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TimeRip496
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Homework Statement
$$g_o=\frac{GM}{R^2}e_R$$
where g0 is the gravitational acceleration, G = 6.67 * 10-11Nm2 /kg2 is the universal gravitational constant, M= 5.98 *1024 kg is the mass of the Earth, and R = 6.38 * 106 m its radius
$$g = g_0-ω*[ω*(r+R)]$$....(8.29)
where gravitational acceleration g that is felt near the surface of the Earth (i.e., on the surface of the Earth we cannot discerned between gravity g0 and the centrifugal acceleration -ω*[ω*(r+R)], we can only feel the resulting acceleration g )
The Earth’s angular velocity vector is given by ω= 7.3 * 10-5ez rad/s (i.e., it is directed along the z'-axis ), and we assume that it is a constant.
It is to be noted that because of the presence of the centrifugal acceleration
-ω*[ω*(r+R)] in this equation for the effective gravity, g and g0 will in general
not point exactly in the same direction. This effect is rather small, but measurable as
ωR^2/g_0= 0.0035 . It should also be clear from the equation (8.29) that the magnitude of the effect is a function of latitude.
Source:http://www.astro.uwo.ca/~houde/courses/PDF files/physics350/Noninertial_frames.pdf
where R is the radius of Earth, r is the position of the object measure from the rotating frame of reference which is moving at a constant velocity and constant angular velocity, r' is the position of the object measure from the center of the Earth
2. Homework Equations
ωR^2/g0= 0.0035 where R is the radius of Earth.
I have no idea how they get this equation. Besides I don't quite get what they mean about "that the magnitude of the effect is a function of latitude.".
The Attempt at a Solution
I have no idea how to get to this equation ωR^2/g0= 0.0035. I try to calculate the above equation but I get is 303234462.4 instead of 0.0035.
$$ωR^2/g_0=ωR^2*\frac{R^2}{GM}=(7.32*10^{-5})(6.38*10^6)^2*\frac{(6.38*10^6)^2}{6.67*10^{-11}*5.98*10^{24}}=303234462.4$$
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