question 16 ? (2 Viewers)

[Mumma]

i back the 0.75MJ answe--did it a diferent way to the pdf posted, but got the same asnwer

Nice

 

[bboyelement]

Yeah, I got 0.75MJ too

yeh i got it too but my friend said it was from the centre or sumthing so you should add a 1 too it....so it will be 1.75 mj but i cant remember what the question asked... i thought i remembered it asking how much more energy is required but he said it was from the centre ...

but yeh its easie to work out just remember that -Gm1m2 is a constant the only bit that changes is the radius ... basically work out what -Gm1m2 is first using the values given then apply to 80000 km

 

[shinji]

..ne1 for .25mj? rofl.

 

[shsshs]

7.5 x 10^5 J is correct

however, interesting to note that it takes less energy to move the object a further distance

 

[Mumma]

7.5 x 10^5 J is correct

however, interesting to note that it takes less energy to move the object a further distance

Not really. Simply, further distance = weaker field.

 

[mojo_rising]

Yeh i got 750,000 J as well. Did a slightly different way to mumma but im pretty sure its right.

 

[zeek]

I did it this way....

ΔE_p=E_p(final)-E_p(initial)
=-(G.m₁m₂)/R₁ + (G.m₁m₂)/R₂
=G.m₁m₂)(1/R₂ -1/R₁)

Then you would substitute the values they gave you to move from 10000km to 20000km and you would find m₁m₂
You would then do apply the same formula but this time, you would change the values of R₁ and R₂

My final answer was 1.75 MJ i think

 

[BoganBoy]

yep, 0.75MJ is correct. and yeah, i was tempted to use the W=Fd formula cos it could of being so much easier . i was tricked by the 45 degree question in multiple choice. but overal the paper was pretty easy.

 

[passion89]

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.

Yeha I got that too

 

[Mumma]

Now way its 1MJ per 10,000KM, as I said, the field gets weaker. Thus it would take less energy further out.

 

[4DOGS]

^ is M = x10^6

 

[red802]

i got 6mj

i think my working out might be right, well, i hope so

i use W=fd
since W = 1x10^6
S = 10000

So i got the force as a 100, which i thought might of been constand, well, i continue,

than since it move from 20000, and is now the new stationary point,

i put W = 100 x 60000

and thats how i got 6MJ, but i also said, from the start, at 10000, it move to 7MJ,

 

[markus123456789]

C = Change
Ep = Grav. potential energy
r = radius

C.Ep = -GMm / C.r

1x10^6 = -GMm/(1x10^4)

-GMm = 1x10^10
...........................

C.Ep = -GMm / C.r

= (1x10^4)/(6x10^5)

=166.67 KJ

Im pretty sure this is how its done

Yeh thats what i got also... at first i looked at it and was thinking... how could it be so much less energy, but the i realized that as the radius increased, the amount of energy was reduced significantly I hope thats right cause thats what i got. I did it a bit differently though where i made M1 and M2 = K cause they were constant, so basically i guess it was just ratios. Lets hope that was the right way

 

[STx]

this wasnt q16 though, im just confused, q16 was the projectile motion?
I got the same as Mumma i think.

 

[klaw]

For the GPE question this is what I put:
KL says:
you know how it said that 1 MJ of work had to be done to get from 10000km to 20000 km?
KL says:
I just said that GPE at 20000 km - GPE at 10000 km was equal to 1 MJ
KL says:
so then -Gmm/r-(-Gmm/2r)=1 MJ, where r = 10000 km
KL says:
so -Gmm/r=2 MJ
KL says:
and at 80000 km, GPE = -Gmm/8r
KL says:
which = -Gmm/r * 1/8 = 2 * 1/8 = .25 MJ
KL says:
and so the work required to get from 20000 km to 80000 km was 1 MJ - 0.25 MJ
KL says:
= 0.75 MJ
KL says:
anything wrong with what I've done?

 

[Sober]

Alot of people are saying 0.75 MJ. This is close to the correct answer.
If you read it carefully it was asking for you to move it from 10,000 to 80,000 (the previous statement of moveing it from 10,000 to 20,000 was just a hypothetical). Hence you have to add 1 MJ.

Correct answer is 1.75 MJ

 

[bboyelement]

Alot of people are saying 0.75 MJ. This is close to the correct answer.
If you read it carefully it was asking for you to move it from 10,000 to 80,000 (the previous statement of moveing it from 10,000 to 20,000 was just a hypothetical). Hence you have to add 1 MJ.

Correct answer is 1.75 MJ

is that the correct wording of the question because i remembered it goes how much more energy is required to move from 80 000... so if its at 20 000km already tthan wouldnt it just be 20 000 to 80 000 ...

 

[Sober]

There was something along the lines of "suppose instead". Perhaps not exactly that, but something to the same effect. I was quite particular about it.

 

[bboyelement]

There was something along the lines of "suppose instead". Perhaps not exactly that, but something to the same effect. I was quite particular about it.

were you sober ? meh lame joke

 

[barker25]

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.