settle a bet

If you could throw an object very high, so high that it would take a few
hours to fall back to earth. Would it land in the same location it was
thrown from or would the fact that the Earth is spinning mean that the
object would land in a new location.
Thanks for any help.

b)

In article , Bear
wrote:

If you could throw an object very high, so high that it would take a few
hours to fall back to earth. Would it land in the same location it was
thrown from or would the fact that the Earth is spinning mean that the
object would land in a new location.
Thanks for any help.



Interesting question, if you negated air effects, then Newtonian
physics would say that what goes straight up comes straight down.
However, if the object went up really high, then the effects of frame
dragging (the effect of any rotating mass on space/time/gravity) would
ensure the object would come down in a slightly different place, as the
Earth would have rotated round under the ejected mass's space/time
location. See http://en.wikipedia.org/wiki/Frame_dragging

And if its proved experimentally then Einstein was right about another
aspect of general relativity!

Clear Dark and Steady Skies

Torcuill

"Torcuill Torrance" wrote in
message
s.com__removelihgt...
In article , Bear
wrote:

If you could throw an object very high, so high that it would take a few
hours to fall back to earth. Would it land in the same location it was
thrown from or would the fact that the Earth is spinning mean that the
object would land in a new location.
Thanks for any help.



Interesting question, if you negated air effects, then Newtonian
physics would say that what goes straight up comes straight down.

No, it would be accellerated towards the centre of mass of the Earth, which
is not quite the same thing with a rotating planet. The point it was thrown
from would continue on its normal path of rotation. The object would have an
initial velocity of the tangential velocity of rotation plus its vertically
thrown velocity, and then would follow a curved path until it touched Earth
again. I'm 99% certain b).

In ,
Andy Guthrie typed:
The
object would have an initial velocity of the tangential velocity of
rotation plus its vertically thrown velocity, and then would follow a
curved path until it touched Earth again. I'm 99% certain b).


Or would miss the earth on its way back down and continue in an elliptical
orbit.

Er...I think.

Jo

Originally Posted by Bear

If you could throw an object very high, so high that it would take a few
hours to fall back to earth. Would it land in the same location it was
thrown from or would the fact that the Earth is spinning mean that the
object would land in a new location.
Thanks for any help.

Try it - just a gentle lob will do - but use something heavy so as not to be deflected by wind - and run fast. Alternatively you could visit Pisa and repeat a famous experiment of old - it wont matter if the object is rising or falling;-)
Nytecam

"Jo" wrote in message
...
In ,
Andy Guthrie typed:
The
object would have an initial velocity of the tangential velocity of
rotation plus its vertically thrown velocity, and then would follow a
curved path until it touched Earth again. I'm 99% certain b).


Or would miss the earth on its way back down and continue in an elliptical
orbit.

Er...I think.


I'm no rocket scientist, ahem, but I reckon that after departing from a
point on the surface it could not achieve an elliptical orbit without some
further powered adjustments.

thanks for the answers guys

Wasn't it Bear who wrote:
If you could throw an object very high, so high that it would take a few
hours to fall back to earth. Would it land in the same location it was
thrown from or would the fact that the Earth is spinning mean that the
object would land in a new location.
Thanks for any help.

Interesting question.

Lets do the experiment on an airless body, otherwise air resitance and
wind effects are going to swamp any real effect.

You throw the object "straight up" in your frame of reference, but your
frame of reference has a sideways velocity due to the fact that you're
standing on a rotating planet. So the object has vertical and horizontal
components to its initial velocity when viewed in a non-rotating frame
of reference.

The object then follows an elliptical orbit around the centre of the
Earth.

I can't think of a quick way to calculate whether this lands it in the
same location when launched from the equator, but it clearly misses if
the launch point is anywhere other than the equator or poles.

The ground track of any orbiting body is always a great circle, but your
launch site moves along a line of latitude. Unless you launch from the
equator, these are not going to be the same thing. The ground track of
the object will follow the great circle that is tangential to your line
of latitude, and land somewhere closer to the equator.

--
Mike Williams
Gentleman of Leisure

Wasn't it Andy Guthrie who wrote:

I'm no rocket scientist, ahem, but I reckon that after departing from a
point on the surface it could not achieve an elliptical orbit without some
further powered adjustments.

No. Any thrown object follows an elliptical orbit. What they told us in
school about a thrown object following a parabola would be true if we
lived on a flat earth where lines of gravity were parallel. For objects
thrown only a short distance, the difference between a parabolic an
elliptical trajectory is negligible because gravity is close to be
parallel.