# What is the equation that relates delta v of a rocket to time taken to complete an orbit?

I wanted to see how the delta-v produced by a rocket affects the time it takes to complete a certain orbit. So what is the equation that relates these two quantities?

• Why do you think these two quantities are related? Dec 16, 2021 at 3:39
• An easy way to see that they're not related is to replace your rocket with just a rock. It'll still orbit just fine, given a sufficient initial velocity, but has approximately zero delta-v. Dec 16, 2021 at 20:14
• Well, there is a kind of relation: the more delta-v you have, the higher - and slower orbit you can reach, starting from the ground below.
– SF.
Jan 4 at 14:09

There is actually no relationship between the two. In a simple 2-body problem, each orbit starts and ends in the same place with the same velocity vector. The "delta" in "delta-v" means change. No change, no delta. The spacecraft repeats the same orbit until the end of this 2-body universe. The spacecraft likely wonders why it was endowed with delta-v which it doesn't need and never uses.

Then one day, Houston wakes up the spacecraft and asks it to change its orbit. This is where the delta-v comes in. Any change in orbital altitude or inclination will cost delta-v. Any change: faster, slower, higher, lower is a change and that's where the "delta" comes in.

Every spacecraft has a limit in how much it can change its velocity before it runs out of fuel. If it only has "X" delta-v and you ask it to go somewhere that requires 2X delta-v, it will say, "I'm sorry, Dave, I can't do that.

So, to answer your question, a rocket's delta-v has no effect on the time to complete an orbit.

Except... a continuous radial burn (rocket continually pointing to the center of it's orbit) would increase the centripetal acceleration. Einstein tells us this would be indistinguishable from being in orbit around a more massive central body, so it would shorten the orbital period.

This burn would be a waste of fuel and would get you fired from Kerbal Space program.

• Oh right, I am actually quite new to orbital mechanics so I thought that a greater delta v would enable the rocket to move at a faster velocity in an orbit. But yeah, I guess the engine is shut off after reaching an orbit
– Ryan
Dec 16, 2021 at 4:32
• One of the weird things to get used to in orbital mechanics is the way your velocity while the rocket is off is determined by your distance from the body you are orbiting. In order to go faster, both in meters per second and time taken to complete an orbit, you have to move your orbit closer to the central body, which means you need to lose energy, which means you need to slow down in order to speed up. The reverse is also true. If that hurts your head, welcome to the club. Keep staring at it long enough, and it might start to make sense. Dec 16, 2021 at 5:21
• @RyanC: It's maybe slightly more intuitive if you note that your angular momentum (which is proportional to your orbital radius times speed) still goes up as you accelerate forward and down as you decelerate. It's just that the speed of an (approximately) circular orbit is inversely proportional to the square root of its radius, so increasing the angular momentum requires increasing the radius, which decreases the speed (but not enough to cancel out the increase in angular momentum from the increased radius). Dec 16, 2021 at 20:29

You need the initial speed, $$|v_i|$$ and the final speed, $$|v_f|$$, and the radius, $$r$$, at which the speed change took place.

The orbital period is directly related to the orbital energy, so you need to find the change in kinetic energy that results from the delta-V. The delta-V can be in any direction with respect to the instantaneous velocity. It may be forward or backward or sideways etc. The direction in which the delta-V is applied makes a huge difference. The initial and final speeds are determined by the vector addition of velocities: $$v_f = v_i + \delta V$$

Following Kepler and Newton, the orbital period is given by $$T = 2\pi \sqrt(a^3/GM)$$, where $$M$$ would be the large central mass (e.g. Earth) and $$a$$ is the semi-major axis of the spacecraft's elliptical orbit. The specific orbital energy is given by $$E = GM/2a$$ per unit mass of the spacecraft. It follows that the period is also given by $$T = \pi GM / \sqrt(2E^3)$$

The instantaneous specific orbital energy is constant and can also be expressed at any point in the orbit as the sum of the (negative) specific potential energy (wrt infinity) plus the specific kinetic energy, $$E = -GM/r + v^2/2$$, (both terms per unit mass).

So the change in period is:

$$T_f - T_i$$
$$= \pi GM / \sqrt(2E_f^3) - \pi GM / \sqrt(2E_i^3)$$
$$= \pi GM[1/\sqrt(2E_f^3) - 1/\sqrt(2E_i^3)]$$
$$= \pi GM[1/\sqrt(2(-GM/r + v_f^2/2)^3) - 1/\sqrt(2(-GM/r + v_i^2/2)^3)]$$

Again, $$r$$ is the radius at which the $$\delta_V$$ is applied and $$|v_i|$$ and $$|v_f|$$ are the initial and final speeds of the spacecraft.