Gravitation Potential energy. (1 Viewer)

k02033

Member
hahahah i like how this guy keeps using tertiary physics to explain hsc physics.
cant wait till you explain electromagnetism in motors and generators where it has all these calculus and vector terms lol.

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i just substituted 2 input values into an hsc functions and subtracted the output values, demonstrating how to properly use the equation. don't think that is uni physics.
this is uni physics :

$\bg_white U=-\int \overrightarrow{F}\cdot \overrightarrow{ds}$

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helper

Active Member
Heard a bit more about the question today, and found out it also included mass loss from the fuel being expelled. So would really like to see the question before comitting.

Member
So there are two things effecting the GPE of the rocket, the mass of fuel loss and the increase in distance from planet. Depending on which is occurring at a faster rate, then the GPE of the rocket may go up or down.
Exactly and including the loss of potential bond energy (from fuel burning) the net potential is going to be less

helper

Active Member
Exactly and including the loss of potential bond energy (from fuel burning) the net potential is going to be less
No, this statement is irrelevant and would be treated as such by the markers.

k02033

Member
So there are two things effecting the GPE of the rocket, the mass of fuel loss and the increase in distance from planet. Depending on which is occurring at a faster rate, then the GPE of the rocket may go up or down.
this is fine, but you and i were analysing something that wasnt gaining in altitude, in which i believe you gave a wrong interpretation

k02033

Member
Exactly and including the loss of potential bond energy (from fuel burning) the net potential is going to be less
Hey Justin!

riseek

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

annabackwards

<3 Prophet 9
Gpe = -gMm/r or Potential energy = mgh
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

k02033

Member
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.

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