# What happens to a satellite's orbit when you speed it up?

If a satellite is initially orbiting the earth with a constant velocity $v$ on a circular orbit.

What would happen, if the velocity is suddenly increased by 5%?

Would the orbit still be circular?

I guess it will be elliptic, but how to prove this and find the eccentricity of the ellipse?

• Hint: Use the vis viva equation $v^2 = GM\left(\frac2r - \frac1a\right)$ and use the fact that $r=a(1-e)$ at periapsis. Jun 30 '17 at 12:35

To be clear we have to understand that "speed it up" means applying thrust in the direction of the orbit. To be complete we also have to pay attention to how hard and for how long you thrust.

Two natural limiting cases are

• $\Delta v$ applied over a time much shorter than the original orbital period (high thrust applied for a short time)

• $\Delta v$ applied over a time much longer than the original orbital period (very low thrust applied continuously over the course of many orbits)

In the former case the circular orbit takes on an elliptical character1 with a apogee2 higher than the original orbit and the perigee remaining at the height of the original orbit. This can be observed in the initial burn of outward going Hohmann transfers.

In the latter case you end up with the satellite in a higher and paradoxically slower but still circular orbit. A natural test of this case is the tidal recession of the moon.

For an in-between case—non-negligible thrust applied for a significant fraction of one orbit—you end with a elliptical orbit where the apogee has also been raised and the perigee has also been raised, but not as high.

1 In extreme cases the orbit can be made parabolic or hyperbolic but let's not bother about that.

2 I'm going to use vocabulary that assumes Earth-centered orbits for concreteness, but of course the results apply to all cases.

The eccentricity of the orbit would increase. Assuming the orbit began as perfectly circular, the point in the orbit where the acceleration took place would become the perigee (nearest approach). The opposite point in the orbit would become the apogee (most distant point from Earth). As far as determining the new eccentricity,more information is needed. The formula would be:

$$e=\sqrt{1+\frac{2Eh^2}{(GM)^2}}$$

$$E=$$ orbital energy
$$h=$$ Specific relative angular momentum
$$G=$$ gravitational constant ($$6.7×10^{−11} \frac{\text{m}^3}{\text{kg}⋅\text{s}^2}$$ or so)
$$M=$$ mass

• thanks, but can you tell why and how the point will become perigee? Jun 30 '17 at 13:32

I am fairly a beginner of Orbital Mechanics, and I can roughly answer to you that on increasing the velocity of a satellite or any object in orbit, the point - say P' opposite to that of the point say P at which the satellite raised its velocity will increase in its orbital altitude and hence the orbit will become elliptic, the more you increase the velocity the more elliptic it becomes. The Apogee of your orbit which is the highest part of the orbit will increase in altitude too, and as you keep increasing the velocity, the apogee of your orbit will take place of the raising part P' of your orbit. As you keep increasing the velocity, you will then reach the Escape Velocity of the Earth and will break out from its orbit escaping into space. In order to make your orbit circular back again, you will have to wait until you reach the raised part P' of your orbit and then increase the velocity so the opposite part P will raise in altitude and your orbit will be circular again.

I dont know how to prove that, but its some basic Orbital Mechanics

• thanks, but can you tell why and how the point will become perigee? Jun 30 '17 at 13:33