Right, but this assumes you want things to stay in orbit. If you don’t, it’s easier/faster to go West than East. Looking at the math though, you don’t gain much at all due to the ICBM’s short flight time.
You’re absolutely right–I didn’t understand your post at the time I read it. For something coming back down, yeah, you want to go West.
Tripler
“Go West young man, go West!”
Since the OP is answered, and it was my OP:
This seems backward to me – the Earth rotates west-to-east, so an object shot “straight up” would have an appreciable eastward velocity, wouldn’t it? To head West, you’d need to overcome that eastward velocity taking off from ground-level. Or perhaps I’m misunderstanding your (literal) viewpoint? Thanks. [/hijack]
Wow, that is awesome. Thanks!
You’re half right. Imagine a giant tower sticking out of the earth. The velocity of the tower at the base is (radius of earth)*(rotational speed of earth) eastward, let’s call that X. The velocity at the top of the tower is (raidus of earth+tower height)*rotational speed of earth) eastward, call that 1.2X. As you can see, the top of the tower is going faster eastward by a factor of the tower height times the rotational speed of earth.
Now imagine you have a rocket at the bast of this tower. When you launch the rocket, it has the speed of the base of the tower i.e. X. Since you’re launching it straight up, you aren’t adding any east/west velocity (ignore the atmosphere). When the rocket gets to the top of the tower, it will still have the same velocity as it did at the base. The top of the tower has a velocity of 1.2X, so from the point of view of the tower, the rocket is moving with a velocity of .2X westward.
Hope that makes sense.
I have a slight nitpick with this: the top of the tower will move at the velocity 1.2X because the entire, rigid tower is fastened to the earth. At the top of the tower, the rocket will be inertially moving in relation to the center of the earth at 1.2X as well, because of the rotational velocity imparted to the rocket at the moment of launch (the angular inertia)–as opposed to 0X velocity in respect to the launch pad.
At the point of view of NASA at Cape Canaveral–as well as the tower–the rocket will appear moving at velocity “X” vertically, but to a stationary Klingon observer watching the earth rotate from orbit, the rocket will move at an angular velocity more and more the higher in altitude it climbs. Once the rocket is launched, it is no longer physically connected with the earth, and must perform a burn to rid itself of that imparted velocity and actually realize that .2X westward velocity.
My analogy: imagine a kid throwing water balloons from a moving car. From the kid’s point of view, if the car is moving at 20 MPH, and he throws the balloon at 10 MPH, he sees the pitch at 10 MPH. Now the cop in the patrol car will see the speed of the car (20 MPH) and the extra velocity of the baloon (10 MPH) and will get hit with a 30 MPH water balloon which will soak his donut. To do what you’re doing with the .2X westward rocket velocity, the kid needs to throw the balloon behind him at 10 MPH. Then, from the kid’s point of view, he’ll see that .2X velocity. The imparted inertia of the balloon (from riding in a 20 MPH car) needs to be lost by some action of the kid (a body in motion tends to stay in that motion).
The bottom line is this–once an object leaves the earth’s surface, the rotation of the earth does not matter anymore. The earth becomes one giant solid and gravity doesn’t give a darn if the rocket is over Europe or Ulaan Batar. All gravity (through the eyes of that observant Klingon) cares is that the rocket is successfully keeping it out of the gravity well of the large ball of whatever underneath through inertia generated by kinetic energy.
Clear as mud, eh? One has to realize that there are two frames of reference. Once the rocket leaves the earth’s surface, it needs to be portrayed in an angular frame. . .
Tripler
For the record: I was never that kid in the car. I swear.