runny4
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http://i-want-to-study-engineering.o...tational_pull/
Why is the effective mass of the ball at c less than at b?
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lerjj
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(Original post by runny4)
http://i-want-to-study-engineering.o...tational_pull/
Why is the effective mass of the ball at c less than at b?
Firstly, although I think I know what you mean by 'effective mass', you should probably be thinking of fixed masses, and variable gravitational fields. Specifically, you know that with gravitaional field g:

F=mg
g=G\dfrac{M}{r^2}

So now the question becomes: where is the field, g, the greatest? Remember, g is a vector so you can have negative contributions as well as positive ones.
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Guy123
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The mass is greatest at B because at C, there is less planetary mass beneath the object so less mass means weaker gravity. Thus, the object is lighter underground. Anywhere above B and the gravitational pull also decreases as the object is moving further away from the gravitational field.
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runny4
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(Original post by lerjj)
Firstly, although I think I know what you mean by 'effective mass', you should probably be thinking of fixed masses, and variable gravitational fields. Specifically, you know that with gravitaional field g:

F=mg
g=G\dfrac{M}{r^2}

So now the question becomes: where is the field, g, the greatest? Remember, g is a vector so you can have negative contributions as well as positive ones.
thanks
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runny4
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(Original post by Guy123)
The mass is greatest at B because at C, there is less planetary mass beneath the object so less mass means weaker gravity. Thus, the object is lighter underground. Anywhere above B and the gravitational pull also decreases as the object is moving further away from the gravitational field.
thanksImage
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runny4
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(Original post by Guy123)
The mass is greatest at B because at C, there is less planetary mass beneath the object so less mass means weaker gravity. Thus, the object is lighter underground. Anywhere above B and the gravitational pull also decreases as the object is moving further away from the gravitational field.
so is the effective mass the weight
and is a's mass greater than b's?
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Guy123
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(Original post by runny4)
so is the effective mass the weight
and is a's mass greater than b's?
Yes, the effective mass is basically the weight, the gravitational pull on an object.
No, b's mass is greatest because the surface is where the gravity is strongest. If you're further away, you're getting further from the centre of gravity, so the weight decreases.
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runny4
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(Original post by Guy123)
Yes, the effective mass is basically the weight, the gravitational pull on an object.
No, b's mass is greatest because the surface is where the gravity is strongest. If you're further away, you're getting further from the centre of gravity, so the weight decreases.
but wouldn't gravity be greater near the core so c's effective mass is greater than b's
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Guy123
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(Original post by runny4)
but wouldn't gravity be greater near the core so c's effective mass is greater than b's
No it wouldn't, because as you already know, objects with greater mass have stronger gravity. However, if you go deep underground, there is much less mass between you and the centre of the earth so you are effectively reducing the earth's mass and its gravity.
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runny4
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(Original post by Guy123)
No it wouldn't, because as you already know, objects with greater mass have stronger gravity. However, if you go deep underground, there is much less mass between you and the centre of the earth so you are effectively reducing the earth's mass and its gravity.
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Guy123
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(Original post by runny4)
thanks
No problem
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