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I'd like to finalize the conclusion from @AaronStevens's great answer. In the truer expression for normal force (on flat ground) that he arrives at,

$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$

Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?

Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, from the top of my head, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.

If we plug in $r$ and $v$, we get something like:

$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$

Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin to the effective gravitational acceleration $(g-\frac{v^2}{r})$ is thus only something like 0.3 %. You can try to calculate a normal force for an object with and without this indluence and see if there is a significant difference within significant figures.

I'd like to finalize the conclusion from @AaronStevens's great answer. In the truer expression for normal force (on flat ground) that he arrives at,

$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$

Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?

Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, from the top of my head, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.

If we plug in $r$ and $v$, we get something like:

$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$

Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin to the effective gravitational acceleration $(g-\frac{v^2}{r})$ is thus only something like 0.3 %. You can try to calculate a normal force for an object with and without this influence and see if there is a significant difference within significant figures.

I'd like to finalize the conclusion from @AaronStevens's great answer. In the true expression for normal force (on flat ground),

$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$

Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?

Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, from the top of my head, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.

If we plug in $r$ and $v$, we get something like:

$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$

Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin to the effective gravitational acceleration $(g-\frac{v^2}{r})$ is thus only something like 0.3 %. You can try to calculate a normal force for an object with and without this indluence and see if there is a significant difference within significant figures.

I'd like to finalize the conclusion from @AaronStevens's great answer. In the truer expression for normal force (on flat ground) that he arrives at,

$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$

Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?

Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, from the top of my head, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.

If we plug in $r$ and $v$, we get something like:

$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$

Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin to the effective gravitational acceleration $(g-\frac{v^2}{r})$ is thus only something like 0.3 %. You can try to calculate a normal force for an object with and without this indluence and see if there is a significant difference within significant figures.

I'd like to finalize the conclusion from @AaronStevens's great answer. In the true expression for normal force (on flat ground),

$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$

Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?

Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.

If we plug in $r$ and $v$, we get something like:

$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$

Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin in the term $(g-\frac{v^2}{r})$ is thus only 0.3 %. You can try to calculate a normal force for an object with and without this indluence and see if there is a significant difference within significant figures.

I'd like to finalize the conclusion from @AaronStevens's great answer. In the true expression for normal force (on flat ground),

$$N=mg-\frac{mv^2}{r}=m\left(g-\frac{v^2}{r}\right)\quad ,$$

Earth's rotation adds the term $\frac{v^2}{r}$ so it deviates from the expected $N=mg$. How much is the influence of $\frac{v^2}{r}$?

Earth's radius is around $r=6400\;\mathrm{km}$. In one day, which is $t=24\,\mathrm{hr}=86400\,\mathrm s$, we move through the entire circumference of Earth, which is $d=40200\,\mathrm{km}$. That gives us a constant speed of around $v=d/t=465\,\mathrm{m/s}$. I am aware that I have used rough numbers here, from the top of my head, mainly fitting from the Equator. You can try to redo the calculations with more accurate values.

If we plug in $r$ and $v$, we get something like:

$$\frac{v^2}{r}=0.0338\,\mathrm{m/s^2}$$

Compare this with $g=9.80\,\mathrm{m/s^2}$. The contribution of Earth's spin to the effective gravitational acceleration $(g-\frac{v^2}{r})$ is thus only something like 0.3 %. You can try to calculate a normal force for an object with and without this indluence and see if there is a significant difference within significant figures.