Orbital Mechanics Questions

[Crixomix]

Hey all. I've played over a thousand hours and studied orbital mechanics as much as I can, but I STILL am having trouble understanding a few things... I will number them and try to organize the questions as best as I can.

1. Example A. I am going to the Mun from Kerbin. Which of the following options would cost less dV total:
   1. Launch into a nice 90km orbit around kerbin, wait until i'm in the right spot for a nice prograde burn, and get a Mun intercept that way.
   2. Wait until the launch site of kerbin is pointing more or less "at" the Mun, and literally launch straight up and never even bother circularizing around Kerbin, get an intercept with the Mun, and then go from there. My gut tells me the intercept with the Mun will take less dV, but it will take more dV to do the insertion burn around the Mun. The question is which one is worth more?
2. Example B. I am going to the Mun from Kerbin, which of the following costs less dV?
   1. I set up a Mun encounter where my craft, once entering the Mun's SOI, passes BEHIND the Mun (relative to the direction the Mun is orbiting Kerbin), and then I circularize at that periapsis.
   2. I set up a Mun encounter where my craft, once entering the Mun's SOI, passes IN FRONT of the Mun, and I circularize at that periapsis.
3. Example C. I'm landed on the Mun with a craft and want to go back to Kerbin. Again, what's the most efficient?
   1. I launch in an eastward orbit and circularize at a low altitude (let's say 30km). Then I burn prograde such that my ejection from the Mun is directly "backwards" from the Mun's direction of travel, thus lowering my periapsis down to Kerbin. This would mean burning prograde when on the "front side" of the Mun, between Kerbin & the Mun.
   2. Same as #1 but in a WESTWARD orbit (which would put me on the far side of the Mun when burning prograde to get back to Kerbin).
   3. Neither 1 or 2, but actually go straight towards "backwards" from the mun. Launching in such a way that I never circularize around the Mun, but I eject from the mun away from the Mun's direction of travel, thus lowering periapsis to Kerbin.

Thanks for your help everyone! Bonus question: what about planets, but the same as in example B. Should you encounter planets in front of them or behind them?

Edited August 16, 2020 by Crixomix

[Entropian]

1 hour ago, Crixomix said:

Example A. I am going to the Mun from Kerbin. Which of the following options would cost less dV total:

1. Launch into a nice 90km orbit around kerbin, wait until i'm in the right spot for a nice prograde burn, and get a Mun intercept that way.
2. Wait until the launch site of kerbin is pointing more or less "at" the Mun, and literally launch straight up and never even bother circularizing around Kerbin, get an intercept with the Mun, and then go from there. My gut tells me the intercept with the Mun will take less dV, but it will take more dV to do the insertion burn around the Mun. The question is which one is worth more?

 

I think 1) is the cheapest burn.

1 hour ago, Crixomix said:

Example B. I am going to the Mun from Kerbin, which of the following costs less dV?

1. I set up a Mun encounter where my craft, once entering the Mun's SOI, passes BEHIND the Mun (relative to the direction the Mun is orbiting Kerbin), and then I circularize at that periapsis.
2. I set up a Mun encounter where my craft, once entering the Mun's SOI, passes IN FRONT of the Mun, and I circularize at that periapsis.

 

I think you mean going from Kerbin to the Mun.  I think both are pretty much equal in dV.

1 hour ago, Crixomix said:

Example C. I'm landed on the Mun with a craft and want to go back to Kerbin. Again, what's the most efficient?

1. I launch in an eastward orbit and circularize at a low altitude (let's say 30km). Then I burn prograde such that my ejection from the Mun is directly "backwards" from the Mun's direction of travel, thus lowering my periapsis down to Kerbin. This would mean burning prograde when on the "front side" of the Mun, between Kerbin & the Mun.
2. Same as #1 but in a WESTWARD orbit (which would put me on the far side of the Mun when burning prograde to get back to Kerbin).
3. Neither 1 or 2, but actually go straight towards "backwards" from the mun. Launching in such a way that I never circularize around the Mun, but I eject from the mun away from the Mun's direction of travel, thus lowering periapsis to Kerbin.

 

1) and 2) both are practically the same thing because the Mun is tidally locked.  3) is really inconvenient because you have to land on a specific point on the Mun.

 

1 hour ago, Crixomix said:

Bonus question: what about planets, but the same as in example B. Should you encounter planets in front of them or behind them?

Doesn't really matter except for gravitational assists, as far as I know.

[jimmymcgoochie]

One thing I would say about questions 2 and 3 is this- for most planets and many moons, it makes a lot more sense to enter a prograde orbit (i.e. come in 'behind' them so your orbit is counter-clockwise when viewed from the north pole) if you plan to land, and to launch into a prograde orbit when leaving*.

For something like the Mun, Ike or the inner 3 moons of Jool, which are all tidally locked to their parent, the rotation of the moon is pretty slow so the difference between surface and orbital velocity is relatively small; however for other moons and (almost!) all planets you'll end up spending more fuel when trying to land retrograde as you have to cancel out twice as much speed as the moon/planet's rotation to bring your surface velocity to zero. Not significant on Gilly, a minor irritation on Minmus, but a bit more of a problem for somewhere like Eeloo or Dres. Moho is the odd planet out in this case as its rotation speed is so slow (much like Mercury).

If that didn't make sense, look at your next rocket on the launch pad. Zero surface velocity, but 175m/s orbital velocity. To launch into an equatorial orbit retrograde will take 350m/s more delta-V than a prograde equatorial orbit because you have to make up for that speed difference twice over- once to 'stop' your orbital rotation relative to Kerbin and the second time to apply it again in a retrograde direction. This effect is reduced the further from the equator your orbit is, but retrograde orbits require more delta-V to land from and launch to on every planet or moon; it makes no real difference when transferring from another planet or moon as the difference in velocity will be negligible and orbital velocity is the same going prograde or retrograde.

Regarding direct vertical launches- they are theoretically less efficient than orbiting and then leaving as you're fighting directly against gravity the entire time rather than coming at it from an angle, plus it's pretty hard to get your landing site right on a tidally locked body or to time your launch right anywhere else. Not impossible, but difficult enough that you should really aim to launch into orbit first.

*The only reason to launch into a retrograde orbit from Kerbin to go somewhere else is to allow ion-powered ships to stay in the sun when aiming towards a destination further out from the sun; prograde orbits mean that your ideal departure point is in the shadow of Kerbin which isn't any use for solar panels, and ion drives use quite a bit of power. The situation is reversed when aiming sunwards- prograde orbits are best for staying in the sun when burning towards a target nearer the sun with ion drives. (The best solution is of course to use a non-ion powered thrust system when leaving Kerbin, as ions are really wimpy and you'll get terrible burn accuracy with loooooong burns as a result.

[Crixomix]

1 hour ago, jimmymcgoochie said:

One thing I would say about questions 2 and 3 is this- for most planets and many moons, it makes a lot more sense to enter a prograde orbit (i.e. come in 'behind' them so your orbit is counter-clockwise when viewed from the north pole) if you plan to land, and to launch into a prograde orbit when leaving*.

For something like the Mun, Ike or the inner 3 moons of Jool, which are all tidally locked to their parent, the rotation of the moon is pretty slow so the difference between surface and orbital velocity is relatively small; however for other moons and (almost!) all planets you'll end up spending more fuel when trying to land retrograde as you have to cancel out twice as much speed as the moon/planet's rotation to bring your surface velocity to zero. Not significant on Gilly, a minor irritation on Minmus, but a bit more of a problem for somewhere like Eeloo or Dres. Moho is the odd planet out in this case as its rotation speed is so slow (much like Mercury).

If that didn't make sense, look at your next rocket on the launch pad. Zero surface velocity, but 175m/s orbital velocity. To launch into an equatorial orbit retrograde will take 350m/s more delta-V than a prograde equatorial orbit because you have to make up for that speed difference twice over- once to 'stop' your orbital rotation relative to Kerbin and the second time to apply it again in a retrograde direction. This effect is reduced the further from the equator your orbit is, but retrograde orbits require more delta-V to land from and launch to on every planet or moon; it makes no real difference when transferring from another planet or moon as the difference in velocity will be negligible and orbital velocity is the same going prograde or retrograde.

Regarding direct vertical launches- they are theoretically less efficient than orbiting and then leaving as you're fighting directly against gravity the entire time rather than coming at it from an angle, plus it's pretty hard to get your landing site right on a tidally locked body or to time your launch right anywhere else. Not impossible, but difficult enough that you should really aim to launch into orbit first.

*The only reason to launch into a retrograde orbit from Kerbin to go somewhere else is to allow ion-powered ships to stay in the sun when aiming towards a destination further out from the sun; prograde orbits mean that your ideal departure point is in the shadow of Kerbin which isn't any use for solar panels, and ion drives use quite a bit of power. The situation is reversed when aiming sunwards- prograde orbits are best for staying in the sun when burning towards a target nearer the sun with ion drives. (The best solution is of course to use a non-ion powered thrust system when leaving Kerbin, as ions are really wimpy and you'll get terrible burn accuracy with loooooong burns as a result.

Thanks for this. I did know about the rotational velocity piece but had kinda forgotten that would be a factor. However, I'm still very curious about even if you didn't land on the planet, and disregarding things like sun location. Does it actually change the dV requirements for ejection + insertion to go in front of vs. behind the mun? The main thinking is this: When you slingshot in front of the Mun, it slows down your orbit, reducing total orbital energy. Whereas when you slingshot behind it, it increases total orbital energy. And you NEED more orbital energy to "catch up" with the Mun when you get near it, since it's going faster than you, relative to Kerbin. So that seems to imply that it's going to be cheaper to get into a mun orbit from behind the mun, rather than in front of it, because you get more energy added to your orbit relative to Kerbin for "free" by being behind it. But I have no evidence of this and need a more complete explanation I think.

[jimmymcgoochie]

2 minutes ago, Crixomix said:

Thanks for this. I did know about the rotational velocity piece but had kinda forgotten that would be a factor. However, I'm still very curious about even if you didn't land on the planet, and disregarding things like sun location. Does it actually change the dV requirements for ejection + insertion to go in front of vs. behind the mun? The main thinking is this: When you slingshot in front of the Mun, it slows down your orbit, reducing total orbital energy. Whereas when you slingshot behind it, it increases total orbital energy. And you NEED more orbital energy to "catch up" with the Mun when you get near it, since it's going faster than you, relative to Kerbin. So that seems to imply that it's going to be cheaper to get into a mun orbit from behind the mun, rather than in front of it, because you get more energy added to your orbit relative to Kerbin for "free" by being behind it. But I have no evidence of this and need a more complete explanation I think.

Orbital slingshots only affect the craft's trajectory relative to the parent body. It makes absolutely no difference to the orbit of the body you're inside the SOI of at the time and braking into orbit prograde and retrograde will cost essentially the same delta-V. If you're getting a gravity slingshot from the Mun, your angular velocity is being changed relative to Kerbin- the Mun doesn't care either way as you'll leave its SOI at the same speed you entered it due to conservation of momentum.

Think of it this way:

If you make a transfer burn from a fixed orbit (e.g. 80km by 80km prograde equatorial LKO) to the Mun, the delta-V required to elevate your apoapsis to intersect the Mun's orbit is the same no matter when you make the burn because the Mun's orbit of Kerbin is perfectly circular. If you change when you make the transfer burn, that will put your periapsis either in front of or behind the Mun when it intercepts; the velocity is exactly the same, so the capture burn will require exactly the same delta-V at the same periapsis altitude.

It's only when you let the craft carry on out of the Mun's SOI that the trajectories will be different- going retrograde will slow you down relative to Kerbin, which is how you get a free return trajectory back to Kerbin without any further fuel use, but going prograde will quite probably slingshot you clear out of Kerbin's SOI.

And the TL;DR version of all that- just stick to prograde orbits unless there's a very specific reason to go retrograde, like free return trajectories, sun exposure for ion craft, contract parameters, or if you're using something like SCANsat where retrograde orbits cover more ground faster.

[bewing]

1) Partially, it depends on your TWR and drag characteristics. Launching straight up is technically simpler, has lower drag (because you increase altitude faster than with a proper gravity turn), and has much lower control requirements. You can achieve higher speeds faster, which helps with oberth. There is almost always a small dV penalty to launching straight up (because your gravity losses never end until you cross the SOI boundary), but that penalty can be countered by, for example, not needing a fairing for your payload.

2) You will find that, as you burn to leave Kerbin, your (constantly changing) projected orbit always encounters the Mun from the back, first. It always takes a little more of a burn to get an encounter that goes around the front. However, there is a name for a trajectory that encounters the Mun from the front. It's called a "free return" trajectory. If you read about the Apollo missions, for example, you will find that they did it this way. The reason is that if your orbit insertion engines fail, then the rest of your trajectory takes you back very close to the home planet, and you can get home again with just a very small burn. Since KSP engines never fail, this is not really a concern, but it is interesting for historical and IRL reasons.

3) One interesting fact often makes a difference, here. If you are prograde orbiting a planet beyond Kerbin, and you want to come home, then your burn point is always in the dark, and usually has no comm link with Kerbin. Similarly with a moon, the burn point for a retrograde orbit will never have LOS comms with Kerbin (orientation with the Sun is random). So, having solar power is sometimes important (especially for ion drives), and having comms and control is sometimes important.

Edited August 16, 2020 by bewing

[king of nowhere]

22 hours ago, Crixomix said:

Hey all. I've played over a thousand hours and studied orbital mechanics as much as I can, but I STILL am having trouble understanding a few things... I will number them and try to organize the questions as best as I can.

1. Example A. I am going to the Mun from Kerbin. Which of the following options would cost less dV total:
   1. Launch into a nice 90km orbit around kerbin, wait until i'm in the right spot for a nice prograde burn, and get a Mun intercept that way.
   2. Wait until the launch site of kerbin is pointing more or less "at" the Mun, and literally launch straight up and never even bother circularizing around Kerbin, get an intercept with the Mun, and then go from there. My gut tells me the intercept with the Mun will take less dV, but it will take more dV to do the insertion burn around the Mun. The question is which one is worth more?

 

I tried option 2. i also had a hunch that it may be cheaper. but it wasn't. i had enough deltaV on my ship to get to mun normally, and i burned it all without ever getting close.

while i am not sure about the reasons for this, i can hypothesize that it is because of gravity drag. as long as you're not in orbit, your rocket is falling down and you have to constantly burn to make up for it. if you keep launching straight up, you never enter orbit, and you keep paying for gravity drag.

Quote

1. Example C. I'm landed on the Mun with a craft and want to go back to Kerbin. Again, what's the most efficient?
   1. I launch in an eastward orbit and circularize at a low altitude (let's say 30km). Then I burn prograde such that my ejection from the Mun is directly "backwards" from the Mun's direction of travel, thus lowering my periapsis down to Kerbin. This would mean burning prograde when on the "front side" of the Mun, between Kerbin & the Mun.
   2. Same as #1 but in a WESTWARD orbit (which would put me on the far side of the Mun when burning prograde to get back to Kerbin).
   3. Neither 1 or 2, but actually go straight towards "backwards" from the mun. Launching in such a way that I never circularize around the Mun, but I eject from the mun away from the Mun's direction of travel, thus lowering periapsis to Kerbin.

Thanks for your help everyone! Bonus question: what about planets, but the same as in example B. Should you encounter planets in front of them or behind them?

 

1 and 2, the only difference in cost is that by orbiting eastward you get the extra boost from the planet rotation. as others pointed out, on mun it is very small, but elsewhere it is significant.

3 does not work, for the same reason i explained before.

[JebIsDeadBaby]

On 8/17/2020 at 4:09 PM, king of nowhere said:

i had enough deltaV on my ship to get to mun normally, and i burned it all without ever getting close.

This may be heavily dependent on TWR of your rocket. Imagine, you have a 100 kg rocket capable of 1200 N of thrust. From the Newton's 2nd law of motion we know it will be capable of initial acceleration of 12 m/s^2. BUT... when your rocket goes straight up, there is gravity which accelerates it in the opposite direction at ~10 m/s^2 (on Earth). This means your rocket accelerates at 2 m/s^2, 6 times slower than expected, while still burning fuel at full rate. A 1000 N engine (which gives TWR of 1.0) not only wouldn't be able to reach the Mun or even 80 km, it wouldn't go anywhere at all. 

So I think a rocket powerful enough to reach the required speed fast enough could in theory be more efficient. 

[king of nowhere]

38 minutes ago, JebIsDeadBaby said:

This may be heavily dependent on TWR of your rocket. Imagine, you have a 100 kg rocket capable of 1200 N of thrust. From the Newton's 2nd law of motion we know it will be capable of initial acceleration of 12 m/s^2. BUT... when your rocket goes straight up, there is gravity which accelerates it in the opposite direction at ~10 m/s^2 (on Earth). This means your rocket accelerates at 2 m/s^2, 6 times slower than expected, while still burning fuel at full rate. A 1000 N engine (which gives TWR of 1.0) not only wouldn't be able to reach the Mun or even 80 km, it wouldn't go anywhere at all. 

So I think a rocket powerful enough to reach the required speed fast enough could in theory be more efficient. 

nope. i can recognize the difference between my ship howering mid-air and my ship getting faster and faster but never getting close to mun.

you can only squeeze TWR so far. even if your acceleration once out of the atmosphere was istantaneous, you still need hours to go to mun, and during those hours you're losing speed all the time by gravity drag, in addition to what you lose by gravity potential. consider this: at 600 km, gravity is 0.25 g. say that in average from LKO to 600 km you are subject to 0.5 g. say that you immediately achieve 3 km/s once out of atmosphere - that's the amount of deltaV it would take to reach mun by normal means. so, at 3 km/s, you  need 3 minutes to cover that distance. and during those 3 minutes, you lose (180*g*0,5)=900 m/s. no, worse, because as you slow down, you need more than 3 minutes. which result in going even slower. and you're losing speed by gravity potential too. and then, once you are to 600 km, you still need to get farther. the 0.125 g (1/8) mark is at 1800 km. sure, you are subject to less gravity along the way, but it's a longer road. and you are going slower. you should see that going straight up, you can never reach mun with a fuel budget comparable to the normal manuever. 

of course, you lose less speed if you start faster, but by now it's already no longer economical.

and don't forget that the faster you go, the faster you will be at mun intercept, and the more fuel you'll need to slow down.

EDIT: oh, i forgot to mention that in practice it would take you several minutes to achieve those 3 km/s with a rocket burn. and all the while, you are still losing speed because you are burning against gravity. sure, a more powerful rocket can help you there, but remember that 1) bigger engine means more dry mass, translating to less deltaV, and 2) engines with high TWR have poor Isp, again translating to less deltaV. as it is, if you had a TWR of 4 (extremely big and overpowered rocket) you could get an upward acceleration equal to 3 g (1 is required to counter the gravity), and it would take you 100 seconds to reach 3 km/s. you'd make 150 km along the way, and you'd have burned 4 km/s of deltaV. already much higher than the normal manuever

Edited August 28, 2020 by king of nowhere

[Spricigo]

13 hours ago, king of nowhere said:

nope. i can recognize the difference between my ship howering mid-air and my ship getting faster and faster but never getting close to mun.

Well,  the discussion is about how to reach the Mun with lower deltaV cost. Those are both cases of not reaching the Mun.

@JebIsDeadBaby is correct when he says it depends on TWR (or rather, Excess Thrust) because a more powerful rocket reach the required speed faster. 

However what he gets wrong is that even with insanely powerful rockets we don't want to go straight up, that is not the direction we are already going. As we know, the most effective way to raise the orbit is to burn prograde at the lower point of it. But there is  inconveniences to do so from the launch pad, such as the lower part of our orbit being somewhere below the surface. Overheating because atmosphere is not quite what we are looking for either

So, what is left for us to do is to try to get that ideal trajectory, that is not shallow enough to suffer with high atmospheric drag and not steep enough to suffer with gravity drag.

 

 

 

 

 

 

 

[Zhetaan]

  

On 8/17/2020 at 10:09 AM, king of nowhere said:

I tried option 2. i also had a hunch that it may be cheaper. but it wasn't. i had enough deltaV on my ship to get to mun normally, and i burned it all without ever getting close.

while i am not sure about the reasons for this, i can hypothesize that it is because of gravity drag. as long as you're not in orbit, your rocket is falling down and you have to constantly burn to make up for it. if you keep launching straight up, you never enter orbit, and you keep paying for gravity drag.

Yes, but not quite the way you think.  So long as you are moving directly up (or, I suppose, at some point it's really more out than up), you are still in some sort of gravitational relationship with the primary, which is to say, in orbit.  This is a special class of orbit called a radial orbit.

Spoiler

The key defining feature of radial orbits is that they have zero angular momentum.  This also means zero semi-minor axes, but they all have eccentricities of 1 even though they are not all parabolic orbits.  This results from a feature of the way eccentricity is calculated:  if taken from the periapsis and apoapsis radii, r_p and r_a respectively, the periapsis is zero and the apoapsis is nonzero, so the equation e = (r_a - r_p) / (r_a + r_p) always gives an answer of 1.  Of course, this is also seen in the eccentricity-based calculations for periapsis and apoapsis radial distance:  r_p = a (1 - e) and r_a = a (1 + e) which gives 0 and 2a, where a is the semi-major axis, respectively.

If using the energy-based equation for eccentricity, e = √(1 + (2εh² / μ²)), ε is the specific orbital energy, μ is the standard gravitational parameter, and h is the specific angular momentum (angular momentum divided by the reduced mass)--but since radial orbits are defined by zero angular momentum, that term drops out, leaving an eccentricity of √1 or simply 1.

For yet another way of looking at them, there is a relationship between specific angular momentum and the velocities at the periapsis and apoapsis:  h = r_pv_p = r_av_a = constant, and if that constant is zero, then it means that all of the other terms must be zero, too.  The apoapsis achieves this with zero velocity.  For the periapsis, it is zero altitude.

Eccentricity is not an informative parameter of radial orbits, so instead they are characterised by their specific orbital energy.

Spoiler

Specific orbital energy is equal to -μ / 2a, which makes things easy for us because 2a is the apoapsis distance in a radial orbit.  Assuming that we want to go to the Mun, 2a is thus equal to 12,000,000 m, and the orbital energy required to go there is therefore -(3.5316 x 10¹² / 12 x 10⁶) or -294,300 joules per kilogram (the energy is negative because it is a closed orbit; escape orbits under this convention are non-negative).

Specific orbital energy is also equal to (v² / 2) - (μ / r).  This reflects the exchange between potential energy and kinetic energy at all points on the orbit, with the spacecraft's particular location expressed as r metres from the planet's centre.  Using the Mun altitude and our known needed specific energy of -294,300 J/kg, it should be plain that because r and 2a are the same at the apoapsis, that term provides the necessary energy on its own and thus the velocity there needs to be zero as shown above--which is exactly what we expect at the apex of a straight-up trajectory.  What, therefore, is the requirement at Kerbin's surface?

ε = (v² / 2) - (μ / r), where ε is -294,300, μ is 3.5316 x 10¹², and r is 600,000, gives this:
-294,300 = (v² / 2) - (3.5316 x 10¹² / 600,000)
-294,300 = (v² / 2) - 5,886,000
5,591,700 = v² / 2
11,183,400 = v²
3,344.2 m/s = v

This velocity is what you need to have at Kerbin's surface in order to reach the Mun--this is not considering atmospheric drag, of course.

Part of understanding this equation is knowing that in normal usage, potential energy is considered to increase with altitude and decrease as an object falls to end at zero when it reaches zero altitude, for whatever altitude you choose to use as your datum.  This works because an object at the same altitude as you has no potential with respect to you, but that won't work for orbits.  This equation complicates that by considering potential energy to be negative:  instead of a lower boundary at zero, the lower boundary is at negative infinity, which opens the possibility of altitudes from which it is physically impossible to ascend.  This interpretation (and one's rocket) breaks down upon impact with the surface, for several reasons, but the point is that when considering an orbit, potential begins at an extremely low value and increases eventually to zero at infinite distance--which is proper, because at infinite distance, an object does not have infinite potential energy so much as it instead becomes gravitationally unbound.

However, the greater part of the problem is that you don't have that velocity at Kerbin's surface--perhaps some kind of cannon would help (that's an interesting thought experiment) but instead you have zero velocity and need to build it up to that value, or its equivalent at whatever altitude you reach when you have the required velocity.  It's a problem in at least three variables:  you have needed velocity as a function of altitude, and you have acquired velocity as a function of time--assuming that rocket acceleration is constant, which it is not--and the answer will change with the rocket.

More relevant is the character of the orbit:  there is another situation in which we might see a spacecraft approach a point on a radial orbit or otherwise zero its angular momentum, and that is in a suicide burn landing--a radial approach to the Mun is essentially a reverse of a suicide burn landing from Mun altitude.

Imagine zeroing velocity at twelve million metres and coasting down to perform a suicide burn landing.  This is a terrible approach because all of that time spent going straight down involves a build-up of velocity that will cause problems if you don't shed it.  At the same time, reducing velocity means spending more time descending, more descent consuming propellant, and more propellant trying to hover.  Although suicide burns are the result of a desire to put off until the last second the necessary wastage of such a stunt in an effort to have as little unnecessary wastage as possible, that does not obviate the fact that wastage occurs.  The same is true of the reverse burn.

To illustrate this wastage, consider a more typical approach to changing one's orbit:  should you want to change the apoapsis of your orbit, where is the best place to do it?  Normally, that would be at the periapsis, but for a radial orbit, the periapsis is unavailable.  You can only change the apoapsis from the apoapsis side of the orbit--in fact, because of the way radial orbits work, it is functionally similar to waiting to reach the apoapsis of an elliptical orbit and then burning radial-out, which is to say, it is among the most inefficient methods possible.  This is why, even for suicide burns, the standard advice is to approach the surface as closely as possible before cancelling one's velocity and proceeding with the landing, which is then timed so that the radial-out burn is as short as it can be while still (hopefully) providing for the survival of the vehicle and its occupants.