# What is the magnitude of the centripetal acceleration

1. Oct 14, 2006

### ubiquinone

Hi, I have a question regarding centripetal acceleration that I'm not certain on how to solve. I was hoping if anyone may please help me out with this question. Thanks.

Question: a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? b) What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration of magnitude $$9.8m/s^2$$

Part a) I think I just use the formula for centripetal acceleration (i.e. $$a_c=\frac{v^2}{R}$$, where $$v$$ is the speed of the object around the circumference of a circle and $$R$$ is the radius of the circle)
Hence, I believe the values that I need to substitute into this formula are $$R=6370 km$$ (the radius of the Earth. However, what would $$v$$ be in this case? Is it the speed at which the Earth orbits around its axis? If so, how may I find this value, is it a constant that I can find in physics books?

Part b) Since, another way to express the centripetal acceleration is $$a_c=\frac{4\pi^2r}{T^2}$$ where r=radius and T=period, we can solve for T, $$T=\sqrt{\frac{4\pi^2 r}{a_c}}$$
We can substitute $$r=6370\times 1000 m$$ and $$a_c=9.8m/s^2$$
We get $$T=\sqrt{\frac{4\pi^2(6370000m)}{9.8m/s^2}}=5065.7s$$

2. Oct 14, 2006

### gunblaze

Unless, well.., you are given the linear speed of the object on the surface of the earth, you can't use the formula above $$a_c=\frac{v^2}{R}$$. What you could do, is to use the other formula to calculate centripetal acceleration.< ie: a= W^2 X R> You will be able to find W since it is given by <2 pi/T> where T is the period of one revolution of the earth which is the time in seconds of 1 day.

Last edited: Oct 14, 2006
3. Oct 14, 2006

### ubiquinone

Hi there gunblaze!! Thanks alot for your help in explaining part a) so it should resemble something like this:
$$a_c=\omega^2r$$, where $$\omega=\frac{2\pi}{T}$$
$$T= 1day\times\frac{24h}{day} \times \frac{60min}{h}\times\frac{60s}{min}=86400s$$
Therefore, $$\displaystyle a_c=\left(\frac{2\pi}{86400s}\right)^2(6370\times 1000m)\approx 0.034m/s^2$$

Is it right now? By the way is part b) correct as well? Thanks again for your time and patience.

Last edited: Oct 14, 2006
4. Oct 14, 2006

### gunblaze

yep. Its correct now for both parts.