How Does Earth's Rotation Affect Acceleration at the Equator?

[Benzoate]

Homework Statement

A body is at rest at a location on the Earth's equator. Find its acceleration due to the Earth's rotations. [Take the Earth's radius at the equator to be 6400 km.]

Homework Equations

Possible relevant equations: a=(dv/dt)*t + (v^2/rho)*n, n is the unit normal vector., F= M*m*G/(R^2(earth))

The Attempt at a Solution

If the body is at rest , then the velocity of the body must be zero. F=M*m*G/(R^2(earth))=> m*(a+g)= M*m*G/(R^2(earth)) => a+g= MG/(R^2(earth)) , M being the mass of the earth. Therefore a= MG/(R^2(earth))-g = (6.0e24)*((6.6e-11))/((6.4e6 m)^2) - 9.8 = -.0294 m/s^2.

The actually answer is .034 m/s^2. What is wrong with my calculations?

 

[dynamicsolo]

The Attempt at a Solution

If the body is at rest , then the velocity of the body must be zero. F=M*m*G/(R^2(earth))=> m*(a+g)= M*m*G/(R^2(earth)) => a+g= MG/(R^2(earth)) , M being the mass of the earth. Therefore a= MG/(R^2(earth))-g = (6.0e24)*((6.6e-11))/((6.4e6 m)^2) - 9.8 = -.0294 m/s^2.

The actually answer is .034 m/s^2. What is wrong with my calculations?

One thing that's going to give you trouble is that none of your data is precise enough to get the answer accurately to three decimal places (you are plainly in the right neighborhood, but you couldn't properly give the result to better precision than 0.03 m/sec^2). You should use at least three significant figures for G and the Earth measurements...

But there's a separate problem -- where is that 9.8 coming from and is it really '9.8'? You're actually making the problem harder than it needs to be. Since the object is at rest right on the Equator, you only need to work out the centripetal acceleration

$$a_c = \frac{v^2}{R_E} = \omega ^2 \cdot R_E$$

It will probably be easier to deal with using the Earth's equatorial radius and its angular speed, $$\omega = \frac{2 \cdot \pi}{T}$$. Note: you'll want to use the length of the sidereal day, the actual period of Earth's rotation (23 hrs. 56 min. 4.091 sec.), rather than the solar day of 24 hours.

 

[Dick]

Whoa, whoa. M_earth*G/R_earth^2 IS g. What you are computing there is the roundoff between the exact value of g and the approximation g=9.8m/s^2. The acceleration due to the rotation of the Earth is the centripetal acceleration v^2/r. Compute that.

 

[dynamicsolo]

Whoa, whoa. M_earth*G/R_earth^2 IS g. What you are computing there is the roundoff between the exact value of g and the approximation g=9.8m/s^2.

You are correct, though there is a small subtlety that comes up if you go to look up values. The relation that you give is what is generally taken as g and is where the number 9.807 m/sec^2 comes from. But if you look up the measured value of Earth's gravitational acceleration (at mean sea level), you find something more like 9.78 m/sec^2 (what I believe some people refer to as g_o). The difference is largely due to what Benzoate is being asked to compute.

In fact, local values of "g" vary about 9.78 by a few hundredths of a m/sec^2 over the Earth's surface at different latitudes and altitudes -- to say nothing of "anomalies" due to variations in the planet's density (the so-called "mass concentrations and deficits": the relics of Earth's formation from planetesimals).

 

[Dick]

Right, right. But what Benzoate is asked to compute is the acceleration due the the Earth's rotation. That's pure v^2/r. Nothing to due with estimated values of g or planetesimals or anything. Benzoate is just doing the wrong calculation.

 

[Benzoate]

One thing that's going to give you trouble is that none of your data is precise enough to get the answer accurately to three decimal places (you are plainly in the right neighborhood, but you couldn't properly give the result to better precision than 0.03 m/sec^2). You should use at least three significant figures for G and the Earth measurements...

But there's a separate problem -- where is that 9.8 coming from and is it really '9.8'? You're actually making the problem harder than it needs to be. Since the object is at rest right on the Equator, you only need to work out the centripetal acceleration

$$a_c = \frac{v^2}{R_E} = \omega ^2 \cdot R_E$$

It will probably be easier to deal with using the Earth's equatorial radius and its angular speed, $$\omega = \frac{2 \cdot \pi}{T}$$. Note: you'll want to use the length of the sidereal day, the actual period of Earth's rotation (23 hrs. 56 min. 4.091 sec.), rather than the solar day of 24 hours.

So when an object at rest near the surface of the earth, I should act as if the object is rotating around the Earth instead of the Earth rotating while the object is motionless near the Earth's equator?

 

[Dick]

If the object is not rotating with the Earth then there is no acceleration due to the Earth's rotation, is there??

 

[dynamicsolo]

Right, right. But what Benzoate is asked to compute is the acceleration due the the Earth's rotation. That's pure v^2/r.
... Benzoate is just doing the wrong calculation.

No argument there. I was pointing out the hazard of using "g" too casually in a problem, as was done in the OP. As you point out, the result in the first post should be exactly zero.

 

[dynamicsolo]

So when an object at rest near the surface of the earth, I should act as if the object is rotating around the Earth instead of the Earth rotating while the object is motionless near the Earth's equator?

Ah, the problem is ambiguously stated: they mean that the object is at rest with respect to the surface of the rotating Earth, which mean it is revolving around the center of the Earth once a sidereal day.