read the question, it said it was statinary, so all u people who did centripital for or sumthing to do with that have a high possibility of being incorrect.

work equals force times distance

1mj for 10k kms distance
therefore 6 mj for 60k kms of distance

thats the only way which it made sense to me, no other equations and working would work, but i could be wrong.

 

[tauren]

god i used F = ma to find out the mass then m of earth and m of the object to find out the work done, i hope i'm right

 

[dhampoet]

I got exactly 750,000J or 750KJ.

I have attached my solution, I hope it makes sense (and that its right)...

It's the right answer
I'm still in year 11 (year 12 technically) and now we're studying projectile motion so my teacher brought the paper and do it together for this question
He did it like 5 minutes and my friend had done it for 1 minutes
LOL

 

[BobbyRocks]

Yourrt All dum Looserss, this mehtod to do it is using the mass dailtation formula cvoz that massz is moving when you move it you fuycking morons. So sub da into f = ma and you knwo da accelaration due to the upward gravity forceQ then you sub F into E = hf and you know H= rydberg constant and hence fidn energy losers

 

[dartm2]

due to the upward gravity forceQ

Upward gravity force Q ???


It don't think you need the Rydberg constant, i don't remember it being in the space syllabus at all, and what has mass dialation got to do with it, it is talking about the work done, or the difference in potential energy, speed isn't important, and therefore neither is mass. Or are you just screwing with us?

 

[machiavel]

According to my phys teacher 0.75MJ is right. Should I be worried that he did it in his head while he was on the phone?

Unless i've remembered the question horribly wrong, or I made a stupid mistake. Wasn't the quetsion asking for us to move something from the centre of the earth to 80000km away. In which case what puzzels me is how we are meant to calculate the poteintual energy of the obect is when it is in the centre as the distance in the equation is equal to zero the potientual energy become an negitive infinate number as it is -MmG/r.

From memory it said that the object was 10 000, then 20 000 then 80 000km above the surface of a planet, not necessarily Earth, so you can't sub in for M. (BTW it has to be from the surface because E_U=GMm/r only when the object is above the Earth, but below it becomes more complicated because you can't consider the planet as a point mass with a nice centre of gravity.) Did anyone else do the satellite moving from 20 000km to 80 000km? Hopefully we'll get marks because the question was dodgy. If it was definitely 10 to 80 then I am screwed and my whole ans is wrong, but I might get a mark for the right concepts.

If that follows it requires an infinate amount of energy to move the object away from the centre. That really confuses me.

This comes from Newton's definition of a gravitational field: F_G = GMm/r² so as r-->infinity, F-->0. What I think you're confused about is where the object is moving: at an infinite distance, there is no F_g so E_U=0 because grav. potential energy is provided by the field. As you move towards infinity, the satellite does work and some of the E_U is converted to the E_U which makes the satellite move. It would take an inifite amout of energy to move out of the gravitational field because you can't go the infinite distance required to get out of the field. You don't need an infinite amount of energy to get the satellite to the surface of the Earth from the centre or something.

In the end I think I said something like its 8 times the distance therefore 8 times the Ep because of a linear relationship.

It's not linear as a function of r (but it is as a function of m, the satellite mass) because it's 1/r = hyperbolic. Also it wants the change in E_U as you go from 20 000km to 80 000, not the loss in E_U from the surface. I'll stick my solution on lj and link it here.

Hopefully no-one else is confused now. If I do have something wrong, yell at me b/c I'll feel terrible misleading these ppl. (Only b/c there are no ends to justify it. )

 

[angmor]

i got 6MG

work done = mgh

 

[machiavel]

Here is my solution. It took me ages to get the (*&@^$*) html right.
http://machiavelli-imp.livejournal.com/2353.html
And for the last time, you can absolutely not use GPE=mgh because:
1. g is not constant when the changes in altitude are so large. The whole point of learning Newton's Law of Gravitation is to point out that g changes and G is constant (Look, Q1 from today's paper).
2. We don't know what m is!!!
Aaargh....

 

[Sgs2006]

i got 6MG

work done = mgh

g isn't constant

 

[Fobio]

I got exactly 750,000J or 750KJ.

I have attached my solution, I hope it makes sense (and that its right)...

QFT