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At the equator, it is said that the weight of the object will be lesser than that at the poles. This is because of the circular motion at the earth. The explanation given is that the Earth’s gravitational field strength at the Equator is responsible for both the centripetal acceleration of the falling ball and its free fall acceleration. As such, the free fall acceleration measured will be less than the gravitational acceleration if there was no centripetal acceleration.

However, how can we say that part of gravitational acceleration provides for the centripetal acceleration when centripetal acceleration is itself the net force on the object?

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  • $\begingroup$ Is this a quote from a textbook? Which page? $\endgroup$
    – Qmechanic
    Jul 17, 2021 at 11:21

2 Answers 2

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The net downward force is what your weight is that you measure. Considering the earth moving frame, there is an inward gravitational force and an outward centrifugal force So that net downward force given by
$$F_\text{net}=mg\hat{r}-mr\Omega_\text{Earth}^2\hat{r}=mg_\text{eff}<mg$$ $$W_\text{equator}<W_\text{pole}$$


How can we say that part of gravitational acceleration provides for the centripetal acceleration when centripetal acceleration is itself the net force on the object?

You can understand it using $$F_r=m(\ddot{r}-r\dot{\theta}^2)=-mg\rightarrow m\ddot{r}=-mg+r\dot{\theta}^2$$ So the gravitational force provide two type of forces, One the inward centripetal and other radial acceleration. And because of cetripetal acceleration, the radial acceleration get lower so does the force that is weight.

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  • $\begingroup$ Hi thanks for your reply, but i don't quite understand the last line. What does (r¨−rθ˙^2) represent? $\endgroup$
    – john
    Jul 17, 2021 at 10:01
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    $\begingroup$ just fyi there are three effects that contribute to the pole/equator difference in the acceleration experienced by a person standing on the surface; differences in $GM_E/r^2$, $r \dot{\theta}^2$ and the Earth's quadrupole moment $$ a_r = J2 \frac{1}{r^4} \frac{3}{2} \left( 3 \sin^2 \theta - 1 \right) $$. See here and links therein, and refer to DavidHammen's second table. $\endgroup$
    – uhoh
    Jul 17, 2021 at 11:11
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    $\begingroup$ @uhoh I'm taking the approximation in which $g$ is constant. $\endgroup$ Jul 17, 2021 at 11:12
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    $\begingroup$ yep, carry on, it's just that the other two effects are just as large, so it's not so much of an approximation as it is a fundamentally wrong assumption. $\endgroup$
    – uhoh
    Jul 17, 2021 at 11:13
  • $\begingroup$ @john In the polar coordinate one writes the force as $$m(\ddot{r}-r\dot{\theta}^2)\hat{r}+m(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}=\mathbf{F}$$ I used the radial part for radial acceleration. $\endgroup$ Jul 17, 2021 at 11:14
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Weight can be measured with a spring balance as long as the object being weighed and the spring balance are not accelerating.

If you take an object and a spring balance that are stationary in space near the surface of the Earth, the spring balance will register a force $mg$. This is the true weight of the object.

But if you accelerate the spring balance and the object downwards (towards the centre of the Earth) the spring balance will register a force less than $mg$ because it and the object are no longer in equilibrium. The spring balance now shows the apparent weight of the object. Its true weight is still $mg$, but its apparent weight is less than $mg$.

Similarly, if you place the spring balance and the object at a fixed point on the equator, they are now rotating with the Earth so once again they are no longer in equilibrium. The reading on the spring balance, which is the apparent weight of the object, will be less than $mg$ because the spring balance and the object are accelerating towards the centre of the Earth (centripetal acceleration). But the true weight of the object is still $mg$.

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