## Gravitational Potential Energy Question

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sonmychest
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### Gravitational Potential Energy Question

Doing some revision for my exams, I need to know how to work out this question

> What would be the Gravitational potential energy of a 200kg satellite on the launch pad on the surface of a planet with a 6000 km radius and an acceleration due to gravity of 8 ms^-2

jmorgan3
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### Re: Gravitational Potential Energy Question

If you're doing non-calc-based physics, they should have given you an equation for this situation.

If you're in calc-based physics, you should have an equation that relates a force field to potential energy. Set the potential energy an infinite distance from the planet to be equal to zero and go from there.

Either way, don't be freaked out if you get a negative answer.
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rpenido
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### Re: Gravitational Potential Energy Question

Potential energy = mass x gravity x height

PE = 200kg * 8ms^-2 * 6.000.000 m = 9.600.000.000 J

I'm missing something ?

Tass
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### Re: Gravitational Potential Energy Question

rpenido wrote:I'm missing something ?

Yes. Gravity changes with height, so you can't just multiply in this case. That formula is an approximation for small h's.

Technically, if the problem formulation is exactly what you have written, you can try this stunt:
Attachments Last edited by Tass on Wed Oct 13, 2010 12:35 pm UTC, edited 1 time in total.

jaap
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### Re: Gravitational Potential Energy Question

rpenido wrote:Potential energy = mass x gravity x height

PE = 200kg * 8ms^-2 * 6.000.000 m = 9.600.000.000 J

I'm missing something ?

Probably, though it depends on the intentions of the question setter.
The E=mgh works fine within a small range of heights so that you can assume g is constant in that range. If you do this, then the point relative to which height is measured does not matter, as long as you do it consistently. In that case 0 would be just as good an answer to the question as any, as you might as well measure height above ground rather than height above the centre of the planet. The actual values don't matter, but you must measure consistently so that differences in height correspond exactly to differences in amount of potential energy.

In this case however the question is about a satellite. This will likely travel far enough away that g cannot be considered constant any more. In that case it is fairly standard to use 'infinitely far away' as a reference point. Things infinitely far away have zero potential energy, and as they lose that energy when they get closer, everything nearby has negative potential energy.
For this question you would first calculate the planet mass from knowing g at the current distance (=radius of planet), and use that in the proper formula for potential energy. This formula you can deduce by looking at how much energy it would take to move the satellite 'infinitely far away'.

I think the question setter intended the latter type of calculation. A bit more context around the given question would be helpful to make sure.

oxoiron
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### Re: Gravitational Potential Energy Question

Tass wrote:Technically, if the problem formulation is exactly what you have written, you can try this stunt:
(see Tass's attachment above)

That was my first thought; if the satellite is sitting on something, it has no gravitational potential relative to that surface.
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Solt
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### Re: Gravitational Potential Energy Question

rpenido wrote:Potential energy = mass x gravity x height

PE = 200kg * 8ms^-2 * 6.000.000 m = 9.600.000.000 J

I'm missing something ?

Probably not. It technically depends on what class the OP is taking but since there isn't more information regarding h, I'm guessing it's a basic physics class so you are right.
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thoughtfully
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### Re: Gravitational Potential Energy Question

Tass wrote:
rpenido wrote:I'm missing something ?

Yes. Gravity changes with height, so you can't just multiply in this case. That formula is an approximation for small h's.

More precisely, it's an approximation for when the height is close to the radius used to find little g. Which they are, in this case!
Assuming you're interested in PE with respect to the center of the planet, which seems the reasonable interpretation to take. Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away.
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BlackSails
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### Re: Gravitational Potential Energy Question

I would have marked Tass down on that actually. Unless otherwise specified, its convention that the potential vanishes at infinity.

Charlie!
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### Re: Gravitational Potential Energy Question

BlackSails wrote:I would have marked Tass down on that actually. Unless otherwise specified, its convention that the potential vanishes at infinity.

Or unless the potential is a log. Or, for theoretical cases, linear. But yeah.
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BlackSails
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### Re: Gravitational Potential Energy Question

Thinking about it more, the answer is still wrong, but I probably would offer extra marks for cleverness.

Tass
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### Re: Gravitational Potential Energy Question

thoughtfully wrote:
Tass wrote:
rpenido wrote:I'm missing something ?

Yes. Gravity changes with height, so you can't just multiply in this case. That formula is an approximation for small h's.

More precisely, it's an approximation for when the height is close to the radius used to find little g. Which they are, in this case!
Assuming you're interested in PE with respect to the center of the planet, which seems the reasonable interpretation to take.

No. Gravity changes with depth to. mgR would wrong in that case. If the planet is uniform in density the result would actually be mgR/2, why is left as an exercise for the reader.

BlackSails wrote:I would have marked Tass down on that actually. Unless otherwise specified, its convention that the potential vanishes at infinity.

So would I It is pretty obvious that the meaning is to find the gravitational energy with respect to infinity, in other words how much energy do you need to put into the satellite to get it off the planet?

I am not going to give the answer to this one.

Antimony-120
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### Re: Gravitational Potential Energy Question

Tass wrote:
thoughtfully wrote:
Tass wrote:
rpenido wrote:I'm missing something ?

Yes. Gravity changes with height, so you can't just multiply in this case. That formula is an approximation for small h's.

More precisely, it's an approximation for when the height is close to the radius used to find little g. Which they are, in this case!
Assuming you're interested in PE with respect to the center of the planet, which seems the reasonable interpretation to take.

No. Gravity changes with depth to. mgR would wrong in that case. If the planet is uniform in density the result would actually be mgR/2, why is left as an exercise for the reader.

BlackSails wrote:I would have marked Tass down on that actually. Unless otherwise specified, its convention that the potential vanishes at infinity.

So would I It is pretty obvious that the meaning is to find the gravitational energy with respect to infinity, in other words how much energy do you need to put into the satellite to get it off the planet?

I am not going to give the answer to this one.

Depending on the rest of the test I may well have given the marks. If the student clearly knew most of the math, then I would probably give the marks, as they clearly listened in class and understood the concepts, they just blanked on this one question. On the other hand if they did uniformly terribly, then I would most likely not give the marks, as then they're just trying to cover for the fact that they didn't study.
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