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Since Gravity has infinte range... watch

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    does that mean I have high GPE for every single planet/star out there? Like I have high GPE when relating to jupiter...

    or something xD
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    Well yes Jupiter will exert a gravitational force on you, as you will on it, although this will be insignificant compared to Earth gravity and all other forces. If I remember right gravitational force falls away with distance squared?

    Disclaimer - I haven't done physics for a very long time so might be wrong
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    (Original post by jk1986)
    Well yes Jupiter will exert a gravitational force on you, as you will on it, although this will be insignificant compared to Earth gravity and all other forces. If I remember right gravitational force falls away with distance squared?

    Disclaimer - I haven't done physics for a very long time so might be wrong
    I think that's Gravitational Field strength. I think I have high GPE though...
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    The force on you here on Earth from the planet Jupiter is about 10^-8 N

    You can work it out from GM1*M2/r^2
    Mass of Jupiter 9 x 10^24 kg approx
    distance approx 600 million km
    G= 6.7 x 10^-11
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    All GPE is negative, 0 at infinite distance from the object. So no you have a very very small negative GPE :p:.
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    (Original post by Stonebridge)
    The force on you here on Earth from the planet Jupiter is about 10^-8 N

    You can work it out from GM1*M2/r^2
    Mass of Jupiter 9 x 10^24 kg approx
    distance approx 600 million km
    G= 6.7 x 10^-11
    But I want to know if the GPE is high. I don't really care about the force :P
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    (Original post by 2710)
    I think that's Gravitational Field strength. I think I have high GPE though...
    He's right, everything with mass is having a gravitational pull on you and vice versa because of your point in their gravitational field, however small it may be - rightly said an inverse square law to distance.
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     \displaystyle F = \frac{Gm_1m_2}{r^2}, if you want to know the specific equation :p:
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    But....

    Argh,

    in Mechanics, isn't GPE greater the higher you go? Well, in this case, it is higher, but it is tending towards 0, so its a different model.

    Wait, if Gravotational Potential the same as Gravitational Pot. Energy?

    Or is GP x m = GPE?

    Yeh, I think the last one is right.

    Thanks

    Sorry, I suck at potential
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    (Original post by 2710)
    But....

    Argh,

    in Mechanics, isn't GPE greater the higher you go? Well, in this case, it is higher, but it is tending towards 0, so its a different model.

    Wait, if Gravotational Potential the same as Gravitational Pot. Energy?

    Or is GP x m = GPE?

    Yeh, I think the last one is right.

    Thanks

    Sorry, I suck at potential
    I was having trouble with this yesterday aswell,

    (Or is GP x m = GPE?)

    Is correct, but is the energy at that point but not the work done to get to that point. I.e not mgh at large distances.
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    (Original post by 2710)
    But I want to know if the GPE is high. I don't really care about the force :P
    The GPE depends on the force because it measures the work done by that force moving you from infinity to where you are now.

    The thing is, for every mass in any direction out in the Universe that is pulling you towards it, there is a mass in the opposite direction pulling you the other way. This is because the universe is more or less uniform in any direction. So the overall force on you due to all the mass in the universe is just about zero. It's only the local mass (Earth, Moon, Sun) that has any real effect, and the Earth is by far the most important.
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    OP: You're pretty much right, though you need to be a little more careful with your wording.

    When we talk about the GPE of an object in a gravitational field, we usually define it as the work done to move that object to it's location from somewhere infinitely far away. That is to say, an object close to the sun, for example, would have a very negative potential energy, and as it moves further away the potential energy increases (it would have to move infinitely far for the potential energy to reach zero).

    So yes, I might have -10 megajoules of GPE with reference to jupiter, but -3000 megajoules of GPE with reference to the earth, because I'm standing right on it.
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    (Original post by Stephens)
    I was having trouble with this yesterday aswell,

    (Or is GP x m = GPE?)

    Is correct, but is the energy at that point but not the work done to get to that point. I.e not mgh at large distances.
    Can you elaborate T__T

    Ok, so

    Any item in say the field of earth, has potential. If they are at the same height, then they have the same gravitational potential. To get the energy, you time the GP by their mass. So, what do you mean in your post?

    EDIT: Ok some people have clarified a bit. I will try and get this all in my head.
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    I think this method would give you an accurate value of what your looking for. (sorry I don't know how to use latex)

    We know the equation of Newton's law of gravitation. We know GPE = mass x gravity x height

    If we take a person as weighing 60kg, the distance to jupiter as ~600million km, and your distance from the centre of jupiter at the surface as ~ 7x10^4 km, then;

    integrating between (7x10^4) and (0.6x10^9) for (3.618 x 10^16)/(r^2) with respect to 'r', we get a value of ~ 500MJ.

    This is the energy you would convert into kinetic energy if you feel freely towards jupiter (obviously thus ignoring any other massive bodies)
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    (Original post by Deno_7951)
    I think this method would give you an accurate value of what your looking for. (sorry I don't know how to use latex)

    We know the equation of Newton's law of gravitation. We know GPE = mass x gravity x height

    If we take a person as weighing 60kg, the distance to jupiter as ~600million km, and your distance from the centre of jupiter at the surface as ~ 7x10^4 km, then;

    integrating between (7x10^4) and (0.6x10^9) for (3.618 x 10^16)/(r^2) with respect to 'r', we get a value of ~ 500MJ.

    This is the energy you would convert into kinetic energy if you feel freely towards jupiter (obviously thus ignoring any other massive bodies)
    But I thought stephen said that you can't use mgh for high distances.... >__>
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    Have you studied calculus? This method uses integration with the mmg/r^2 equation to do mgh. When you use mgh it assumes that the 'ground' is an infinite plane, therefore the gravity doesnt decrease in strength. With spherical bodies in space it is necessary to observe the inverse square law.

    My calculations basically is doing mgh, but takes into account the fact the as the distance increases, the force of gravity varies in proportion to the inverse square of the distance.
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    Basically, yes, GPE = mgh should be increasing as h increases all other things equal. However, g = 9.8 m/s^2 is an assumption you make in mechanics and assume that it's always 9.8. However, in reality, as h increases, that g will become smaller and at a faster rate than h increases, so that mgh isn't increasing after a certain point.

    Most posters have missed the point.
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    (Original post by Swayum)
    Basically, yes, GPE = mgh should be increasing as h increases all other things equal. However, g = 9.8 m/s^2 is an assumption you make in mechanics and assume that it's always 9.8. However, in reality, as h increases, that g will become smaller and at a faster rate than h increases, so that mgh isn't increasing after a certain point.

    Most posters have missed the point.
    Yes that is correct, that is why it is necessary to use integration. The force of gravity tends to zero for very large values of 'r' so past a certain point any gains in GPE are insignificant. If you plot the function I integrated as a graph of force against distance, it will be assymptotic to the x-axis.

    edit: check this link, it will give you a better representation of the integral, and a picture of the graph I was describing. I hope this makes it more clear.

    http://www.wolframalpha.com/input/?i...80.6*10%5E9%29
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    (Original post by Deno_7951)
    I think this method would give you an accurate value of what your looking for. (sorry I don't know how to use latex)

    We know the equation of Newton's law of gravitation. We know GPE = mass x gravity x height

    If we take a person as weighing 60kg, the distance to jupiter as ~600million km, and your distance from the centre of jupiter at the surface as ~ 7x10^4 km, then;

    integrating between (7x10^4) and (0.6x10^9) for (3.618 x 10^16)/(r^2) with respect to 'r', we get a value of ~ 500MJ.

    This is the energy you would convert into kinetic energy if you feel freely towards jupiter (obviously thus ignoring any other massive bodies)
    Can I just ask how you got (3.618 x 10^16)? G x M x m doesnt give you that...

    Thanks
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    Sorry my apologies, I think I got a bad value for mass of jupiter lol (was in imperial tons or somthing ) the correct integrand would be then in fact;

    (8x10^18)/(r^2)

    It's a factor of 2 off, but the method is exactly the same. I'll post another link here for a correct integration and value.

    (also note that in this case G is the gravitational constant, not g (ie local field strength))

    link:

    http://www.wolframalpha.com/input/?i...8600*10%5E6%29

    That's around 100 Terra Joules. That's a lot of energy
 
 
 
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