Heres a highschool physics question thats got my g/f stumped Need help

[auslander]

Heres a high school physics question that's got my g/f stumped Need help:)

Here is the question.
Because the Earth rotates once per day, the effective acceleration of gravity at the equator is slightly less than it would be if the Earth didn’t rotate. Estimate the magnitude of this effect. What fraction of g is this?

Here is the answer= 7. 3.44 x 10-3

What she needs is the work that gets that answer so like the necessary equation to come up with it. Any help would be highly appreciated. THANKYOU!

 

[Curious3141]

Is the "7" in your answer a typo ?

I get the effect as being around 3.45*10^(-3)*g = 3.38*10^(-2) m/s/s.

Consider the centrifugal "force" and the resulting "acceleration" a mass on the Earth's equator feels as a result of the Earth's rotation. The equation for the (real) centripetal acceleration is $$a = r\omega^2$$. The (fictitious) centrifugal acceleration is of the same magnitude but directed outward. The radius of the Earth is approximately 6400 km. Figure out $$\omega$$ in terms of the period of the Earth's rotation, and you should be able to get the answer.

If you want a more rigorous treatment avoiding inertial "forces", then consider that the supposed force of gravity "felt" by a mass is actually the normal reaction force from the ground. Then equate $$F_N = F_G - F_C$$

where the subscripts N, G and C refer to "normal reaction", "true gravitational" and "centripetal" respectively.

 

[gulsen]

$$g_{observed} = g_{"actual" one} - \omega^2 r_{equator}$$
$$\omega$$ is the angular speed of earth, which is of course $$2 \pi$$ per 24 hours, or 7.27e-5 rad/sec approximately.
$$r_{equator}$$ is approximately 6378 meters. Taking $$g_{observed}$$ = 9.8

$$\frac{ \omega^2 r_{equator}} { g_{observed}} = 3.44 \times 10^{-6}$$

Are you sure you don't have a typo?

 

[Curious3141]

$$g_{observed} = g_{"actual" one} - \omega^2 r_{equator}$$
$$\omega$$ is the angular speed of earth, which is of course $$2 \pi$$ per 24 hours, or 7.27e-5 rad/sec approximately.
$$r_{equator}$$ is approximately 6378 meters. Taking $$g_{observed}$$ = 9.8

$$\frac{ \omega^2 r_{equator}} { g_{observed}} = 3.44 \times 10^{-6}$$

Are you sure you don't have a typo?

That cannot be right, it's 3 orders of magnitude too small. The radius of the Earth is 6400 kilometers, not meters !

 

[gulsen]

That cannot be right, it's 3 orders of magnitude too small. The radius of the Earth is 6400 kilometers, not meters !

Whoops! Another over-night :yuck: Yeah, my world is pretty small at this time of dawn!
BTW, to OP, how about posting such stuff in Homework & Coursework Questions -> Introductory Physics next time?

 

[auslander]

Thankyou guys for all your help :)

 

[Klizja]

Now why exactly is this force directed outwards ? From what I've learned in school , the force , which makes an object rotate is directed inward to the center of the trajectory.

 

[Hootenanny]

Now why exactly is this force directed outwards ? From what I've learned in school , the force , which makes an object rotate is directed inward to the center of the trajectory.

As curious states, the centrifugal force is ficticous. It originates from analysing motion in a non-inertial frame, i.e. using a reference frame fixed on the rotating body.

~H

 

[Klizja]

If so ,then the g should increase due to the Earth's rotation , not diminish.The gravitational force and the centripetal force should sum up , but instead you substract the centripetal force from the true gravitational.Why ?

 

[Curious3141]

If so ,then the g should increase due to the Earth's rotation , not diminish.The gravitational force and the centripetal force should sum up , but instead you substract the centripetal force from the true gravitational.Why ?

There are two ways to look at it. One is a physically non-standard way (teachers discourage you from using it) that's easier from an intuitive perspective. The other is a physically standard, conventional way that's also a little harder to explain intuitively.

The easier to explain way is using the concept of centrifugal force when considered from the non-inertial frame of reference of the object on the Earth's surface. Centrifugal force is felt as an outward push in this frame. There's nothing inherently unnatural about this, you would experience the same in a car or bus going around a corner, for example.

The harder (but physically more standard) way is to use Newtonian forces from the inertial frame of reference outside the Earth. Viewing from this frame and disregarding the Earth's rotation for a moment, there are two forces acting upon the body : an inward gravitational pull and an outward normal reaction force exerted by the Earth on the body. In the case of a nonrotating Earth, these two forces would balance exactly. The reaction force is what a person would perceive as "weight" (for example, in a lift going up or down, you perceive your apparent weight to change, purely based on the reaction force of the floor of the lift on you).

If we now get the Earth rotating and the body remains in place on the Earth, there has to be something to provide the centripetal force pulling the body in toward the center of rotation (center of the Earth). This is accounted for by the resultant of the other two forces acting on the body. The centripetal force, therefore, is equal to the gravitational pull inward minus the reaction force from the ground.

What is the net effect on the "apparent weight" perceived by the body ? Well, if you do the algebra, you'll find that the normal reaction force has now been reduced by a small amount to bring everything back into balance. This is why the body now feels lighter.

 

[Klizja]

Thanks a lot !