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    http://www.physicscentral.com/poster-earth.cfm

    I've seen this problem presented a few times, where a hole is drilled through the centre of the earth and you have to work out how long it would take to reach the other side.

    I think it's all very straight forward, apart from 1 bit that is really annoying me, I'm sure it's something stupid that I'm missing but I'm tired and it's really pissing me off.

    At the point where he starts talking about density

    p=M/V etc V= 4/3(pie)r^3

    How does he manage to get

    F=

    from

    a=

    Can someone show a step by step walkthrough? I've written down all the equations and can see how they would fit but it just doesn't seem to be happening for some reason....

    I feel retarded.
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    Watch "The Core" its science will tell you everything about getting to the centre of the earth.
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    (Original post by atheistwithfaith)
    Watch "The Core" its science will tell you everything about getting to the centre of the earth.
    Thanks for the useful advice.......
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    (Original post by suneilr)
    That's just the Law of gravitation.  \frac {r}{R^3} just gives the vector direction. The r should be underlined or bold to show it's a vector

    Mmm...

    What was the point in the whole paragraph about subsituting for density and volume etc?

    I don't see how he's made it any simpler >.< surely he still has the same variables and stuff...

    /confused.
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    Well the only mass that has a net gravitational pull on an object is the mass "below it". So the total mass of the earth is:

    \displaystyle M_E = \frac{4}{3} \pi \rho R_E^3 where R_E is the radius of the earth.

    If you are at a radius r from the centre of the earth where r < R_E then the mass "below you" is:

    \displaystyle M = \frac{4}{3} \pi \rho r^3 = M_E \frac{r^3}{R_E^3}

    so the force acting on a mass m at a radius r < R_E is:

    \displaystyle |F| = \frac{GMm}{r^2} = \frac{G M_E m r^3}{R_E^3 \,r^2} = \frac{G M_E m r}{R_E^3}

    Hope this helps
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    Yup ignore what I wrote...
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    reach the other side? you wouldnt... unless u assume no losses :awesome:
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    (Original post by div curl F = 0)
    Well the only mass that has a net gravitational pull on an object is the mass "below it". So the total mass of the earth is:

    \displaystyle M_E = \frac{4}{3} \pi \rho R_E^3 where R_E is the radius of the earth.

    If you are at a radius r from the centre of the earth where r < R_E then the mass "below you" is:

    \displaystyle M = \frac{4}{3} \pi \rho r^3
    Just after this is where I lost you =/

    2 hours sleep in 2 days must be really taking it's toll on me...

    I can see there is some sort of relationship between the first equation you wrote and then how you got to here = M_E \frac{r^3}{R_E^3} but I can't seem to see just how you did it...

    I think once I get that small bit the rest should slot into place, again, sorry for my stupidity.
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    (Original post by You Failed)
    Just after this is where I lost you =/

    2 hours sleep in 2 days must be really taking it's toll on me...

    I can see there is some sort of relationship between the first equation you wrote and then how you got to here = M_E \frac{r^3}{R_E^3} but I can't seem to see just how you did it...

    I think once I get that small bit the rest should slot into place, again, sorry for my stupidity.

    Well the part  \frac{4}{3} \pi \rho = \frac{M_E}{R_E^3} = \frac{M}{r^3} , this is in both equations.

    so multiply both sides by r^3 and you have:  M = \frac{M_E r^3}{R_E^3} .

    Any better?
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    (Original post by div curl F = 0)
    Well the part  \frac{4}{3} \pi \rho = \frac{M_E}{R_E^3} = \frac{M}{r^3} , this is in both equations.

    so multiply both sides by r^3 and you have:  M = \frac{M_E r^3}{R_E^3} .

    Any better?
    Ah I got it, it didn't occur to me to use  \frac{4}{3} \pi \rho as a common thing and re-arrange and sub stuff in using that.

    I really do feel bad though that I've finished my A Level Physics course and am going to do Physics in Uni and yet I couldn't do that entire process by myself without having to make some stupid post on TSR...
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    (Original post by You Failed)
    Ah I got it, it didn't occur to me to use  \frac{4}{3} \pi \rho as a common thing and re-arrange and sub stuff in using that.

    I really do feel bad though that I've finished my A Level Physics course and am going to do Physics in Uni and yet I couldn't do that entire process by myself without having to make some stupid post on TSR...

    I wouldn't worry too much about it to be honest, we all have mental blanks on things that can seem very obvious to others. At Uni you'll learn a method of doing this calculation using surface integrals (a form of Gauss' law) which is much easier than the above calculation, it basically does all the work for you.
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    I'm probably being dumb.
    but cant you multiply both sides by m, then asumming r=R, times the denominator and numerator by r?
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    (Original post by lm_wfc)
    I'm probably being dumb.
    but cant you multiply both sides by m, then asumming r=R, times the denominator and numerator by r?
    R and r are different variables.
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    (Original post by Scipio90)
    R and r are different variables.
    standing for?
    distance to centre and radius of the earth?
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    (Original post by lm_wfc)
    standing for?
    distance to centre and radius of the earth?
    Yup:yes:
 
 
 
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