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#### claudia.augustine

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« Reply #1620 on: January 12, 2017, 05:08:20 pm »
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How is the gravitational potential energy formula (E= -GMm/r) derived? I get why it's negative, im just confused on why the r is not squared like that in Newton's universal gravitation law

#### RuiAce

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« Reply #1621 on: January 12, 2017, 05:13:07 pm »
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How is the gravitational potential energy formula (E= -GMm/r) derived? I get why it's negative, im just confused on why the r is not squared like that in Newton's universal gravitation law
Go back to prelim.

W=Fs. What was that formula again?
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#### Rathin

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« Reply #1622 on: January 12, 2017, 05:17:33 pm »
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How is the gravitational potential energy formula (E= -GMm/r) derived? I get why it's negative, im just confused on why the r is not squared like that in Newton's universal gravitation law

Equate F(G)=GMm/r^2 and F(G)=mg
to get E(p)=GMm/r

BUT We take a reference point at infinity where E(p)=0 so any finite distances will be trivially negative..THUS E(p)=-GMm/r
« Last Edit: January 12, 2017, 05:27:47 pm by Rathin »
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#### RuiAce

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« Reply #1623 on: January 12, 2017, 05:24:00 pm »
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Equate F(G)=GMm/r^2 and F(G)=mg
to get E(p)=-GMm/r
Not too sure how that one works... that just gives g=GM/r^2
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#### Rathin

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« Reply #1624 on: January 12, 2017, 05:26:46 pm »
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Not too sure how that one works... that just gives g=GM/r^2

Yes that is correct, now we take a reference point at infinity at which E(p)=0 so any finite distances (aka below our reference point) is negative..thus the whole quantity 'Gravitational Potential Energy' is always negative.
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#### RuiAce

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« Reply #1625 on: January 12, 2017, 05:28:31 pm »
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Yes that is correct, now we take a reference point at infinity at which E(p)=0 so any finite distances (aka below our reference point) is negative..thus the whole quantity 'Gravitational Potential Energy' is always negative.
And how does g=GM/r^2 magically become Ep=-GmM/r?

Look again. I don't believe you answered the question
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#### claudia.augustine

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« Reply #1626 on: January 12, 2017, 05:36:35 pm »
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W=Fs. What was that formula again?

Ahh so it would be,

w= Fs
= GMm/r^2 x r
= GMm/r

#### RuiAce

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« Reply #1627 on: January 12, 2017, 05:37:14 pm »
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Ahh so it would be,

w= Fs
= GMm/r^2 x r
= GMm/r
The basic idea, yep.
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#### Rathin

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« Reply #1628 on: January 12, 2017, 05:39:37 pm »
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And how does g=GM/r^2 magically become Ep=-GmM/r?

Look again. I don't believe you answered the question

Oh damm.. didn't read question properly..I thought OP was asking for why its negative..soz
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#### Iminschool

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« Reply #1629 on: January 12, 2017, 05:53:33 pm »
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I genuinely don't know how to do this question
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#### RuiAce

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« Reply #1630 on: January 12, 2017, 05:57:36 pm »
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I genuinely don't know how to do this question
This is related to the motors and generators topic. Have you been taught stuff regarding electromagnetic induction and Lenz's law?
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#### Iminschool

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« Reply #1631 on: January 12, 2017, 06:04:49 pm »
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I haven't covered them at school however i have covered both electromagnetic induction and Lenz's law as extra study
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#### RuiAce

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« Reply #1632 on: January 12, 2017, 06:30:00 pm »
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I'm not gonna be able to offer a complete and/or fully accurate explanation as I'm ridiculously rusty at this, and the question isn't all that simple.

Recall that by the principle of electromagnetic induction, a change in magnetic flux is required to induce a force into an object. For this response, I will assume that the iron rod is just sitting there, and not a part of the circuit. This is because the spring is what's submerged in the mercury solution.

At the instant the switch is turned on, conventional current will flow out of the positive terminal and upwards. The current will then flow into the spring. Recall from studies in preliminary that if an electrically conductive substance (wire or something) has a charge flowing through it, it will produce its own magnetic field.

By using the right-hand grip rule, if the fingers point in the direction of the spring (the spring is a COILED wire), the thumb points in the direction of the magnetic field produced. Hence, the magnetic field travels through the spring from top to bottom (so the bottom is a north).

This suddenly newly produced magnetic field invoked a change in flux around the iron rod. As per the principle of electromagnetic induction, the iron rod will produce its own magnetic flux (briefly), and as per Lenz's law this will be to counteract the change in flux imposed by the spring.

This change in flux caused by the iron rod then affects the spring. And thus as per electromagnetic induction, the spring is now affected by a force, which causes it to bounce.

JAMON you need to fix up my response
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#### jamonwindeyer

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« Reply #1633 on: January 12, 2017, 07:40:16 pm »
+1
I genuinely don't know how to do this question

Here's my explanation, use it with Rui's

We need to know first about electromagnetic induction. A conductor that experiences a changing magnetic flux will have a current induced through it. This current is a manifestation of the conservation of energy (but that isn't necessary to explain here). The current acts to establish a new magnetic field (remember all currents are surrounded by a magnetic field), and according to Lenz's Law, this magnetic field will act in the opposite direction to the one that created it. That is, it opposes the change that created it.

Now when the switch is closed, it is a direct current that flows through the spring. Normally, we don't associate direct currents with changing magnetic fields and electromagnetic induction, since the current flows in a single direction and so establishes a constant magnetic field, not a changing one. However, that's long term - There is still a brief moment where the magnetic field around the spring changes. As the current goes from nothing, to something, that causes a changing magnetic field. This is the tough bit to understand - If it's a little confusing let me know!

We don't really care about direction in this scenario. The spring acts as a solenoid, and as the current through it increases, that creates a changing magnetic field. The iron rod experiences this changing magnetic field and will briefly have currents induced through it (eddy currents). These currents create a new magnetic field, and this interacts with the magnetic field of the spring to exert a force which causes it to bounce. Kind of like a bar magnet pushing on another bar magnet, sort of. This interaction is very quick, it only happens as the current begins flowing through the spring. Once it is flowing properly the force on the spring is gone.

If you want to get super technical, the movement of the bouncing spring may cause the rod to experience another small change in magnetic flux which could cause another force. However, this process wouldn't sustain itself (you'd lose energy to heat in the rod and the bouncing would fade away anyway)

#### Iminschool

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« Reply #1634 on: January 12, 2017, 08:57:58 pm »
+1
Here's my explanation, use it with Rui's

We need to know first about electromagnetic induction. A conductor that experiences a changing magnetic flux will have a current induced through it. This current is a manifestation of the conservation of energy (but that isn't necessary to explain here). The current acts to establish a new magnetic field (remember all currents are surrounded by a magnetic field), and according to Lenz's Law, this magnetic field will act in the opposite direction to the one that created it. That is, it opposes the change that created it.

Now when the switch is closed, it is a direct current that flows through the spring. Normally, we don't associate direct currents with changing magnetic fields and electromagnetic induction, since the current flows in a single direction and so establishes a constant magnetic field, not a changing one. However, that's long term - There is still a brief moment where the magnetic field around the spring changes. As the current goes from nothing, to something, that causes a changing magnetic field. This is the tough bit to understand - If it's a little confusing let me know!

We don't really care about direction in this scenario. The spring acts as a solenoid, and as the current through it increases, that creates a changing magnetic field. The iron rod experiences this changing magnetic field and will briefly have currents induced through it (eddy currents). These currents create a new magnetic field, and this interacts with the magnetic field of the spring to exert a force which causes it to bounce. Kind of like a bar magnet pushing on another bar magnet, sort of. This interaction is very quick, it only happens as the current begins flowing through the spring. Once it is flowing properly the force on the spring is gone.

If you want to get super technical, the movement of the bouncing spring may cause the rod to experience another small change in magnetic flux which could cause another force. However, this process wouldn't sustain itself (you'd lose energy to heat in the rod and the bouncing would fade away anyway)

Wow, you and Rui went in hard on this one and its much appreciated. Mb i forgot to think of the little details  in my answer. Your explanation was really good and helped me further understand electromagnetic induction.
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