August 18, 2019, 12:44:05 pm

### AuthorTopic: When to use what (energy)  (Read 783 times) Tweet Share

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#### /0

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##### When to use what (energy)
« on: June 13, 2008, 03:40:07 am »
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I've got two questions that they don't ask in the textbook lol

1. Is the kinetic energy in orbit the same as the kinetic energy of any moving object?

i.e. $\frac{GMm}{2R} = \frac{1}{2}mv^2$ ?

2. Why and when should you use gravitational potential energy as opposed to 'normal' potential energy?

I've heard people say you use $mgh$ when you're close to the ground, but that isn't enough of a reason for me... after all, one is positive and one is negative! A varying gravitational field can't possibly answer a difference that large!

$-\frac{GMm}{R} \ (close \ to\ earth) = mgh$ ?  ?  ?

Thanks

#### Neobeo

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##### Re: When to use what (energy)
« Reply #1 on: June 13, 2008, 07:54:31 am »
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Assuming I interpreted both questions correctly,

Note that these are both quantitative arguments. I can't give a completely definite answer as to when you should use what type of energy. I'm just laying out the concept behind the different formulas and how they are derived, so that a more conscientious decision can be made.
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##### Re: When to use what (energy)
« Reply #2 on: June 13, 2008, 02:59:52 pm »
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Wow thanks Neobeo! Looks like you put a lot of work into that. I think I understand now how $-\frac{GMm}{R}$ is "absolute" and how $mgh$ is relative.

But I don't get how you came to $\frac{dU_g}{dR} = \frac{GMm}{R^2}$ ?

#### Mao

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##### Re: When to use what (energy)
« Reply #3 on: June 13, 2008, 04:51:01 pm »
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fundamental theorem of calculus:

if $U_g$ is relative to earth:

$U_g=\int_{R_E}^{R} \frac{GMm}{R^2}\; dR$

$\implies \frac{d}{dR}U_g=\frac{d}{dR}\; \int_{R_E}^{R} \frac{GMm}{R^2}\; dR=\frac{GMm}{R^2}$
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#### Neobeo

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##### Re: When to use what (energy)
« Reply #4 on: June 13, 2008, 05:14:33 pm »
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$U_g(R) = -\frac{GMm}{R}$
Then just take the derivative with respect to R.
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#### /0

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##### Re: When to use what (energy)
« Reply #5 on: June 13, 2008, 05:19:10 pm »
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Ok thanks Mao and Neobeo! ^^