question 16 ? (2 Viewers)

[casomerville]

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....

I agree. I also took the question to mean from 20000km to 80000km.

Now, for all those people out there. You can't use GPE = mgh, you can't use F=ma and you certaintly can't use mass or time dilation.

You had to use the GPE formula. The question was ambiguous but if you took it from 10000 to 80000, the answer was 1.75MJ. If you took it from 20000 to 80000, the answer was 0.75MJ. The solution contained in the link is the right one. No other answer beside those two could possibly be true and until someone produces the exact wording of the exam question, it is impossible to know whether it was supposed to be 10000 or 20000km.

 

[shinji]

okay, here's the exact wording of the question:

An object is stationary in space and located at a distance 10,000km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20,000 km from the centre of the planet.

Calculate how much more work needs to be done to move the object to a stationary point 80,000 km from the centre of the planet.

emphasized the MORE part and hence, i believe .75MJ is correct

 

[casomerville]

okay, here's the exact wording of the question:



emphasized the MORE part and hence, i believe .75MJ is correct

Thankyou. In that case the answer is 0.75MJ and everything else is wrong. (1.75MJ will probably get you some marks.)

BTW: I got the 4u joke about de moivre. very funny...

End of discussion.

 

[bboyelement]

http://www.esnips.com/nsdoc/f1c7bb8b-9980-4f23-b916-9758d018bc7d


thats the paper credit to helper

if you want to double check ... basically 0.75mj is the correct answer.

 

[angmor]

ok i fucked this up badly. i used the MGH method fuck yall.

 

[s2indie]

oh crap. But i expected to get it wrong anyway.

 

[raikkonen_rulez]

okay, here's the exact wording of the question:



emphasized the MORE part and hence, i believe .75MJ is correct

The bit I emphasized on was "80000km...FROM THE CENTRE blah blah"

At first I did it 20000 - 80000, then I changed it.

But anyway, I did state in front of the solution that I'm taking it to be 10000 - 80000.

WHY are they trying to trick us with the wordings/english in a PHYSICS exams? I'm really unhappy with the BOS.

 

[dunno04]

Yeh i think your wrong cuz it should be grater than 1MJ to lift it to a height like greater than before.

I got 6MJ too and so did a few ppl

it's just W=fs
and u end up getting 7MJ to take it to 80 000km so since it's from 20 000km then it's 6MJ i.e. 1MJ per 10 000km lifted.

Well that's what i think is the way. fuk i'm not real confident though.

i guess this is right.
W=Fs.
That's as simple as that.

I THINK.

 

[theodore]

I got exactly 750,000J or 750KJ.

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

yeah i got this, turned out nice at 0.75 x 10 ^ 6 MJ

 

[charismo]

gravitational potential energy approaches 0 as you approach inifity.. therefore in thsi question gravitational potential energy must increase in the sense that it become more positive, but in terms of magnitude.. it should be less then 1MJ

I think we had to use ratios and recognise that in the Ep formular- Gm1m2 is a constant... and go from there.. 'r' changes..

 

[angmor]

the question was talkikng about work done...shouldnt the work required to lift and object higher in the gravitational field require more work than an object closer to the surface?

 

[casomerville]

You cannot use:

W = Fd
W = mgh
F = ma
Time dilation
Mass dilation

You had to use the GPE formula.

0.75MJ is the correct answer.

 

[sinist4]

i guess this is right.
W=Fs.
That's as simple as that.

I THINK.

thats wat i got hahahahahahaha ....

 

[nath_quam]

Well i got 0.75MJ below is all my working, lucky guess worked

10000-->20000 = 1MJ

therefore double distance and half engery 20000--->40000 = 0.5MJ
and, 40000--->80000 = 0.25MJ

20000--->80000 = 0.50MJ + 0.25MJ = 0.75MJ

 

[behemoth100]

I formed 2 equations and solved them simultaneously:
-(Gm1m2) - (-G(m1m2)) = 1000000
a 20000 aaaa 10000

= -G(m1m2) +2G(m1m2)= 2000000

=G(m1m2)= 2000000 (1) Therefore letting G(m1m2) equal a constant 2000000
and;

-G(m1m2) - (-G(m1m1)) = x aaa sub in (1)
a 80000 aaaa 10000

=-2000000 + 2000000 = x
a 80000 aaaa 10000

=-2000000 +16000000 = 8x

=14000000 = 8x

= x = 14000000
aaaaaaa 8
= x = 1750000 J
= 1.75 MJ
And for all you people out there spouting about how brilliant your method was about using mass dilation and E=mc^2, sometimes the simpler answer is better. So for the majority of the people out their who are freaking about if they got the wrong or right answer just remember it doesnt matter how fancy your method is, if you get the right answer you get the marks, and if you get the WRONG answer but your method is sensible and logical, whether its sophisticated or not, you will get some marks.
Good luck guys.

 

[Jono_2007]

I got exactly 750,000J or 750KJ.

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

Yeah thats the right answer 0.75MJ, but it is much easier to let 10,000km=r, then the question becomes algebraic rather that numerical. It's so tedious carrying lines of working with those extra numbers.
We went throught this question in class after we finished section 1 in the space core modual.

 

[MisterPresident]

I got as far as 80 000 - 10 000 = 70 000
Then put 7MJ
I was entirely unsure about this question.