7.2.4 True g and Apparent g

In certain situations, there is a difference between gravitational acceleration g, aka the true g, and the acceleration of free fall g.

Let’s revisit the elevator-man problem. Let’s have the man experiencing a gravitational pull of mg and the elevator accelerating downward at a.

Applying N2L on the man, we have


We say that the man is experiencing an apparent weight of mg-ma, which is less than his true weight of mg.

We can write N as mg’, where g’ denotes the apparent g. We then have


So another way to interpret this scenario is that the observable value of g in the elevator is not the true g, but the apparent g, g'=g-a. If the man were unaware that he is in an accelerating elevator and therefore assume that he is stationary, then as far as he is concerned, the acceleration of free fall is g-a. But to people outside the elevator, that’s only the apparent g.

Realize that it also takes a longer time for a “free-falling” object dropped from rest in the elevator to accelerate and reach the elevator floor (because the floor itself is accelerating downward at a). Again, as far as the man is concerned, the measurable acceleration of free fall is g-a. Again, to people outside the elevator, what he measured is only the apparent g.

A similar effect is caused by the Earth’s rotation. This time, it is our turn to be in the elevator. And instead of a downward acceleration, the elevator is having a centripetal acceleration.

We must remember that the Earth is rotating. This means that objects which seem to be stationary are actually moving along circular paths. The radius of circular motion depends on the latitude; Objects on the equator undergo circular motion with the largest radius and speed. The North and South poles are the only two locations where there is no circular motion.

Let’s consider a mass m “resting” on the equator. We know that this mass is actually in circular motion. And the radius of the circle is the radius of the Earth, R.

Applying N2L on m, we have


Writing N as mg’, we obtain

\displaystyle g'=\frac{N}{m}=g-{{a}_{c}}

So one way to interpret the math is that a fraction of the gravitational acceleration (true g) is “used up” to provide the required centripetal acceleration for circular motion. What’s left is thus the acceleration of free fall (apparent g or g’).

Since the Earth rotates one round every 24 hrs, the centripetal acceleration on the Equator is

\displaystyle {{a}_{c}}=R{{\omega }^{2}}=(6400\times {{10}^{3}}){{(\frac{{2\pi }}{{24\times 3600}})}^{2}}\approx 0.03\text{ m }{{\text{s}}^{{-2}}}

So g=9.81\text{ N k}{{\text{g}}^{{-1}}} is the true g whereas g'=9.81-0.03\approx 9.78\text{ N k}{{\text{g}}^{{-1}}} is the apparent g on the equator. As such, a 1−kg mass weighs 9.81 N at the poles but 9.78 N at the equator. Similarly, all masses free fall at 9.81 m s-2 at the poles but 9.78 m s-2 at the equator. It is only a small 0.3% difference, which explains why it is hardly noticeable.

Apparent Weightlessness

The difference between the true g and apparent g in an orbiting spacecraft however is HUGE. Since the gravitational force is providing the required centripetal force for the spacecraft to orbit the Earth,

Notice that \displaystyle \frac{{GM}}{{{{r}^{2}}}} is the true g, whereas r{{\omega }^{2}} is the required centripetal acceleration ac.

So the apparent g is

\displaystyle g'=g-{{a}_{c}}=\frac{{GM}}{{{{r}^{2}}}}-r{{\omega }^{2}}=0

That’s why everything is “weightless” in the ISS! The true g (about 8.7 m s-2 at ISS’s orbital altitude) is completely “used up” to provide the required centripetal acceleration to keep the ISS (and all its contents) in circular motion, leaving an apparent g of g'=0. When a ball is released from rest in the ISS, it simply hovers. It does not even start to fall! When a 1−kg mass is placed on a weighing scale, the scale reads zero. The ball does not even remain in contact with the scale.

Video Explanation

Rotational Speed of the Earth

How Earth’s Rotation Affect g?

Apparent Weightlessness

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