Consider a simple model for the interior of the Earth: there is a spherical iron

Question

Consider a simple model for the interior of the Earth: there is a spherical iron core with constant mass density po and radius a: outside the core is 'rock* with constant density p. Use these values for the densities: pu = 8.90 x 103 kg/m3 and pi = 3.90 x 103 kg/m3. The radius of the Earth is R = 6.40 x 10 degree m. (a) First, calculate the mass of the Earth. [Hint: Use G and g.] Tries 0/25 (b) Calculate the radius a of the iron core. Tries 0/25 (c) Derive the gravitational field g(r) as a function of r. Hand in (in Room 1312) a numerically accurate graph of g(r). The value at r = R must be 9.81 m/s2. (d) Calculate g(a). [5J Currently operating in the short run with labor the variable input, a law firm faces the following hourly circumstances:

Explanation / Answer

( a ) g = GM / R^2 ;

9.8 = ( 6.673 e -11 ) M /   [   ( 6.4 e 6 ) ^2 ] ;

or M = 6.015 e 24 kg ;

( b ) M = 1 V1 + 2 V2 ;

M = ( 8.9e 3 ) ( 4 /3   a ^3 )    +   ( 3 .9 e 3 ) [ 4 /3     ( R ^ 3 -   a^3 )   ] ;

6.015 e 24 = ( 8.9e 3 ) ( 4 /3   a ^3 ) - ( 3.9 e 3 ) [ 4 /3 a^3 ]   + ( 3 .9 e 3 ) [ 4 /3    ] * R ^ 3 ;

or 1.732 e 24 = ( 5 e 3 ) ( 4 /3   a ^3 ) ;

or a = 4356754.529 m = 4356.754 km ;

( c ) g ( r ) =   G( menc ) / r^2   ;

m ( r ) = 0 * ( 4 /3   r^3 )

g ( r ) = G . 0 ( 4 / 3 ) r    for r < = a ;

g( r ) = G .   [    0 ( 4 /3 ) a + 1 * ( 4 /3 ) ( R^3 - a^3 ) ]   /    r ^ 2   ;      for    r >a   and r < R

the graph is linear from the origin , till point a , [ since force at centre is zero ] ,   after point r= a , it decreses according to the above given relation for r >a and r < R ;

( d )

g ( r ) = G . 0 ( 4 / 3 ) r    for r < = a ;

so g ( a ) = G . 0 ( 4 / 3 ) a = ( 6.673 e -11 ) * ( 8.9 e 3 ) ( 4 /3 ) ( 4356754.529 ) ;

g ( a ) = 10.83834 m /s ^2

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