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    (Original post by philo-jitsu)
    Been trying to get an answer for this....Im still at a loss lol
    maybe no one knows xD such a vague topic oh well
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    (Original post by abro1089)
    hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?
    (Original post by philo-jitsu)
    Been trying to get an answer for this....Im still at a loss lol
    Gravitational potential is a scalar quantity. When considering a point between two planets, work has to be done in both directions to move it to that point, so you essentially add the potentials at a particular point.

    So basically, at the midpoint between two planets of equal mass M, separated by distance r

    the potential in moving from one planet is -GM/(0.5r) = -2GM/r
    the potential in moving from the other is identical, -2GM/r

    so the net potential is the sum, V = -4GM/r


    hope this helps, if you need more ask
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    (Original post by Ayaz789)
    Well is the picture you attached on page 22?
    Yes it is.
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    Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

    cheers
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    (Original post by alevelstresss)
    Gravitational potential is a scalar quantity. When considering a point between two planets, work has to be done in both directions to move it to that point, so you essentially add the potentials at a particular point.

    So basically, at the midpoint between two planets of equal mass M, separated by distance r

    the potential in moving from one planet is -GM/(0.5r) = -2GM/r
    the potential in moving from the other is identical, -2GM/r

    so the net potential is the sum, V = -4GM/r


    hope this helps, if you need more ask
    Thanks for this, I have a question though :-)

    If the potential at distance r from earth is Xjoules, and then you add another earth at a distance of double r, so r is now the midpoint.

    why would the potential add? because wouldnt you now use less energy to get to point r, becaus now you have an attractive force from the other earth, what I mean is no work would have to be done to move from a distance towards a single earth because the earth is attracting the mass, so when there are 2 earths wouldnt the amount of energy required to get to the mid point be less because you now have a force working 'with' the mass that is moving away from earth 1.


    If you can make sense of what I just said you are quite brilliant lol
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    (Original post by philo-jitsu)
    Thanks for this, I have a question though :-)

    If the potential at distance r from earth is Xjoules, and then you add another earth at a distance of double r, so r is now the midpoint.

    why would the potential add? because wouldnt you now use less energy to get to point r, becaus now you have an attractive force from the other earth, what I mean is no work would have to be done to move from a distance towards a single earth because the earth is attracting the mass, so when there are 2 earths wouldnt the amount of energy required to get to the mid point be less because you now have a force working 'with' the mass that is moving away from earth 1.


    If you can make sense of what I just said you are quite brilliant lol
    The attractive gravitational force is not the same thing as potential. Potential is the work done per unit mass in moving an object from infinity to a particular point in the field.

    The effect of adding another Earth essentially just adds another gravitational field - work still needs to be done for this field, even if the gravitational force acts in the other direction, so the potential is doubled.

    Just consider that potential is a scalar quantity, they will never cancel out like vectors do, its a hard concept to understand - but basically potential is kind of like the work done in the field, but there are now two fields so the work done has do be doubled.
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    List of probable six marker topics?
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    (Original post by WillRose)
    Right hand rule he meant, that's left :' don't think iv ever seen right come up in an exam tbh
    My bad 😂
    In answer to the actual question I have had to use it once in section B to gain a mark and for two questions in section A before
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    (Original post by bag2gies)
    Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

    cheers
    Gravitational field strength is a vector quantity, given by g = GM/r2

    Make the two field strengths equal and find the distance:

    let x be the distance of equal field strength from M

    GM/x2 = 4GM/(d-x)2
    4x2 = (d-x)2
    4x2 = d2 -2xd +x2
    3x2 + 2xd - d2 = 0
    d = 1/3, d = -1 (using quadratic formula)

    but d > 0, so d = 1/3 only

    the distance is d/3 from the planet of mass M

    so the answer is y/d = 1/3
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    (Original post by bag2gies)
    Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

    cheers


    nvm answer above me makes waaayyy more sense
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    Name:  Screen Shot 2016-06-19 at 16.02.32.png
Views: 91
Size:  116.0 KB Was wondering if anyone could help me out with the question, for (a)(i) the answer is from P to S, but wouldn't that make QR remain still, as it is parallel to the magnetic field?
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    Name:  ImageUploadedByStudent Room1466348745.206486.jpg
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    Just to clarify. Is no work done when a charge moves perpendicular to a uniform electric field? Is work only done when the charge moves parallel to the field lines?


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    (Original post by spanisholive)
    Name:  Screen Shot 2016-06-19 at 16.02.32.png
Views: 91
Size:  116.0 KB Was wondering if anyone could help me out with the question, for (a)(i) the answer is from P to S, but wouldn't that make QR remain still, as it is parallel to the magnetic field?
    Yes, but only a section of the wire experiences the force, not all of it, the force is experienced by the two longer sides
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    (Original post by alevelstresss)
    Yes, but only a section of the wire experiences the force, not all of it, the force is experienced by the two longer sides
    But it says in the question that PQ and RS are parallel to B? The mark scheme also says that these two sides don't experience a force:confused:
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    (Original post by bag2gies)
    Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

    cheers
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    Does anyone have any predictions as to what the six marker might me?
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    (Original post by spanisholive)
    But it says in the question that PQ and RS are parallel to B? The mark scheme also says that these two sides don't experience a force:confused:
    Hey can i see the paper?
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    (Original post by SirRaza97)
    Hey can i see the paper?
    Sure, here it is Question 6 btw
    Attached Images
  1. File Type: pdf Unit_4_-_Section_B__Waves__Fields_and_Nuclear_Energy_Question_Paper-3.pdf (212.2 KB, 102 views)
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    (Original post by alevelstresss)
    The attractive gravitational force is not the same thing as potential. Potential is the work done per unit mass in moving an object from infinity to a particular point in the field.

    The effect of adding another Earth essentially just adds another gravitational field - work still needs to be done for this field, even if the gravitational force acts in the other direction, so the potential is doubled.

    Just consider that potential is a scalar quantity, they will never cancel out like vectors do, its a hard concept to understand - but basically potential is kind of like the work done in the field, but there are now two fields so the work done has do be doubled.
    HMMM the 'like 2 fields' thing has almost got me to understand it lol...I'll have a think

    thanks alot for this!
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    Can someone help me with this question: Question 9 http://filestore.aqa.org.uk/subjects...1-QP-JAN13.PDF thanks
 
 
 
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