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# Related to Earth's rotation?

If I launch i spaceship straight up in the atmosphere from Canada, will in say 8 hours, it will be in a different country? This is because the earth rotates, so straight up from Canada would soon be straight up from Europe?

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I think that we should first isolate the system to study, it would be the earth, the rocket and the atmosphere.

Assumption: The rotation of the earth around the sun is neglected for our system. the 8 hrs is almost nothing in

comparison with 360 days (a lap).

So we have:

w, the angular velocity that the earth moves, it is roughly,

w = 7.3 *10^(-5) rad/sec

s(t), the distance that it takes every time t,

h(t), the perpendicular height of the rocket from

the surface.

a_rocket, acceleration of the rocket,

v_rocket(t), velocity of the rocket at time t,

v_earth, velocity of the earth,

v_earth = 465.1 m/sec

Then for the earth,

s(t) = v_earth / w = 465.1 / (7.3 *10^(-5))

= 6,371* 10^3 km

And for the rocket,

h(t) = a_rocket * t^2 / 2 + t * v_rocket(t)

8/24 = 4/12 = 2/6 = 1/3

By going on with the rotation of the earth,

The Latitude will be considered like constant and

the longitude variable, so

the 8 hours over 24 hrs, a day is one-third,

8/24 = 4/12 = 2/6 = 1/3

For a total of 180 degrees longitude

the third part is 60 degrees.

Say, if the rocket is launched at Montréal at 0h

then at 8h the position of the earth will turned

So Montréal is at: 45° 28' N 73° 45' W

Then 73° + 90° - 2*(60°)= 43°

Then the place to see would be:

45° 28' N 43° 45' W

It is somewhere around in the former Soviet Union,

at Don Astrakhan between the Black Sea and

the Caspian Sea.

About the height of the rocket or spaceship will

depend on its acceleration and velocity at time t.

Hope it helps some.

• When the space ship lifts off, it retains it's angular momentum with respect to the center of rotation of the earth. That value is given by r x p (vector cross-product of radius from the center and it's linear momentum perpendicular to that radius). As the ship rises, r increases, and therefore p must decrease; p = m*vp. The mass is constant (ignoring fuel consumption) so vp (the perpendicular component of v with respect to r) must decrease. The earth surface velocity does not change, so there will be a different country under the rocket after a while.

• It would make sense that your thinking is right, but to complicate your questions even more, what role does the earth's revolving take in a launch to begin with? If you notice a lauch doesn't go straight up it goes horizontally slanted.

Edited.

the first answer wasn't there when I started typing mine. LOL

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