According to Fn=Fg-Fc , if earth's angular velocity is big enough to make Fc equal to Fg, would we feel weightless? I did some basic calculation earth has to rotate at 1 rev/5064s at equator for Fc to equal Fg.
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Qmechanic
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1Are you taking into account that Earth gets more deformed with more rotation? See e.g. this Phys.SE post and links therein. – Qmechanic Oct 16 '16 at 20:31
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@Qmechanic no, assuming earth is a rigid body and radius in fixed. – alend ahmed Oct 16 '16 at 20:35
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One problem you'll run into is that above a certain rotation rate objects tend to not remain spherical, or even oblate, but to start separating into two objects that are orbiting each other. The intermediate shape is kind of dumb-bell shaped. – Sean E. Lake Oct 16 '16 at 22:25
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Possible duplicate of How fast would the Earth need to spin for us to feel weightless? – sammy gerbil Oct 17 '16 at 12:22
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If the Earth spins with an period of 88 minutes the centripetal acceleration at the Earth's surface along the Equator is about $9.8\; \rm ms^{-2}$.
This means that your weight $mg$ is just the required force to keep you rotating with the Earth.
$mg - N = m R_{\rm E} \left ( \dfrac {2 \pi}{T}\right ) ^2$ where $N$ the normal reaction on you due to the Earth is zero $R_{\rm E}$ is the Earths radius
So if you drop an object it does not fall down towards the Earth but rotates with you.
Note that at the North and South Poles you would still have a weight of $mg$ and the normal reaction on you due the Earth would be $mg$ upwards.
However there are many other things which will happen amongst them the Earth losing its atmosphere and the Earth itself would "fly apart" because of the brittle nature of the Earth's crust.
Farcher
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Your weight is the normal reaction from the ground. When this normal reaction becomes zero, you are weightless.
sammy gerbil
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