# 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?

Relevance

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,

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.

• gp4rts
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