If I am walking, the ground is stationary, and I move $X \frac{\text{units}}{\text{hour}}$ relative to the ground.

If I am in low-Earth orbit, then I am still moving $X \frac{\text{units}}{\text{hour}}$ relative to the ground, only it's a much bigger $X \frac{\text{units}}{\text{hour}}$.

If I am in geo-stationary orbit, I am moving $0 \frac{\text{units}}{\text{hour}}$ relative to the ground. But relative to the axis of the Earth, I am moving as fast as the Earth spins. So am I moving at $0 \frac{\text{units}}{\text{hour}}$ or the speed of the Earth's rotation (plus a little since it's away from the surface)?

If I go to the Moon, the Earth will be rotating so my speed would have to be relative to the distance from the axis of the Earth, and not the position around the axis like in geo-stationary orbit.

Now imagine I am leaving Earth's orbit, my speed can't be measured relative to the distance I am from the Earth's axis because it's moving around the Sun. So at what point does my speed change from being relative to Earth to being relative to the Sun and is there a spike in the numbers from this transition?

I guess I am confused because from everything I read and hear about spacecraft, the speed is always brushed over like it's really obvious and intuitive. But I don't really think it is.

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    $\begingroup$ You get to pick the frame of reference to use, depending on what kind of calculation you are doing. $\endgroup$ Commented Jul 20, 2019 at 11:31
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    $\begingroup$ "If I am walking, the ground is stationary, and I move X units/hour relative to the ground." – The ground is only stationary because you define it to be. It actually moves at ~40000km/24hr around the axis of the Earth (at the equator), it also moves at a considerable speed around the Sun. $\endgroup$ Commented Jul 20, 2019 at 14:34
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    $\begingroup$ At the "transition" there isn't a "spike in the numbers." It's just a change in which set of numbers you're using. $\endgroup$ Commented Jul 20, 2019 at 15:27

3 Answers 3


Your concerns are all perfectly valid - giving just a number doesn't tell a lot. So, in all proper publications, the reference system has to be mentioned.

Typically, the reference is the body the spacecraft is mainly influenced by. I.e. in a Earth orbit it is the center of Earth, in lunar orbit the Moon. For interplanetary probes in transit it's usually the center of the Sun. Close to lift-off and landing one usually switches to a reference frame fixed on the surface of the planet. For most calculations it is the best to have a fixed, non-accelerating coordinate system (or at least one with a negligible acceleration, like Earths center in respect to LEO)

In other cases, but usually not when talking about velocity, the reference system might be a rotating one - e.g. the typical 8-shaped plot of the trajectory of the Apollo missions is drawn in a coordinate system centered on Earth, but rotating with the Moon.

  • $\begingroup$ Some images would be nice, for instance the trajectory of the Apollo missions in differentt coordinate systems, rotating with the Moon and rotating with Earth as well as not rotating. $\endgroup$
    – Uwe
    Commented Jul 20, 2019 at 14:02
  • $\begingroup$ @Uwe Please feel free to add some - I don't have any in stock and don't have the software to create them. On the other hand rotating reference frames are slightly out of scope of this question. $\endgroup$
    – asdfex
    Commented Jul 21, 2019 at 11:50
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    $\begingroup$ As an addotional point the body doesnt have to be a planet either! Even in a redevouz frame of reference you can express the speed of your spacecraft in terms of the other spacecraft. For instance- once a module is docked to the ISS its moving 0mph relative to the ISS. During docking its flying around the earth blisteringly fast but its approaching the ISS at a relative speed of 5-10m/s. Speed can be defined relative to anything! $\endgroup$ Commented Jul 21, 2019 at 14:39
  • $\begingroup$ i guess i'm most confused about LEO and GSO. Are GSO typically referenced as traveling 0 units/hour relative to a fixed point on the axis of the earth, or ~1000 mph relative to the axis of the earth. When/how does their speed matter $\endgroup$
    – zoplonix
    Commented Jul 24, 2019 at 8:29

This is a supplementary answer in addition to the excellent answer from @asdfex, with an attempt to explain things in simplified terms, as, judging from the comments, OP is still confused with Geostationary Orbit.

An object placed in Geostationary orbit (the orbit lies in Earth equatorial plane) would appear motionless to ground observer.

If I am in geo-stationary orbit, I am moving 0 units/hour relative to the ground. 

This is true only when the "units" are angular units, i.e. degrees or radians. This means that observer on the ground and the object in GSO are rotating around the center of the Earth with the same angular speed.

When the "units" are distance units, i.e. meters or miles etc., the above statement is not true, because linear velocities of observer on the ground and the object in GSO are different due to the fact that their corresponding radii are different.

I guess this is where confusion might be coming from. enter image description here

Now, to address ground/LEO/GSO relative speed:

If I am walking, the ground is stationary, and I move X units/hour relative to the ground.

If I am in low-earth orbit, then I am still moving X units/hour relative to the ground, only its a much bigger X.

The key misconception here (comparison of apples to oranges) lies in the fact that when a person is walking on the ground, he/she is not orbiting, whereas in LEO he/she is orbiting. This is why the relative-to-the-ground linear velocity seems that much bigger.

For the discussion below, let's assume everything happening in equatorial plane, orbits are circular and prograde, no perturbations, drag, solar wind etc. and all numbers are approximate/rounded.

  • Ground speed. The Earth makes one full revolution around its axis in 1,436 minutes, hence the Earth's rotational speed is 0.25 degrees/min. This makes linear velocity of an observer on equator (at 6371 km radius) equal to 460 m/s (1,029 mph)

  • a) Orbiting at the ground level. Assuming perfectly spherical Earth without hills/mountains and absence of atmosphere (so that there's no drag), in order to orbit the Earth at 1 meter above its surface, one needs to move with much higher rotational speed compared to earth: 4.27 degrees/min (17 times faster than earth). This corresponds to linear speed of 7,910 m/s (17,694 mph).

The relative orbital speed of the "ground-level" orbiting person to another observer on the ground is 4.27 - 0.25 = 4.02 deg/min.

Relative linear speed is 7,910 - 460 = 7,450 m/s.

  • b) Orbiting in LEO. Let's assume 400km altitude. Orbital angular speed is 3.66 degrees/min, linear speed is 7,672 m/s (17,162 mph).

Relative angular speed is 3.66 - 0.25 = 3.41 deg/min,

Relative linear speed is 7,672 - 460 = 7,212 m/s.

  • c) Orbiting in GSO. The altitude for this orbit is defined as 35,786km. Orbital angular speed is 0.25 deg/min, linear speed is 3,075 m/s (6,879 mph)

Relative angular speed is 0.25 - 0.25 = 0 deg/min,

Relative linear speed is 3,075 - 460 = 2,615 m/s.

If the orbital radius gets further than GSO radius, the object in that orbit would appear (to observer on the ground) to move "backwards" whilst an object in LEO would appear to move "forward" (because in angular terms Earth would rotate quicker than object in the higher-than-GSO orbit) although objects in both orbits (LEO and higher than GSO) and the Earth are rotating in the same direction. Therefore, from observer on the ground point of view, comparison of relative to the ground linear speeds (between LEO, GSO and orbit beyond GSO) are not as relevant as comparison of angular speeds.

P.S. Motion of a person in GSO relative to a person on the ground can be described by the following very simplified analogy:

Imagine a "person A" sitting in a seat of a merry-go-round; Consider another "person B" standing right in the center of the merry-go-round (therefore spinning with it) with his/her arm raised horizontally and pointing a finger to the person in the seat. The person A is analogy to a person in GSO orbit, and the finger of the person B becomes analogy to an observer on the Earths equator.

Whilst the merry-go-round is spinning, the person's B finger and the person A (in the seat) appear motionless between each other, but in terms of distance, for a given time of few seconds the finger moves few inches, whilst the person A moves few feet.


Speed is indeed relative. It depends upon the co-ordinate system in reference. If someone says the velocity is X m/s around earth, it will be incorrect. Though, implicitly we can assume they are talking about norm of velocity in Inertial frame maybe J2000 or ECI.

But sometimes it also necessary to discuss velocity relative to ground, for example when SpaceX does its landing, the terminal inertial velocity must be around ~500m/s ! This is to match the ground speed. So actually landing on a moving ship is nothing but extended version of landing on the moving ground.

Now, the interesting question is when say spacecraft leaves earth orbit and enters orbit around sun the reference is changed to barycentre of solar system or the inertial coordinate frame of target celestial body.

Now when somebody says that helios 2 achieved 70 km/s. One must bear in mind out of that 30 km/s(Earth orbital velocity around sun) is just because it started from earth


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