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Gravitational pull problem

http://i-want-to-study-engineering.org/q/gravitational_pull/
Why is the effective mass of the ball at c less than at b?
Reply 1
Avatar for lerjj
lerjj
13
Original post by runny4
http://i-want-to-study-engineering.org/q/gravitational_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 gg:

F=mgF=mg
g=GMr2g=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.
Reply 2
Avatar for Guy123
Guy123
16
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.
Reply 3
Avatar for runny4
runny4
OP
6
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 gg:

F=mgF=mg
g=GMr2g=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
Reply 4
Avatar for runny4
runny4
OP
6
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.


thanks
Reply 5
Avatar for runny4
runny4
OP
6
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?
Reply 6
Avatar for Guy123
Guy123
16
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.
Reply 7
Avatar for runny4
runny4
OP
6
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
Reply 8
Avatar for Guy123
Guy123
16
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.
Reply 9
Avatar for runny4
runny4
OP
6
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.

thanks
Reply 10
Avatar for Guy123
Guy123
16
Original post by runny4
thanks


No problem :biggrin:

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