Orbit circularisation/insertion spreadsheet

[Treva31]

I am trying to create a spreadsheet that will calculate:

The minimum delta-v required to circularise the orbit of a projectile that has been fired horizontally from the Earths surface at a given velocity.
Completely ignoring atmospheric drag.

See the attached spreadsheet.
The steps are detailed on the second sheet.Can anyone see where I have gone wrong or suggest a better way?It is based on a previous post here: https://www.physicsforums.com/threa...lready-at-ideal-velocity-and-altitude.851982/

Janus provided an excellent answer and equations that I attempted to use when creating my spreadsheet.
However my spreadsheet does not give the same result as his example.

There was also an excellent answer by Jenab2 but I wouldn't know where to start making a spreadsheet based on it.

 

[mfb]

You fire from a known height and in horizontal direction. You can calculate the necessary velocity for a circular orbit, and the difference to the launch speed is your answer. Why do you want to make a spreadsheet for a single subtraction of two numbers?

 

[Treva31]

You fire from a known height and in horizontal direction. You can calculate the necessary velocity for a circular orbit, and the difference to the launch speed is your answer. Why do you want to make a spreadsheet for a single subtraction of two numbers?

But it needs to change the trajectory right? Not just the speed.

Can you give an example of what you mean?
Launching from sea level and entering a circular orbit at 400km.

 

[mfb]

Launching from sea level and entering a circular orbit at 400km.

Ah okay, so you want to go up first.

Why don't you calculate the apogee from the known perigee and the semi-major axis? Your calculated values don't fit.

 

[Treva31]

Ah okay, so you want to go up first.

Why don't you calculate the apogee from the known perigee and the semi-major axis? Your calculated values don't fit.

No not up first.

So you shoot from sea level horizontally. Let's say you shoot at the correct speed for a 400km orbit.
As the Earth curves away the projectile gains altitude.
So it reaches 400km, and let's say it hasn't slowed down at all so it as at the correct speed for a 400km orbit.
But it doesn't just start orbiting right? It's going straight and the Earth (and the orbit trajectory) is still curving away. Before long it is at 401km altitude and still the 400km orbit speed.
It needs to change direction a little bit to follow the curve instead of going straight. So how much delta-v does it need to do that?

 

[Bandersnatch]

To clarify:
are you trying to do this kind of transfer:

Or this:

(in your scenario the 'initial orbit' is the Earth's surface, and the '1st burn' is the projectile being shot)
The first type of transfer requires lower delta V, if that's what you're looking for. It's also easier to calculate.

 

[Treva31]

To clarify:
are you trying to do this kind of transfer:
View attachment 224368
Or this:
View attachment 224369
(in your scenario the 'initial orbit' is the Earth's surface, and the '1st burn' is the projectile being shot)
The first type of transfer requires lower delta V, if that's what you're looking for. It's also easier to calculate.

Minimal delta-v on the "second burn" certainly is the aim.
The delta-v of the "first burn" doesn't matter at all, happy for it to be inefficient.
So does that still mean the first type is better for me? If so let's go with that.

Are there names for those 2 types?

Thanks.

 

[Bandersnatch]

Yea ok let's go with the first one, minimal delta-v on the "second burn" certainly is the aim.

Alright, then this is the case Janus calculated for you in the other thread in his first response:

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)

You don't need to change the direction in this manoeuvre.
The projectile is fired horizontally w/r to the surface, with initial velocity higher than what is needed for a circular orbit at this radius from the centre of the gravity field. I.e. greater than the 'first cosmic velocity':
$$V=\sqrt{\frac{GM}{r}}$$
G - gravitational constant
M - mass of the Earth
r - radius of the circular orbit (here, radius of the Earth)

Launch velocity higher than that makes the orbit elliptical, where the projectile starts horizontally to the Earth's surface at the periapsis. It then climbs higher, exchanging some of its kinetic energy for potential energy (i.e. slowing down), and arrives at the apoapsis of the ellipse - again horizontally to the surface. At this point, it has too low a velocity to stay at this distance, so it starts descending back towards the periapsis.
That's why there's no need to change direction, and a simple insertion burn at apoapsis, making the velocity at that point equal to the velocity needed for a circular orbit, is all that is needed.

You can calculate the velocity at a given height of the apoapsis from the conservation of energy equations.
The total orbital energy is the sum of the potential and kinetic energies, and it stays constant throughout the orbit. It also equals half the potential energy at the distance equal to the semi-major axis.

$$E=const=\frac{mV^2}{2}-G\frac{Mm}{r}=-G\frac{Mm}{2a}$$
E - total energy
M - mass of Earth
m - mass of the smaller body (cancels out anyway, so use 1)
r - distance from the centre of the Earth
a - semi major axis
V - speed of the orbiting body

You can calculate the total energy using the $-G\frac{Mm}{2a}$ part, and deduct from it the potential energy at the desired orbital distance (i.e. apoapsis; assuming here: radius of the Earth+400 km). The difference is the kinetic energy at apoapsis. Knowing kinetic energy it's trivial to get velocity at apoapsis.
It will be lower than the velocity needed for the circular orbit at that same distance (use first equation above). The difference between the two is the required Delta-V for insertion.
In the same way, you can calculate velocity at periapsis, i.e. the launch velocity needed for raising the orbiter to the apoapsis distance. The sum of the two calculated velocities is the total Delta-V needed to put the craft on a circular orbit at a given height.

I'm getting the same numbers Janus got from his calculations in the quoted paragraph, btw.

Are there names for those 2 types?

First one is called Hohmann transfer orbit. It is a special case of the second type, which is a more generic two-impulse transfer. I would swear there was some name attached to the non-Hohmann type, or at least to its flight time-minimising case, but for the love of me I can't remember what it was.

Anyway, here's some further reading:
https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec17.pdf
Wikipedia articles on 'orbital mechanics' and 'orbital maneuver' are also well-written, with lots of relevant equations, but arguaby less pedagogical.

 

[Janus]

Alright, then this is the case Janus calculated for you in the other thread in his first response:You don't need to change the direction in this manoeuvre.
The projectile is fired horizontally w/r to the surface, with initial velocity higher than what is needed for a circular orbit at this radius from the centre of the gravity field. I.e. greater than the 'first cosmic velocity':
$$V=\sqrt{\frac{GM}{r}}$$
G - gravitational constant
M - mass of the Earth
r - radius of the circular orbit (here, radius of the Earth)

Launch velocity higher than that makes the orbit elliptical, where the projectile starts horizontally to the Earth's surface at the periapsis. It then climbs higher, exchanging some of its kinetic energy for potential energy (i.e. slowing down), and arrives at the apoapsis of the ellipse - again horizontally to the surface. At this point, it has too low a velocity to stay at this distance, so it starts descending back towards the periapsis.
That's why there's no need to change direction, and a simple insertion burn at apoapsis, making the velocity at that point equal to the velocity needed for a circular orbit, is all that is needed.

You can calculate the velocity at a given height of the apoapsis from the conservation of energy equations.
The total orbital energy is the sum of the potential and kinetic energies, and it stays constant throughout the orbit. It also equals half the potential energy at the distance equal to the semi-major axis.

$$E=const=\frac{mV^2}{2}-G\frac{Mm}{r}=-G\frac{Mm}{2a}$$
E - total energy
M - mass of Earth
m - mass of the smaller body (cancels out anyway, so use 1)
r - distance from the centre of the Earth
a - semi major axis
V - speed of the orbiting body

You can calculate the total energy using the $-G\frac{Mm}{2a}$ part, and deduct from it the potential energy at the desired orbital distance (i.e. apoapsis; assuming here: radius of the Earth+400 km). The difference is the kinetic energy at apoapsis. Knowing kinetic energy it's trivial to get velocity at apoapsis.
It will be lower than the velocity needed for the circular orbit at that same distance (use first equation above). The difference between the two is the required Delta-V for insertion.
In the same way, you can calculate velocity at periapsis, i.e. the launch velocity needed for raising the orbiter to the apoapsis distance. The sum of the two calculated velocities is the total Delta-V needed to put the craft on a circular orbit at a given height.

I'm getting the same numbers Janus got from his calculations in the quoted paragraph, btw.

Which can be simplified down to the Vis-viva equation:
$$ V^2= GM \left ( \frac{2}{r}- \frac{1}{a} \right )$$
where a is the semi-major axis of the orbit.

 

[Bandersnatch]

Yes. I thought it would be easier to understand if I went through energy conservation step-by-step.
I'm only now looking at OP's spreadsheet (sorry!). I can see @Treva31 used vis-viva to get the semi-major axis. This bit looks fine.
The apoapsis distance calculated later on doesn't match, however (it's ~60 km short).
I think there might be an error in the eccentricity calculations, since that's what is used to get the apoapsis.

Taking:
$r_p=a(1-e)$
$e=1- \frac{r_p}{a}$
I get e=0,030 for the first row vs the calculated value: 0,021

Yes, that's definitely it. Substituting the above formula gives the correct values in the reminder of the spreadsheet.
I don't recognise the equation used for eccentricity. Why is it there?

 

[Janus]

Yes, that's definitely it. Substituting the above formula gives the correct values in the reminder of the spreadsheet.
I don't recognise the equation used for eccentricity. Why is it there?

What he did was take two expressions for Areal velocity (the volume swept out by the radial line of the orbit per time unit), equated them and solved for e

The two expressions are:

$$A = \frac{r cos{\theta}v}{2}$$
where r is the radial distance at a given point of the orbit
v is the velocity at that point
theta is the angle of the velocity relative to the perpendicular to the radial direction.
and
$$A= \frac{\pi a^2 \sqrt{1-e^2}}{T_{per}} $$
The second equation being the area of the oribit's ellipse divided by the orbital period.

 

[Bandersnatch]

What he did was take two expressions for Areal velocity (the volume swept out by the radial line of the orbit per time unit), equated them and solved for e

A-ha!
@Treva31 when calculating eccentricity from the two equations Janus listed above, you were squaring both sides to eliminate the square root of $\sqrt{1-e}$, but you actually forgot to square the other side of the equation. It should correctly read:
$$e=\sqrt{1-(\frac{PV_iR_ecos{\phi}}{2\pi a^2})^2}$$
But if you're launching the projectile horizontally, as you're doing in your spreadsheet, and which you do want to do if your goal is to minimise Delta-V, this is unnecessary busywork. With horizontal launch, if you have the launch distance and the semi-major axis, then you already have eccentricity from the simple relationsip shown in post #10 above.
In any case, with correct eccentricity, the rest of the spreadsheet seems to work just fine.

 

[Janus]

A-ha!
@Treva31 when calculating eccentricity from the two equations Janus listed above, you were squaring both sides to eliminate the square root of $\sqrt{1-e}$, but you actually forgot to square the other side of the equation. It should correctly read:
$$e=\sqrt{1-(\frac{PV_iR_ecos{\phi}}{2\pi a^2})^2}$$
But if you're launching the projectile horizontally, as you're doing in your spreadsheet, and which you do want to do if your goal is to minimise Delta-V, this is unnecessary busywork. With horizontal launch, if you have the launch distance and the semi-major axis, then you already have eccentricity from the simple relationsip shown in post #10 above.
In any case, with correct eccentricity, the rest of the spreadsheet seems to work just fine.

While a bit of overkill for this particular scenario, it is a nifty little equality when dealing with non-horizontal launches.

 

[Treva31]

Awesome thank you so much guys!

 

[mfb]

Not so fast. The Hohmann transfer is the lowest delta_v maneuver (if the ratio of semi-major axes is not too high) if we include both burns. But we do not, we only consider the second burn. A faster initial launch speed leads to a higher horizontal velocity at the right height in addition to a small radial velocity. It is not immediately clear if this is always worse.

To simplify calculations, let's define the radius of Earth as R=1, and the specific gravitational potential is 1/r. The circular orbital velocity is then given by 1/2 v² = 1/(2r) or v² = 1/r. At the surface of Earth this leads to v=1.

We launch with a velocity v>1 with a target radius r_t>1. At this target radius the horizontal velocity dropped to $v_{th} = \frac{v}{r_t}$ from conservation of angular momentum, while the total velocity v_t comes from conservation of energy: $v^2 - 2 = v_t^2 - \frac{2}{r_t}$ => $v_t^2 = v^2 -2 + \frac{2}{r_t}$. Its radial component $v_{tr}$ is then given by $v_{tr}^2 = v_t^2 - v_{th}^2 = v^2 -2 + \frac{2}{r_t} - \frac{v^2}{r_t^2}$.

Our orbital burn has to remove the radial component and increase the horizontal component to $\frac{1}{\sqrt{r_t}}$, the latter means a change of $\frac{1}{\sqrt{r_t}}-\frac{v}{r_t}$. The total delta_v comes from both added in quadrature.

A minimum in this quadratic sum is also a minimum of the square of it, so we can simply add the squares and look for a minimum of them as function of v.
$$(\Delta v)^2 = v^2 -2 + \frac{2}{r_t} - \frac{v^2}{r_t^2} + \left(\frac{1}{\sqrt{r_t}}-\frac{v}{r_t}\right)^2 \\
= v^2 -2 + \frac{2}{r_t} - \frac{v^2}{r_t^2} + \frac{1}{r_t} -\frac{2v}{r_t^{3/2}} + \frac{v^2}{r_t^2} = v^2 - 2 +\frac{3}{r_t}-\frac{2v}{r_t^{3/2}}$$
Calculate the derivative:
$$\frac{d(\Delta v)^2}{dv} = 2v-\frac{2}{r_t^{3/2}}$$
The second derivative is positive so we actually have a minimum. Setting the first derivative equal to 0 and solving for v gives $v=r_t^{-3/2}$. This result is clearly unphysical as it does not satisfy v>1, which means our minimum is indeed at the lowest possible velocity.

This changes for very wide target orbits, where a bielliptic transfer becomes cheaper, especially (!) if you don't have to "pay" for the first burn.

As the Earth curves away the projectile gains altitude.

That is what I meant with "up first" (=not an orbit at surface level).

 

[Treva31]

Not so fast. The Hohmann transfer is the lowest delta_v maneuver (if the ratio of semi-major axes is not too high) if we include both burns. But we do not, we only consider the second burn. A faster initial launch speed leads to a higher horizontal velocity at the right height in addition to a small radial velocity. It is not immediately clear if this is always worse.

To simplify calculations, let's define the radius of Earth as R=1, and the specific gravitational potential is 1/r. The circular orbital velocity is then given by 1/2 v² = 1/(2r) or v² = 1/r. At the surface of Earth this leads to v=1.

We launch with a velocity v>1 with a target radius r_t>1. At this target radius the horizontal velocity dropped to $v_{th} = \frac{v}{r_t}$ from conservation of angular momentum, while the total velocity v_t comes from conservation of energy: $v^2 - 2 = v_t^2 - \frac{2}{r_t}$ => $v_t^2 = v^2 -2 + \frac{2}{r_t}$. Its radial component $v_{tr}$ is then given by $v_{tr}^2 = v_t^2 - v_{th}^2 = v^2 -2 + \frac{2}{r_t} - \frac{v^2}{r_t^2}$.

Our orbital burn has to remove the radial component and increase the horizontal component to $\frac{1}{\sqrt{r_t}}$, the latter means a change of $\frac{1}{\sqrt{r_t}}-\frac{v}{r_t}$. The total delta_v comes from both added in quadrature.

A minimum in this quadratic sum is also a minimum of the square of it, so we can simply add the squares and look for a minimum of them as function of v.
$$(\Delta v)^2 = v^2 -2 + \frac{2}{r_t} - \frac{v^2}{r_t^2} + \left(\frac{1}{\sqrt{r_t}}-\frac{v}{r_t}\right)^2 \\
= v^2 -2 + \frac{2}{r_t} - \frac{v^2}{r_t^2} + \frac{1}{r_t} -\frac{2v}{r_t^{3/2}} + \frac{v^2}{r_t^2} = v^2 - 2 +\frac{3}{r_t}-\frac{2v}{r_t^{3/2}}$$
Calculate the derivative:
$$\frac{d(\Delta v)^2}{dv} = 2v-\frac{2}{r_t^{3/2}}$$
The second derivative is positive so we actually have a minimum. Setting the first derivative equal to 0 and solving for v gives $v=r_t^{-3/2}$. This result is clearly unphysical as it does not satisfy v>1, which means our minimum is indeed at the lowest possible velocity.

This changes for very wide target orbits, where a bielliptic transfer becomes cheaper, especially (!) if you don't have to "pay" for the first burn.

That is what I meant with "up first" (=not an orbit at surface level).

Ah ok, VERY interesting!
So I might need to make a spreadsheet using this and compare the two to find the lowest second burn value for a reasonable LEO.. (I'm thinking between 300 and 600km)