How much mass with the centrifugal forces of earth to break gravity?

[jbriggs444]

Hold up, you just told me mass has nothing to do with centrifugal force, then you correct your equation, then you correct your statement saying, "yes, [increased mass does increase force]"

Then you proceed to brow beat me and offer no solution to a math problem that has to have a solution because MASS DOES MATTER. Please do not respond to this post anymore unless you have the answer. Thank you.

Struggling to find an interpretation of the problem that results in something meaningful...

What mass would the Earth have to have for the centrifugal force on an object resting on its surface to match the gravitational force on that object?

Let M be the mass of the earth, m be the mass of the object on its surface, r be the Earth's radius and G be Newton's universal gravitational constant.
Let $\omega$ be the rotational speed of the Earth (in radians per unit of time).

Gravitational force = $\frac{GmM}{r^2}$
Centrifugal force = $m\omega^2r$

Solve for M.

$\frac{GmM}{r^2} = m\omega^2r$
$\frac{GM}{r^2} = \omega^2r$
$GM = \omega^2r^3$
$M = \frac{\omega^2r^3}{G}$

 

[Orodruin]

So you agree that if the radius and velocity remain constant, that the size of the mass increasing will increase the force though, right?

Yes, but so will the force of gravity, which is also proportional to the mass. What is constant is he gravitational acceleration and the centripetal acceleration, not the gravitational force or the centripetal force. You cannot just make an assumption and then stubbornly claim that it is correct when it is in contradiction to experimental observations. Physics is an empirical science, not philosophy.

Thread closed.

 

[Janus]

Hi, thank you also.

According to Newton's second law the centripetal force can be expressed as

Fc = m ac

= m v2 / r

Are you saying that the centrifugal force is not derived from the same equation and centripetal force? Or are you saying that the result of centrifugal force is not the actual acceleration multiplier on the mass? I get that F has the mass/acceleration mixed in, but even "mass" has mass/acceleration mixed in because the "mass" is a result of gravity on the mass. Gravity is a constant so they use that constant to form weights. I am saying the energy expressed in this equation is a rate of energy transfer. Newtons are really a rate of energy transfer, but it has to have time or distance for it to express the rate.

So far this is what I think, centrifugal force is on mass, so it isn't "independent" of mass. Centrifugal force expresses the acceleration on the weight in Newtons. Gravity force is expressed as weight in Newtons as 143 N. Since I am expressing gravity this way, I expressed centrifugal force the same way. Both my equations are expressing the effects of acceleration as force, no?

I am saying that when comparing the two, you have to stick to acceleration or force in both cases. I don't know where you got that 143 N figure from, as gravitational force would vary from object to object and does depend on mass. The 143 N force you give would be the gravitational force acting on a ~14.56 kg mass sitting on the surface of the Earth, but an object with a different mass would experience a different gravitational force. When we measure the weight of something we are actually measuring the gravitational force acting on it.

Gravitational force between two masses is
F_g = \frac{GMm}{d^2}

Where G is the universal gravitational constant
M and m are the masses involved
d is the distance between the centers of the masses.

If M is the mass of the Earth, then GM has the value of 3.987x10^14 m³/s²
this is sometimes expressed as \mu

Since F=ma and ergo a=F/m

The acceleration due to gravity is
F_g/m
or
A_g = \frac{\mu}{d^2}

which depends of the distance from the center of the Earth where it is measured.

Using the figure for the radius of the Earth you gave, this gives an answer of 9.82 m/sec² this holds for any mass near the surface of the Earth.

As you already noted, centripetal force can be found by

F_c = v^2m/r

And as above centripetal acceleration is

A_c = v^2/r

Which again holds for any mass.

So there is no critical mass for an object resting the surface of the Earth for which the centripetal force needed to hold object on the surface against the spin of the Earth exceeds the force of gravity acting on it. The centripetal force does increase with mass, but so does the gravitational force (weight).

And as I pointed out earlier, the gravitational acceleration far exceeds the required centripetal acceleration.