If there is no atmosphere on a planet that is rotating on an axis and a rocket is launched "Straight up" from its surface, If there is no atmosphere on a planet that is rotating on an axis and a rocket is launched "Straight up" from its surface, won't the rocket still have an angular/orbital velocity because the planet surface's rotation has imparted it's rotation to it? 
Also, would that orbital/angular velocity increase the higher rocket goes straight up?  
(Kind of like a person walking straight out from the center of a merry-go-round covers more distance in each rotation the further out they walk -- thus the faster they go on the outer arms of the merry-go-round.)
Then also would not the rocket be in Stationary orbit straight up above that same point on the planet?  
Please help me to understand this not so "straight up" concept.
 A: I will assume that being launched "straight up" means that the force exerted on the rocket is along a line between the rocket and the center of the planet. I will also assume that the rocket starts on the "equator" of the planet, i.e. it's position vector relative to the center of the planet is perpendicular to the planet's axis of rotation.
What this means is that there is no net torque acting on the rocket about the center of the planet, which means that its angular momentum about the center of the planet must be constant. Also, the rocket will continue to orbit about the same axis as it was before being launched.
Angular momentum in this case is given by
$$L=mr^2\omega$$
where $m$ is the mass of the rocket, $r$ is the distance the rocket is from the center of the planet, and $\omega$ is the rocket's angular velocity about the center of the planet.
The rocket starts at the surface of the planet a distance $R$ from the center of the planet with an angular velocity of $\omega_0$. Therefore, the initial angular momentum is 
$$L_0=mR^2\omega_0$$
Once the rocket turns on, $r$ will be some increasing function of time $r(t)$. Since angular momentum is conserved, we have
$$L=L_0$$
$$mr^2\omega=mR^2\omega_0$$
Or, solving for $\omega$ as a function of time
$$\omega(t)=\left(\frac{R}{r(t)}\right)^2\omega_0$$
Therefore, as the rocket moves upwards, it's angular velocity will decrease. Therefore, it cannot stay at the same place above the planet. There would have to be some additional force tangential to the orbit in order to cause the angular velocity to remain constant, thus increasing the angular momentum as the rocket moves upwards.
A: It all depends on what "straight up is", and also the location of the launch complex. Let's put it on the equator on a prograde planet. A the rocket rises, the guidance will have to account for the Coriolis force pushing the rocket westward in order to keep it above the launch pad. That will require increasing the eastward speed so that the angular velocity remains fixed at one rev per (sidereal) day.
If the rocket reaches the height of the geosynchronous orbit (and has no vertical speed), it will remain orbiting directly above the pad.
Note that presence or lack of atmosphere has no conceptual bearing on this problem.
