# B Delta-v for orbit if already at ideal velocity and altitude?

Tags:
1. Jan 12, 2016

### Treva31

Lets assume you could fire a 100kg projectile using a coilgun at 1 degree above the horizon and have it arrive safely at 400km altitude travelling at 7.8km/s (or slightly more if that helps).

How much delta-v would it need to correct its trajectory into a stable orbit?

If it matters lets assume it is at and along the equator.
And also assume it can withstand the heating and drag of the atmosphere.

2. Jan 13, 2016

### Janus

Staff Emeritus
Some quick calculation gives me an answer of 0.681km/sec. Upon reaching 400 km in altitude, it would be moving at an angle of ~5 degrees to horizontal and ~130 m/sec faster than circular velocity for that altitude. You would have to fire it at ~8.26 km/sec. Total delta-v from launch to orbit 8.94 km/sec

However if you just fired it horizontally at 8.03 km/sec, it would arrive at a 400 km altitude moving horizontally at just 117.5 m/sec slower than orbital speed. Meaning you only need 117.5 m/sec to achieve circular orbit. Total delta v ~8.15 km/sec. (assuming all velocities are measured with respect to the Earth's center and not its surface and not accounting for the additional initial speed needed to overcome atmospheric drag)

On the other hand, if you want to keep your 1 degree launch angle, it would make more sense to fire at 7.633 km/sec, and have it arrive at 400 km traveling horizontally at 7.113 km/sec and needing an additional 556.6 m/sec to attain a circular orbit. This works out to a total delta-v of ~8.19 km/sec ( still slightly more than a horizontal launch)

3. Jan 13, 2016

### Treva31

Great info thanks!

4. Jan 13, 2016

### Treva31

Is there a formula/spreadsheet I can use to work out other launch angles?

5. Jan 13, 2016

### Janus

Staff Emeritus
I can walk you through the steps.

Assuming that Vi is the initial launch velocity, Phi is the angle from horizontal of the Launch and Re is the radius of the Earth:

Then from
(1) $$V_i = \sqrt{k \left ( \frac{2}{R_e}- \frac{1}{a} \right )}$$

where k = 3.987e14 m^3/s^2

you can solve for 'a', the semi-major axis of the orbital trajectory.

Then, with

(2)$$P = 2 \pi \sqrt {\frac{a^3}{k}}$$

you find P, the period of the orbit.

Then from the relationships:

(3)$$A = \frac {V_i R_e \cos \phi}{2}$$
and
(4)$$A = \frac{\pi a^2 (1-e^2)}{P}$$

Where A is the Areal velocity, you can solve for 'e' the eccentricity.

Now, from
(9)$$R_{apo} = a(1=e)$$

This minus R_e gives you the altitude of apogee.

using equation (1) from above and substituting R_apo for R_e and using the above arrived at value for a, we get the orbital speed at apogee (V_apo)

The circular orbital speed at the apogee altitude is

(10) $$V_o = \sqrt{\frac{k}{R_{apo}}}$$

and the difference between this and V__apo is the delta v needed to circularize the orbit at this altitude.

To work out the delta v needed to circularize the orbit at some lower altitude than apogee takes a bit more work.

Assuming that 'r' is the radial distance to the center of the Earth where you want to circularize the orbit, then you need to solve

(11) $$r=a \frac{1-e^2}{1+e \cos \theta}$$

for the theta, the angular distance from perigee.

Then with
(12) $$\tan f = \frac{e \sin \theta}{1+ e \cos \theta}$$

you solve for 'f' the angle to the horizontal for the object's velocity at that point.

Again equation (1) using 'r' gives the magnitude of the velocity(v) at this point. and equation (10) gives the circular orbital speed at this altitude ( again using 'r')

a little vector addition between the present orbital velocity (speed and direction) and the desired circular orbit velocity gives you the required delta v.

The other way to find 'f' is to note that the Areal velocity at apogee is
(13) $$A = \frac{V_{apo} R_{apo}}{2}$$

And since Areal velocity is constant throughout the orbit,

equations (13) and (3) can be equated by using r and v and solving for phi (which in this case will be 'f')

In your original scenario, you gave phi, the initial launch angle, r, the altitude of the object at a point of its trajectory and v, the magnitude of the velocity at that point.

To work out the information you wanted, you do the following:

With r and v solve equation (1) for a
With R_e and a solve for V_i again using equation (1)
With phi and V_i use equation(3) to solve for the Areal velocity
With the Areal velocity, v and r, use equation (3) to solve for phi(f)
Solve for the circular orbital velocity at r using equation (10)
Use vector addition to get the difference between circular and present orbital velocities.

One last point, the reason the total delta v to reach a circular orbit at 400 km was higher with your original set up than the other two was that with the original parameters, you were "overshooting the mark" and the projectile was moving too fast when it got to 400 km and you used up delta v shedding the excess.
Now if you let the projectile continue on its way to its apogee (1298 km altitude) it would have only taken a delta v of 346 m/s to attain a circular orbit at that altitude. This means for a total cost of 8.6 km/sec you get a higher orbit than you would trying the circularize the same launch trajectory at 400 km at a cost of 8.94 km/sec

6. Jan 13, 2016

### Treva31

Awesome, thank you so much

7. Jan 20, 2016

### Jenab2

The delta vee for any orbit change is the vector difference between the velocity that the rocket would have if it were in the desired new orbit and the velocity that the rocket has in its temporary old orbit, calculated for the same position in space.

Obviously, delta-vees are easiest to calculate if the position at which they occur is an apside of both the old orbit and the new one (unless the new orbit is circular). But the instantaneous velocity of an object in an elliptical orbit can be found for any value of the true anomaly as follows:

k = √{ GM / [ a (1−e²) ] }
Vx''' = −k sin θ
Vy''' = k (e + cos θ)
Vz''' = 0

Where GM = 1.3271244e20 m³ sec⁻² (for orbits going around the sun)
a = the semimajor axis of the orbit in meters
e = the orbit's eccentricity
θ = the true anomaly

The true anomaly is the angle subtended at the sun, measured in the plane of the object's orbit in the direction of the object's motion, from the perihelion to the planet's current position. It is zero at the orbit's periapsis and π radians at its apoapsis.

The triple prime velocity is referred to the coordinate system natural to the orbit: heliocentric, with the orbit defining the XY plane, with the x-axis extending through the orbit's perihelion. The velocity is corrected to heliocentric ecliptic coordinates as follows:

Rotate the triple-prime velocity vector by the argument of the perihelion, ω.

Vx'' = Vx''' cos ω − Vy''' sin ω
Vy'' = Vx''' sin ω + Vy''' cos ω
Vz'' = Vz''' = 0

Rotate the double-prime velocity vector by the inclination, i.

Vx' = Vx''
Vy' = Vy'' cos i
Vz' = Vy'' sin i

Rotate the single-prime velocity vector by the longitude of the ascending node, Ω.

Vx = Vx' cos Ω − Vy' sin Ω
Vy = Vx' sin Ω + Vy' cos Ω
Vz = Vz'

The unprimed velocity vector [Vx,Vy,Vz] is the sun-relative velocity in ecliptic coordinates.

Delta-vees enable orbit change. They are very easy to calculate if you know both the velocity in the orbit being left behind and the velocity required by the new orbit.

ΔVx = Vx₂ − Vx₁
ΔVy = Vy₂ − Vy₁
ΔVz = Vz₂ − Vz₁

Transfer orbits are intermediate orbits that connect an initial orbit with a final orbit, with a delta-vee occurring at both ends. In the event that the gravity of massive objects located near the positions of departure or arrival complicates the calculations, you'd need to use a patched conic method and allow for a third delta-vee as a mid-course correction.

(Note: in fact, it is possible in this case to dispense with the mid-course correction, if the correct departure delta-vee is calculated; however, the calculation is complicated and perhaps the easiest way to find it is through a full method of Gauss to find the delta-vee that results in a zero MOID at the arrival end. If you don't mind mid-course corrections, you can just use the third delta-vee to patch the escape hyperbola, relative to Earth, with the transfer orbit ellipse relative to the sun. The math is simpler, anyway.)

Last edited: Jan 20, 2016
8. Sep 27, 2016

### Treva31

OK I have finally put together a spreadsheet using your excellent instructions.
I am getting slightly different figures than your example but its close.
Perhaps its rounding errors, your example was a different flight-path than my calculations use, or I have made a mistake somewhere?

I would be very grateful if someone could check my spreadsheet's accuracy.

Examples on the first sheet and breakdown of the formulas on the second sheet.

File size:
374 KB
Views:
72