Measuring g

The magnitude of the acceleration of gravity at the equator will be reduced. This is because the person at the equator is rotating along with the Earth's surface. Let $a_c$ be the centripetal acceleration and $a_g$ be the apparent gravitational acceleration at the equator. We have $a_c=r\omega^2$ , $\omega=\frac{2\pi}{T}$ and $a_g=9.81-a_c$ . Thus,

$a_g=9.81-r\left(\frac{2\pi}{T}\right)^2$

$a_g=9.81-6370000\left(\frac{2\pi}{24\times3600}\right)^2$

which yields $a_g=9.78$ .

(pi x d/ 24h)(pi x d/ 24h) = v squared v squared/r=a a+g= ans

The block on the equator has a centripetal acceleration while the block on the north pole does not. Therefore, while the force balance at the north pole is $N_{spring}-mg=0$ , where $N_{spring}$ is the normal force the spring exerts on the block, the force balance at the equator is $N_{spring}-mg=ma_c$ , where $a_c$ is the centripetal acceleration of the block. The experimenter measures the value of g by $g=N_{spring}/(1~kg)$ and so gets a different result at the equator. The centripetal acceleration is $a_c=-v^2/r=-\omega^2 r$ , where $\omega$ is the angular velocity of the earth. It's negative because it points towards the center of the earth (i.e. in the negative radial direction). The period T of the rotation is 24 hours or 86400 seconds, so $omega=2 \pi/T=7.27\times 10^{-5}~rad/s$ . Substituting in numbers, we have $N_{spring}=(1)(9.81)-(1)(7.27 \times 10^{-5})^2(6370000)$ . The measured value of g at the equator would therefore be $9.776~m/s^2$ .