2006 Question 18 - Space (1 Viewer)

[iRuler]

An object is stationary in space and located at a distance 10 000 km 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.

(3 Marks)

If someone could please explain the steps involved in this question it would be very helpful. I have the whole worked out answer, but still cant seem to understand all the steps.

 

[ilikebeeef]

An object is stationary in space and located at a distance 10 000 km 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.

(3 Marks)

If someone could please explain the steps involved in this question it would be very helpful. I have the whole worked out answer, but still cant seem to understand all the steps.

GPE=-GMM/r

At 20 000km: -1x10^6=-GMM/2x10^7
At 80 000km: GPE=-GMM/8x10^7
Since 80 000km is 4x as much as 20 000km, it will have -0.25x10^6 J GPE
How much more work needs to be done = -0.25x10^6 - (-1x10^7) = 0.75x10^6 J aka 0.75MJ

 

[iRuler]

Yeah, that answer is right, but I'm lost again at:

it will have -0.25x10^6 J GPE
How much more work needs to be done = -0.25x10^6 - (-1x10^7) = 0.75x10^6 J aka 0.75MJ

more help please?

 

[ilikebeeef]

Yeah, that answer is right, but I'm lost again at:



more help please?

Well to get from -1x10^6=-GMM/2x10^7 to GPE=-GMM/8x10^7, you multiply both sides by 1/4

I'll expand:
How much more work needs to be done = GPEfinal - GPEinitial = GPE at 80000km - GPE at 20000km = -0.25x10^6 - (-1x10^7) = 0.75x10^6 J aka 0.75MJ

 

[iRuler]

Well to get from -1x10^6=-GMM/2x10^7 to GPE=-GMM/8x10^7, you multiply both sides by 1/4

I'll expand:
How much more work needs to be done = GPEfinal - GPEinitial = GPE at 80000km - GPE at 20000km = -0.25x10^6 - (-1x10^7) = 0.75x10^6 J aka 0.75MJ

Wonderful, thanks

 

[weirdguy99]

Heres the way I did it.

Ep=-Gm1m2/r

1000000=-Gm1m2/20000000
.: -Gm1m2=2x10^13

Ep=(2x10^13)/80000000 (at 80000km)
Ep=250000J

Work=Change in Ep
|250000-1000000|=750000 (0.75MJ)

 

[nat_doc]

i couldn't help but notice that this was an "iThread"

heheh look at the first 5 posts