On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole and g = ago at the equator (with 0 < a < 1). find g(9), the freefall acceleration at colatitude 9 as a function of 9.
The reason why there is a difference between free-fall acceleration is a centrifugal force. I attached a diagram that shows how this force aligns with the force of gravity. From the diagram we can see that: [tex]F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\[/tex] Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. [tex]\alpha[/tex] is the latitude. We can calculate g' and wr^2 from the given conditions in the problem. [tex]g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
[/tex] Our final equation is: [tex]g=g_0-g_0(a-1)cos^2(\alpha)[/tex] Colatitude is: [tex]\alpha_c=90^\circ-\alpha[/tex] The answer is: [tex]g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)[/tex]
