Assume an object falls towards Earth (I've drawn a hyperbolic orbit, but this would apply to any orbit). The object starts at $A$, and at this point it is not rotating i.e. an observer on the object would measure no fictitious forces. Will the object rotate as it passes the Earth and moves away?

My Newtonian intuition tells me that the object will not rotate so we will see all sides of the object depending on its orbital position. This is because the object had no spin prior to orbiting Earth. The motion through the orbit would look like:

No rotation

My Einsteinian intuition tells me that the object should "rotate" from our perspective because it's following the curvature of space and it is moving in a straight line. The motion through the orbit would look like:


Which one is correct, and why?

  • $\begingroup$ Note that with spin in physics we usually refer to the intrinsic angular momentum of elementary particles, nuclei and other objects following the laws of quantum mechanics. Is this what you are referring to or do you mean the classical notion of angular momentum? See e.g. this wikipedia article for the differences between the types of angular momentum. $\endgroup$
    – glS
    Feb 22, 2015 at 18:53
  • $\begingroup$ It sounds like he means classical rotation. It's fine to use the word "spin" classically, just not as a noun. "To spin" is a commonly used concept, but "the spin" as a noun, as @glance says, implies a quantum quantity. $\endgroup$
    – Brionius
    Feb 22, 2015 at 19:00
  • $\begingroup$ I'm not sure I understand the assumptions behind this question. Can you explain why you would expect the object to follow a straight line as it is attracted by Earth's gravitational pull? I wouldn't expect that, and it makes me think I don't understand the situation you're setting up. $\endgroup$
    – Brionius
    Feb 22, 2015 at 19:01
  • 4
    $\begingroup$ Point of fact: things do not "fall into orbit" in two body systems. If you were coming from far away (i.e. on a parabolic or hyperbolic track) you are going to leave again unless you brake somehow. $\endgroup$ Feb 22, 2015 at 19:15
  • 1
    $\begingroup$ Related (but not a lot of upvoting there): physics.stackexchange.com/q/98055/17609 $\endgroup$
    – Řídící
    Feb 22, 2015 at 20:00

2 Answers 2


The answer to your question is that to a first approximation the direction of the spaceship will not change, so the upper diagram is the correct one. However the direction of the spaceship will change very slightly due to a phenomenon called the geodetic effect.

The easiest way to see this is to replace the spaceship by a gyroscope, and make the gyroscope sufficiently small that we can ignore tidal effects. Then we can measure the direction of the axis of rotation of the gyroscope. In fact this experiment has been done - it was done by Gravity Probe B. The results are nicely summed up by this diagram from the linked article:

Gravity probe B

Gravity Probe B used a circular orbit, but the same applies to any orbit.

  • $\begingroup$ this is very interesting.. thanks for posting, i will look further into the geodetic effect $\endgroup$ Feb 23, 2015 at 13:09
  • $\begingroup$ "The answer to your question is that to a first approximation the direction of the spaceship will not change, so the upper diagram is the correct one." If we think of the spaceship consisting of two unconnected parts (maybe balls) closely following each other (on the same trajectory), then the direction or the pair will change. So, how's that different? $\endgroup$
    – Řídící
    Feb 23, 2015 at 17:57
  • $\begingroup$ Because the two parts aren't connected. If you connect them they now exert a torque on each other. It's an interesting question and worthy of it's own post if you want to pursue it. $\endgroup$ Feb 24, 2015 at 6:56

Let me assume that the object has spherical symmetry, however, for solving the present problem it is painted on its surface with different colors. So, imagining a plane section that contains the orbit of the object around the earth. The section of the plane through the object is a circle and we will see different points of the circumference painted in different colors.

At every time during its movement the object has a velocity consisting in two components: the original velocity, and the velocity toward the center of the Earth. So, at each time the object undergoes an inertial translation, as a whole (recall what Einstein said that that an accelerated system can be considered at each time as an inertial system with the instant velocity). Thus, at each instant another color will be in the direction of the Earth.

enter image description here

In the picture, one can see two instanciations of the object that approaches the Earth. The thin black line connects the 1st instanciation with the Earth center. The initial velocity, blue, is $v_0$, the instant velocity towards the Earth is the light-green arrow, and the black arrow is their resultant. Thus the object passes to a 2nd position, as shown in the 2nd instanciation. The dotted black line connects the center of the object with the center of the Earth. If one looks very attentively, one can see that the read segment in the 1st instanciation crosses the line towards the Earth, while in the 2nd instanciation it is completely below the line to the Earth center.

But these are two close instanciations. In continuation, as the object moves, the red segment will get more and more to the left of the line towards the Earth center, and the blue segment will be seen from the Earth. Thus, from the Earth, the colored segments will be all seen, step by step.

  • $\begingroup$ so does this mean the moon isnt actually rotating on its axis? $\endgroup$ Feb 22, 2015 at 21:19
  • $\begingroup$ @user2914191 No, the moon doesn't come from afar with an initial velocity. But, I am going to place a picture, and you'll see. $\endgroup$
    – Sofia
    Feb 22, 2015 at 21:49
  • $\begingroup$ i did an experiment by simulating an object in orbit and after 1/4 way around earth my object was facing 90 degrees up, 1/2 way around earth it was upside down. so, the Newtonian illustration is correct. i just dont have the correct understanding of how matter behaves in curved space $\endgroup$ Feb 22, 2015 at 23:09
  • $\begingroup$ @user2914191 what is up? You mean with respect to the Earth? Or you mean with respect to the page, as in my picture? $\endgroup$
    – Sofia
    Feb 22, 2015 at 23:19
  • $\begingroup$ respective to earth's surface (center) $\endgroup$ Feb 22, 2015 at 23:21

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