Question about Delta-V

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Polishguy

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I'm trying to figure out how to calculate the delta-v necessary to go from earth orbit to any asteroid. What formula can I use to calculate the delta-v necessary to go from earth to, say, the Apollo asteroid Hermes?
 
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SpaceTas

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The delta V is just that: the difference in orbital velocity between the origin and the destination. (works if both are in the in same plane otherwise much more tricky).
So to get the velocities from first principles you'll need to find out about orbits, orbital mechanics.
But there should be values for the orbital velocities of most targets on the web or in textbooks.

If you ultimately want to find out what sort of engines you need then

Momentum = mass times velocity (assume constant mass) = mv
Impulse = I = change in momentum = mass * (deltaV = mass (V2 - V1) sign doesn't you still need to change the momentum slowing down or speeding up (getting further out or closer in).

The impulse is also equal to force times time interval force is applied = F*(deltaT)
its actually the integral of applied force over time.

So a small force over a long time gives you the same impulse (change in momentum) as a large force over a small time interval ie an ion engine versus a chemical engine.


The total energy required is the force times the distance traveled (the integral of the force with respect to distance) this is equal to the difference in total (kinetic and gravity potential energy) between origin and destination.

Hope this gives you are start.
 
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theridane

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There's no general equation to calculate maneuver dv, because it depends on the mutual position and velocity of the two objects you want to transfer between.

A transfer generally consists of two impulses (or burns): the ejection burn, which puts you on the trajectory that intersects the trajectory of the target, and the insertion burn, which inserts you into your target's orbit.

You can either download Orbiter with IMFD and get some hands-on experience to help you build the mental picture of the trajectory and the impulses needed to traverse it, or you can run the numbers yourself.

Suppose you want to go from Earth to 1 Ceres:
  • You're starting in a circular Earth orbit of 350x350 km (~ISS), so your orbital velocity is 7.7 km/s. Escape velocity (the speed at which your orbit breaks into a hyperbolic sling-shot trajectory) is always greater or equal to circular_speed*√2, in this case about 10.9 km/s. Simple subtraction tells us that the delta-v needed for the spacecraft to leave Earth's orbit is about 3.2 km/s. Let's do the burn.
  • Now you're in heliocentric (around the Sun) orbit roughly in the same trajectory as the Earth is, your orbital speed in Sun's reference frame is roughly 29.7 km/s. You need to change the orbit so that it intersects the orbit (and position!) of Ceres. Here's where the fun starts - launch windows. Because the relative positions of all bodies change constantly the distances, relative velocities and delta-vs change accordingly. The differences can span two orders of magnitude (100s of km/s versus couple km/s). I'll be assuming that at the time of arrival Ceres will be on the opposite side of the Sun as you launched from - that would be called a Hohmann transfer, a keplerian transfer orbit that takes the least amount of delta-v and the most time to traverse. So, we need to increase the eccentricity of our orbit to touch Ceres. We need to raise the apoapsis (see this) to reach Ceres. Our apoapsis is currently 150 mil. km from the Sun and we need it to reach say 400 mil. km. From the equation in the wiki article you get that you need to accelerate to around 35.8 km/s, delta-v for this impulse is 6.1 km/s.
  • You're now coasting to Ceres. You're in an orbit similar to a planet, so your voyage is going to take over half a year.
  • Upon arrival to Ceres you need to lower your velocity so that you enter the orbit. If you want say a circular orbit 128 km above the surface, you need to slow down to 322 m/s relative to Ceres. But how fast are you going now? At perihelion you accelerated to 35.8 km/s relative to the Sun. You're meeting Ceres at your transfer orbit's apohelion, the farthest point from the Sun and the slowest point of your orbit. Another wiki equation tells you that you're going 13.4 km/s. Ceres itself is circling the Sun at 17.9 km/s, so you'll in fact need to accelerate to match the speed and add another 322 m/s to circularize your orbit. 17.9 - 13.4 + 0.3 = 4.8 km/s delta-v for this impulse.
  • You're in circular orbit around your target, 1 Ceres. Total delta-v was 3.2+6.1=9.3 km/s ejection burn and 4.8 km/s insertion burn, that's 14.1 km/s to get from LEO to Low Ceres Orbit.

Hope it helps :)
 
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MeteorWayne

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There is a further complication in reaching Ceres; it's orbit is inclined almost 11 degrees compared to that o earth, so some addition delta V will be required.
 
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Polishguy

Guest
theridane":251fqugu said:
There's no general equation to calculate maneuver dv, because it depends on the mutual position and velocity of the two objects you want to transfer between.

A transfer generally consists of two impulses (or burns): the ejection burn, which puts you on the trajectory that intersects the trajectory of the target, and the insertion burn, which inserts you into your target's orbit.

You can either download Orbiter with IMFD and get some hands-on experience to help you build the mental picture of the trajectory and the impulses needed to traverse it, or you can run the numbers yourself.

Suppose you want to go from Earth to 1 Ceres:
  • You're starting in a circular Earth orbit of 350x350 km (~ISS), so your orbital velocity is 7.7 km/s. Escape velocity (the speed at which your orbit breaks into a hyperbolic sling-shot trajectory) is always greater or equal to circular_speed*√2, in this case about 10.9 km/s. Simple subtraction tells us that the delta-v needed for the spacecraft to leave Earth's orbit is about 3.2 km/s. Let's do the burn.
  • Now you're in heliocentric (around the Sun) orbit roughly in the same trajectory as the Earth is, your orbital speed in Sun's reference frame is roughly 29.7 km/s. You need to change the orbit so that it intersects the orbit (and position!) of Ceres. Here's where the fun starts - launch windows. Because the relative positions of all bodies change constantly the distances, relative velocities and delta-vs change accordingly. The differences can span two orders of magnitude (100s of km/s versus couple km/s). I'll be assuming that at the time of arrival Ceres will be on the opposite side of the Sun as you launched from - that would be called a Hohmann transfer, a keplerian transfer orbit that takes the least amount of delta-v and the most time to traverse. So, we need to increase the eccentricity of our orbit to touch Ceres. We need to raise the apoapsis (see this) to reach Ceres. Our apoapsis is currently 150 mil. km from the Sun and we need it to reach say 400 mil. km. From the equation in the wiki article you get that you need to accelerate to around 35.8 km/s, delta-v for this impulse is 6.1 km/s.
  • You're now coasting to Ceres. You're in an orbit similar to a planet, so your voyage is going to take over half a year.
  • Upon arrival to Ceres you need to lower your velocity so that you enter the orbit. If you want say a circular orbit 128 km above the surface, you need to slow down to 322 m/s relative to Ceres. But how fast are you going now? At perihelion you accelerated to 35.8 km/s relative to the Sun. You're meeting Ceres at your transfer orbit's apohelion, the farthest point from the Sun and the slowest point of your orbit. Another wiki equation tells you that you're going 13.4 km/s. Ceres itself is circling the Sun at 17.9 km/s, so you'll in fact need to accelerate to match the speed and add another 322 m/s to circularize your orbit. 17.9 - 13.4 + 0.3 = 4.8 km/s delta-v for this impulse.
  • You're in circular orbit around your target, 1 Ceres. Total delta-v was 3.2+6.1=9.3 km/s ejection burn and 4.8 km/s insertion burn, that's 14.1 km/s to get from LEO to Low Ceres Orbit.

Hope it helps :)
Thank you! That's very helpful! :D
 
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