The newly discovered spherical planet has a radius of R = 4*10^7 m and a variabl

Question

The newly discovered spherical planet has a radius of R = 4*10^7 m and a variable density. Geological evidence indicates it is most dense at its center (core) and least dense at its surface. The functional relationship describing it's density has been determined as follows: rho(r) = A - Br with A = 14000 kg/m^3 B = 2 kg/m^(7/2) where 0<= r<= R. Modeling the planet as a series of concetric spherical shells, determine the following physical quantities..with including the computational results: B) the surface gravity g(R):

Explanation / Answer

STEP 1: Determine the mass function for the planet:

Start with a little dimensional analysis:

rho(r) is a mass-density function and should have units: Mass / Volume.

However, we need to normalize Br to have units: Mass / Volume. What are the units? Well they must be 1/m^(-7/2 + 3) = 1/m^(-1/2), because 7/2 - 1/2 = 3 which gives us Mass / Volume.

So, we will choose tau^(-1/2) instead of r. Now, we have a rho(tau):

( ho( au) = A-B au^{-1/2})

So to get the mass function, M(r) for the sphere, it is necessary to integrate over it's volume:

(M(r) =int ho(r) dV = int (A-Br^{-1/2})dV)

(dV = r^2sin( heta)dr d heta dphi)

(0 le r)

(-{pi over 2 }le heta le {pi over 2})

(0 le phi le 2pi)

(M(r) = int_{0}^{2pi} int_{-{pi over 2}}^{pi over 2} int_{0}^{r} {(A-Br^{-1/2})r^2 sin( heta) dr d heta dphi })

(=int_{phi} int_{ heta} ( {A over 3}r^3 - {2B over 7}r^{7/2} )sin( heta) d heta dphi)

(= int_{phi} 2( { A over 3}r^3 - {2B over 7}r^{7/2} ) dphi )

(= 4pi ( {A over 3}r^3 - {2B over 7}r^{7/2} ) = { { 4pi r^2} over 21 }(7Ar - 6Br^{3/2}))

(g(r) = -G{M over |r|^2})

(G = 6.67300 imes 10^{-11} { m^3 over {kg s^2} })

However, our M varies with radius, so the g(r) we will use is defined as:

(g(r) = -G{ M over |r|^2 } = { {-4Gpi r^2} over 21 }{ { 7Ar - 6Br^{3/2} } over {|r|^2} } = { { -4Gpi } over { 21 }}{ (7Ar - 6Br^{3/2} )})

STEP3: Compute g(R)

Now all we have to do is plug in r = R = 4*10^7, A=14000, B=2, G=6.673 * 10^-11

(g(R) pprox -3.99311*10^{-11}( 7(14000)(4*10^7) - 6(2)(4*10^7)^{3/2} ) pprox -35.307) meters/s^2

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