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Q & A: Rotational speed of the earth at the equator

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Q:
Lets assume for simplification that the earth is a huge uniformly dense sphere spinning around an axis through its centre, and we are particles on its surface(rough enough to hold us in position when we are in contact with it)exactly at the equator.We know that the linear(not angular)speed of rotation of a point on the earth's surface is very fast(not sure but maybe around 3000km per sec).Then why doesn't the earth move with this tremendous speed beneath us when we jump? If its because we, as well as the earth, are rotating around the same axis, then WHAT provides the centripetal force for our rotation?It cannot be the gravitational force because its direction, during the jump, does not change(always along the same line connecting the 2 centres of mass) and a centripetal force must constantly change direction to face the axis of rotation. I know that there's a trick somewhere or I'm missing something because the textbook I'm using talks very briefly about this saying only that this centripetal force is supplied by part of the gravitational force(the other part being the observed weight). Please clarify and thanks in advance.
- Mohammed (age 17)
A:
First of all, the rotational speed of the surface of the surface of the earth is more like  v =  465 meters per second, not 3000 kilometers per second. The fundamental principle here is the conservation of angular momentum.  At the surface of the earth the angular momentum of a body of mass m is L = mvR where R is the radius of the earth.  As you ascend to a height, h, your tangential velocity decreases by a factor of δv in order to have Rv = (R+h)*(v- δv).   The earth seems to move a bit faster than you and you would land a very small distance to the west.  An integral is involved in order to get the exact answer.
Newton predicted in 1679 that if you dropped a ball from a height of 8 meters it would land about a half millimeter to the east (the algebraic sign of h changes).

LeeH

I think perhaps you are asking a little different question than the one Lee answered. Let's look at the situation in a non-rotating reference frame at rest with respect to the center of the earth. Initially you and the surface of the earth are moving in the same direction (east) at the same speed. When you jump, you pick up some upward velocity, but that does nothing to reduce your sideways velocity. So you continue to east. The direction in which gravity pulls you does change, because as you move east the direction from you to the center of the earth changes. Then Lee's calculations take care of the small effects that arise from your changing distance from the earth.

Mike W.

(published on 11/10/11)

Follow-Up #1: frictionless orbits

Q:
In your reply you said:- When you jump, you pick up some upward velocity, but that does nothing to reduce your sideways velocity. My question is :- If somehow an object remains up at some height from the Earth's surface without any attachment with the surface, like for example if Earth's equator were wrapped by a magnetic belt with N polarity and a magnet with N polarity put above it floating freely, then would it come to rest after sometime while Earth continues to move below?
- Indrajit Kuri
New Delhi, India
A:
I think I see what you're getting at. Pure magnets won't work for the sort of support you're describing. Earnshaw's theorem proves that there is no way to stably float that magnet just using other magnets. However, there is a physical way to make what you're interested in. Say that the floater is a superconductor that, like most, expels magnetic fields, and you make a sort of magnetic groove to support it.Now we have something supported with (almost) no friction.

The answer is that without friction the mechanical energy of the floater won't change. If it stays at the same height, its kinetic energy won't change. It'll keep moving at the same speed as it started. If the frictionless guide is at fixed latitude, if the floater is initially moving just fast enough to keep up with the earth's rotation, it will continue to do so.

Mike W.

(published on 11/14/11)

Follow-up on this answer.