Gravitation Potential energy. (3 Viewers)

[kwabon]

I'm talking about the first one, because we are talking about a rocket that is moving a long distance away from eath and out of Earth's gravitational field.

The second equation only works for objects that remain inside the Earth's gravitational field

i did mention that equation,
but isnt the first stage of the rocket within the earth's gravitational field?

 

[kwabon]

i just checked my answer, i said it increases (which might be wrong), n got 3/3

then wtf was chippendale smoking when he said that it decreases?
damn we have a really dodgy physics teacher.

 

[annabackwards]

i did mention that equation,
but isnt the first stage of the rocket within the earth's gravitational field?

It is in the gravitational field, but as it's moving towards a point infinity out of the field you have to use the other formula. That's my understanding of it anyway as i was told to pretty much never use GPE = mgh unless it's an object that is being moved a small distance vertically above the Earth's surface.

then wtf was chippendale smoking when he said that it decreases?
damn we have a really dodgy physics teacher.

Go and get your extra marks

 

[helper]

wait i realized what i said is a contradiction, in that how can something split into 2 distinct objects and then defining those 2 to have the same location. And after the rock has split, how can we say the GPE of the rock has changed? which part do we define to be the "rock"? i think some of the initial proposal doesnt make sense.

If you go back to my first post on the thing below, I was talking about the rocket losing fuel.

http://community.boredofstudies.org/263/space/219987/gravitation-potential-energy/2.html#post4559987

Based on that I was saying the GPE would decrease because of the loss of mass of the fuel. I.E. Catholic interpretation the rocket is ship plus fuel.

That is why I said the question should have clarified what they meant by rocket, if they wanted fuel loss included as well.

 

[helper]

It is in the gravitational field, but as it's moving towards a point infinity out of the field you have to use the other formula. That's my understanding of it anyway as i was told to pretty much never use GPE = mgh unless it's an object that is being moved a small distance vertically above the Earth's surface.

mgh only works when g is constant. IE

So, while the changes in d are small, the approximation is valid. It isn't wether it is in the field or out of it.

 

[annabackwards]

So, while the changes in d are small, the approximation is valid. It isn't wether it is in the field or out of it.

Oh wow, i never knew that. Thanks

 

[Bank$]

My best advice to all HSC students -

• 
  For school exams write what your teachers teach,

• 
  For the HSC use the model answers from past papers

When you finish school and go to uni...

FORGET EVERYTHING from the past 2 years (except maths )

Hey Justin!

Hey Yan : )