It is said in Wikipedia, that

On the first orbit, McDivitt attempted to rendezvous with the spent Titan second stage. This was unsuccessful for a number of reasons:

NASA engineers had not yet worked out the idiosyncrasies of orbital mechanics involved in rendezvous,[citation needed] which are counter-intuitive. Simply thrusting the spacecraft toward the target changed its orbital altitude and velocity relative to the target. When McDivitt tried this, he found himself moving away and downward, as the retrograde thrust lowered his orbit, increasing his speed.

I don't understand this.

Is there any explanation, given in local reference frame? Referring "orbital altitude" referrers global reference frame and is OK. But any set can be possibly regarded in any reference frame. Local reference frame should be inertial with tidal, Coriolis and other forces.

How to describe the situation with this?


I need explanation WITHOUT notion of "orbit".


Suppose we are inside giant closed spacecraft like Rama or O'Neill cylinder. This spacecraft is on Earth orbit, but we are inside and don't know this. We feel weightlessness. Now, If Rama is rotating, we can feel some non-inertial effects like centrifugal or Coriolis forces.

But suppose Rama is not rotating.

Then, the only strange thing we will feel is Earth tidal force. The tidal force mean that all objects will periodically distracted along axis, directed to (invisible) Earth.

So, you want to say, that McDivitt failed due to tidal forces?

Hard to believe.


The possibility to regard the task from within any reference frame I like -- is the basic physical principle. You won't convince me it is wrong in the case of orbital movement.

  • 7
    $\begingroup$ It sounds like you are imagining being in an inertial reference frame in flat spacetime. In Newtonian terms, you are not, since you are accelerating. In General Relativity terms, you are not, since you are in curved spacetime. So either way, your assumption that your local reference frame is "inertial" is flawed. Your local reference frame will however appear inertial on time scales short compared to the orbital period. So over a few minutes or less, things will behave about as you expect. $\endgroup$
    – Mark Adler
    Commented Oct 20, 2013 at 22:29
  • $\begingroup$ The question is: how to describe McDivitt fault in LOCAL reference frame. I don't insist it is inertial. If it is non-inertial then explain how does non-inertiality play here. $\endgroup$
    – Suzan Cioc
    Commented Oct 21, 2013 at 7:31

6 Answers 6


The explanation in the frame of the body T that McDivitt was trying to approach is this. When he turned the thrust on, the spacecraft acquired velocity $\mathbf v$ towards the body. The Coriolis force $-2m\mathbf{\Omega}\times \mathbf v$ acted on the spacecraft, where $\mathbf \Omega$ is the angular velocity of the rotation of the body T in the orbit around the Earth. If the two bodies were initially at the same height, the Coriolis force acted upwards, thus giving the spacecraft some velocity in the direction away from the Earth. This necessarily got the spacecraft to higher altitude and there the tidal force plus Coriolis force changed the velocity of the body further so it actually began to recede from the body T (tidal force vector at higher altitude points away from the body T).

  • $\begingroup$ I went to conclusion that this is a correct answer. I forgot, that even if Rama is not rotating, Coriolis force still exist -- because the entire orbital path is curved. When you try to move along path, Coriolis will try to move you "upper" -- this is the same as orbit change, but in local reference frame. $\endgroup$
    – Suzan Cioc
    Commented Oct 21, 2013 at 7:38

This is more of a physics question, but here goes:

Both the piloted spacecraft and the rendezvous target are objects each in their own orbit, even though they may be separated by minimal distance and have minimal relative velocity one to the other.

An object in an ideal orbit always travels in a plane of fixed absolute orientation around the center of mass of the object it is orbiting. The orbit is always an ellipse, such that the center of mass of the orbited object occupies one of the focal points of the ellipse. A perfectly circular orbit is simply a special case where both focal points coincide. As articulated by Kepler, an orbiting object covers equal areas in equal times - travelling most slowly at the "highest" point, most rapidly at its "lowest" point.

Any "push" given to an orbiting object, no matter the direction, will change its orbit in some way - increasing or decreasing its mean altitude, increasing or decreasing eccentricity (how elliptical or circular the orbit is), or changing the orientation of the plane in which it is travelling.

Consider an object in a perfectly circular orbit. Giving it an extra push along its orbit will increase its orbital energy, increasing its mean altitude. That will cause its orbit to become elliptical, "climbing" to some greater altitude and losing speed until it arrives at point opposite to where it got the push, then descending and accelerating back to the altitude and speed it just after the push.

So, imagine you've successfully placed yourself on exactly the same orbital path as an object you want to rendezvous with, just some distance "behind". You fire your thrusters to push you "forward". Well, you start to move toward your target, but because you are moving faster, you start to gain altitude and move "up" above your target... but moving up slows you down and causes you to fall behind!

Instead, you could fire a thruster "downward". Forcing yourself into a "lower" point with the same orbital energy causes you to move faster along your orbit, so you catch up. Then you fire a thruster "upward" to return you to your original energy state (and orbit), closer to your target.

Counter-intuitive, but it comes down to working with orbital energy states.

  • 5
    $\begingroup$ Right. Thrust down, go forward. Thrust up, go backward. Thrust forward, go up. Thrust backward, go down. These effects are seen on time scales of a large fraction of an orbital period. Once you know how it works, rendezvous gets much easier. $\endgroup$
    – Mark Adler
    Commented Oct 20, 2013 at 18:55
  • $\begingroup$ Your description is again in geocentric reference frame. I need explanation without any "orbits". Being in orbit, I am in free fall state. I don't know that Earth exists. Imagine that me and my goal is inside giant closed spacecraft, like Rama (en.wikipedia.org/wiki/Rama_%28spacecraft%29). Why won't I fly to where I thrust? $\endgroup$
    – Suzan Cioc
    Commented Oct 20, 2013 at 18:59
  • 6
    $\begingroup$ It seems like excluding the geocentric reference frame will exclude the correct answer along with it. Yes, you're weightless, but you're in the curved space of a gravity well. Changing kinetic energy must change your potential energy, which is relative to the geocentric frame of reference. $\endgroup$ Commented Oct 20, 2013 at 21:14
  • 7
    $\begingroup$ You do know the Earth exists. Even if somehow you ignore the giant blue ball beneath you. If you imagine that you and another object are in a box and can't see the Earth or fixed stars for inertial reference, simply the act of maneuvering two objects relative to each other inside the box will give you a very good idea that the Earth is there, its mass, how far away its center is and in what direction, and your orbit's eccentricity. Further measurements will reveal the J2 of Earth, telling you the inclination of your orbit relative to the Earth's axis. Yes, you know that Earth exists. $\endgroup$
    – Mark Adler
    Commented Oct 20, 2013 at 22:23
  • 1
    $\begingroup$ Suzan - no you can't. $\endgroup$
    – Rory Alsop
    Commented Nov 10, 2013 at 23:19

The simplest answer is that McDivitt thrusted the Gemini Spacecraft towards the Titan second stage in the same direction that both two objects were travelling. The two were separated far enough that it would take several minutes for Gemini to reach the Titan second stage with amount of thrust generated.

If they were in a zero-g field with nothing else imparting a force.

But they were in orbit. Because thrusting towards the Titan second stage was also the same as thrusting in the direction of their orbit around earth this had the effect of raising the orbit of the Gemini spacecraft. Because the Gemini was now in a higher orbit it was causing the spacecraft to move a slower relative to the Titan's second stage.

You can see this effect by the fact a spacecraft orbiting the earth at a 200 miles altitude takes roughly 90 minutes to orbit the earth and a spacecraft at 26,200 miles takes roughly 24 hours to orbit the earth (otherwise known as geosynchronous orbit).

The distance between the two spacecraft was enough that the effect of the Gemini shifting orbit dominated how the two objects interacted. McDivitt various attempts at approaching the Titan kept shifting the Gemini into higher orbits causing the counter intuitive effect of moving away from the Titan second stage even though he was thrusting straight at it.

The same effect does happens when the range is short. However your orbital altitude doesn't raise or lower fast enough to make a difference. Thus it can be largely ignored.

Ultimately NASA solved this by correctly timing the final approach to a target object. The target was allowed to rise to a certain number of degrees above the horizon relative to the spacecraft. The spacecraft was in a lower orbit and starting out moving faster than the target. Then the commander pointed the Gemini straight at the target. The radar locked on giving the rate of range closure.

The commander then applied thrust toward the target. There was an instrument that showed distance to target and rate of range closure on a scale. The commander job was to keep the rate of range closure at the right spot for the current distance. Generally by thrusting forward at periodic intervals.

If you were looking from the outside what you would see is the Gemini spacecraft rising up and towards the target, passing underneath, and then curling upwards to come to a stop a few hundred meters forward of the target's orbital position.

During the entire maneuver the Gemini spacecraft would be pointed at the target.

When the radar failed on Gemini 12, it killed the crucial rate of range closure. Pilot Buzz Aldrin then used handheld instruments (sextant, etc) and charts to tell Command Jim Lovell when to apply forward thrust. They ultimately successfully completed the rendezvous.


Sorry, but it's impossible to explain this without referring to orbits. When you are in orbit, your altitude and your linear velocity (speed in the direction of the orbit) are inextricably linked: your linear velocity is proportional with the square root of the altitude (radius).
As a result, any change in speed inevitably changes your orbit.

The Wikipedia article does not say this explicitly, but logically, the Gemini would have been ahead of the Titan stage in the same orbit. Then the rest of the description makes sense: McDivitt used the thrusters pointed in the direction of the orbit to try and reduce the distance (i.e. retrograde thrust). So he reduced his linear speed, which had to result in a lower orbit. In a lower orbit, it doesn't take as long to complete one orbit (360 degrees), so as a result he moves away from the Titan stage.

  • $\begingroup$ Everything can be described in any reference frame. The complexity of description can vary, but it always exist. $\endgroup$
    – Suzan Cioc
    Commented Oct 21, 2013 at 10:40
  • 3
    $\begingroup$ You can change the reference frame, but you cannot decide to ignore the influence of Earth's gravity on the two spacecraft. $\endgroup$
    – Hobbes
    Commented Oct 21, 2013 at 10:51
  • $\begingroup$ Earth gravity influence turns into tidal force. If orbital period is 90 mins, then tidal force vector rotates with that period. If we act much faster than 90 mins, then we can ignore tidal force vector rotation and can think about it just as a constant tidal force. $\endgroup$
    – Suzan Cioc
    Commented Oct 21, 2013 at 11:11

The question asked an answer in a local non inertial frame as inside Rama (the large hollow cylinder from Arthur C. Clarke's novel "Rendezvous with Rama"). Other answers are right, but I'll try to add an answer in this particular reference frame.

Let's suppose we are on the center of one base of the cylinder and we want to jump to the center of the opposite base. To make this movement similar to the Gemini 4 setting, Rama is orbiting the Earth, moving in the direction of its longer axis and always parallel to the surface of Earth - that is, Rama is rotating with center of rotation in the center of Earth.

In the novel, Rama is rotating around its long axis to create artificial gravity, but since it doesn't play a role in our setting, we would stop that rotation.

Going back to Gemini, in this setting we have just put a giant cylinder with the center of the forward base at the target and the center of the trailing base at Gemini 4 - assuming their orbit are circular.

In the rotating reference frame of Rama, we would not feel gravity, because gravity would be exactly balanced by centrifugal force. Interestingly, that wouldn't change if we would move along the axis of Rama because both centrifugal force and gravity depend only on distance to the centre of the Earth and that is change (assuming the cylinder is just a few km long).

Then, in our reference frame, we are weightless in one end of a cylinder and we just need to jump straight in direction to the other end - as Gemini 4 did. However, we are in a rotating reference frame and we need to take in account centrifugal and Coriolis forces.

As said, centrifugal force doesn't matter because distance to centre of Earth doesn't change and keeps balancing gravity.

However, since now we are moving in the rotating reference frame, Coriolis force does matter. When moving forward in the tangential direction, Coriolis force is directed outward, and our jump would deviate to the outer wall of the cylinder, just as when a trailing spacecraft accelerates to catch a target and ends in a higher orbit.

In conclusion, the problem can be solved in a rotating reference frame (at least for small distances) and Coriolis force deviates outward any spacecraft accelerating along its orbit and deviates inward any spacecraft reducing orbital velocity.


A good analogy is a bicycle race on a straight track v.s. on a circular track.

To catch up with a bike in front of you, you just pedal harder. But if you are on a circular track, you should also endeavor to "cut the corner" and take the short-cut, because the closer you get towards the center of the turn, the faster you pass the other bike. If you try to pass him on the outside, you'd have to go much faster than him to compensate for the larger arc.

  • 1
    $\begingroup$ This a bad analogy. On a bicycle you can choose any combination of circle diameter and speed, as long as the speed is lower than your perosnal limit. But in an Earth orbit, the orbit diameter and the speed ratio is given by Earth's gravity. If you change speed you change the orbit height. $\endgroup$
    – Uwe
    Commented Jun 6, 2018 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.