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This is the radius that was given to us

r= 6400km

What I've tried is figuring out what is the force that would push the people off the equator,and what is the force that would "pull" people in. The force that would be pushing people off the equator would be the centrifugal force.

Now to calculate the force I'd use this

$$ F = m * \frac {v^2} {r} $$

For the velocity I've simply calcuated the speed that the people at the equator feel when the earth is spinning (Note: I've taken the radius to be the given 6400 km) and it is about v = 1674,6 km/h.Now I'm not sure what I should put in as my mass,since I'm not given a certain mass to calculate with.This is kind of a problem,since I cannot get the centrifugal force.What do I do here?Also after getting the centrifugal force I'm not quite sure what to do next. I'd pressume I'd have to calculate the gravity force that is keeping us in, but than how would I, from these two forces be able to determine what is my maximum speed that can still keep me on the earth?

Thank you!

EDIT: So after some more help this is what I've got, a suggestion said that I can use the mg formula where my g is this

$$ g = \frac {G M} r^2 $$

where M is the mass of the earth. So I did that

$$ g = 9,7 m/s^2 $$

Note(r was 6400 * 10^3 m and M was 5,972 *10^24 kg according to google)

$$ m * \frac {v^2} {r} = m * g $$ If we neglect the mass

$$ \frac {v^2} {r} = g $$ We multiply everything with r to get rid of the fraction

$$ v^2 = g * r $$

$$ v = \sqrt {g *r} $$

v = 7897,08 m/s

Thoughts?

G. Smith
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codeisfun
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  • Okay if the mass shouldnt matter than my equation should look like this $$F = \frac {v^2} {r} $$? – codeisfun Nov 15 '20 at 15:04
  • Terminology - centrifugal force is an apparent force in a rotating frame of reference. Let's not view things in the earth's rotating frame, so let's focus on the centripetal force - the real force that allows someone on the earth's surface to move in a circular path. You realize that it's gravity that is providing this force, so equate the force of gravity at the earth's surface to the formula for centripetal force and solve for v. What would happen if the velocity exceeded this value of v? – Not_Einstein Nov 15 '20 at 15:05
  • The equation you wrote is not dimensionally correct. What @Lewis Miller meant is that the mass cancels out on both sides of the equation, not just one side. – Not_Einstein Nov 15 '20 at 15:07
  • You are getting your masses muddled up. There are two masses: $m$ the mass of a person, and $M$ the mass of Earth. Write $m v^2/r = G M m/r^2$ and take it from there. Or, for the right hand side you can also use $g m$ where $g = G M/r^2$. – Andrew Steane Nov 15 '20 at 15:47
  • Your second equation does not follow from your first. m1 should be m, the mass of the people. Do your algebra carefully! – S. McGrew Nov 15 '20 at 15:48
  • Possible duplicates: https://physics.stackexchange.com/q/136486/2451 and links therein. – Qmechanic Nov 15 '20 at 15:57
  • No,that does not answer my question, that is a lot more advanced than my problem. – codeisfun Nov 15 '20 at 16:07

2 Answers2

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The earth's rotational speed would have to increase to where at the surface the centripetal force required to keep an object on the surface would exceed the force of gravity. This neglects any friction between the object and the earth's surface. That is, $mg < mv^2/r$ and $v = \omega*r$ where $\omega$ is the rotational speed.

With the earth's current speed of rotation, the centripetal force is small compared to the force of gravity, even at the equator. If you consider the problem from a rotating non-inertial reference frame, the inward centripetal force is balanced by the outward centrifugal force, and you weigh a little less at the equator than at the poles.

As others have commented, the centrifugal force is present only in a non-inertial reference frame rotating with the object. You can evaluate the problem from either an inertial or a rotating (non-inertial) frame of reference. In the inertial frame there is only the force of gravity to provide a centripetal force; in the non-inertial frame there is also the centrifugal force.

This is the same problem as figuring out the speed of a satellite in orbit except the distance here is the earth's radius instead of the satellite's distance.

See a basic physics text, or a physics mechanics text for more details. Also, you can find a good overview on Wikipedia, Orbital Speed.

For further study, you may wish to consider what happens to an object thrown up from the surface of the earth; from the point of view of the observer rotating with the earth this involves "fictitious" forces including the Coriolis force. This is discussed in physics mechanics textbooks.

John Darby
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You can do a back-of-the-envelope calculation using the fact that the centrifugal force at the equator with a $24$ hour day is about $0.3 \%$ of the force of gravity, and centrifugal force is proportional to the square of $v$ or $\omega$,.

gandalf61
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