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timofeeve [1]
4 years ago
7

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.

Physics
1 answer:
ohaa [14]4 years ago
5 0
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:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
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. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\&#10;g_0=g'\\&#10;g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\&#10;ag_0=g_0-w^2r\\&#10;w^2r=g_0(a-1)&#10;
Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

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