Reality check on SF action scenes

[Janus]

This is true. I read somewhere that there are hundreds of asteroids that size in the belt. A good percentage are believed to be nickel-iron that I need.
After arriving at the Earth rendezvous, if he slows it to about 7 km/s viewed from the Earth's frame, do you think the orbit will most likely be circular? Or elliptical with a periapsis of several hundred kms?

It depends upon what altitude it's at when it reaches 7 km/sec. At an altitude of 1759 km it would enter a circular orbit. Higher and it will be elliptical with an apogee at a higher altitude. A lower insertion point results in an elliptical orbit with a lower perigee altitude (The lower limit for the insertion will be ~830 km. Lower than that and the orbit would intersect the Earth's surface.

 

[GTOM]

It depends upon what altitude it's at when it reaches 7 km/sec. At an altitude of 1759 km it would enter a circular orbit. Higher and it will be elliptical with an apogee at a higher altitude. A lower insertion point results in an elliptical orbit with a lower perigee altitude (The lower limit for the insertion will be ~830 km. Lower than that and the orbit would intersect the Earth's surface.

Great, i also thought about orbital mechanics.

For my story, i have fusion ships. I estimated that with 100 MW to 100 ton, they could provide 4 miliG acceleration with 50 km/s exhaust velocity, reach Mars from Earth in a month, with two weeks of acceleration and deceleration. Longer ranges means higher exhaust velocity higher delta-V lower acceleration. (Estimated three months from Mercury to Ceres with 2 miliG and 100 km/s exhaust velocity)

I wonder, could such a ship alter its course in order to evade a threat and still land on its destination, just slower? (For example don't do full acceleration phase, but start decelerate, and change a rather straight vector to a more spiralic course. )
I plan a scene where a cargo ship in the asteroid belt is actually looted (a pretty rare event), the pirates (with a much smaller and agile cargo ship) land on a different asteroid.I also wonder about a situation at Mercury. The fighters take off at a base on the night side (not very far from North Pole), and want to attack defence towers of the North Pole, coming from the sunny side.
I can think about two ways, go around the entire planet with low orbital speed, but how about the shorter way, take a ballistic course from the night side to bright side, than stop and take orbital speed toward North? Would it be very energy consuming? I think the fighters delta-V could be about 10km/s with 10km/s exhaust velocity (nuclear battery powered thrusters)

 

[chasrob]

...(The lower limit for the insertion will be ~830 km. Lower than that and the orbit would intersect the Earth's surface.

Calcs much appreciated, Janus. I would have never been able to come up with them by myself.
With the plot I had before (the body braked to 7km/s at 500km altitude), instead of a nice, stable orbit, it would have augured into the ground with a force of several billion megatons of TNT! Dinosaur-killer-class catastrophe!

 

[DaveC426913]

I think the fighters delta-V could be about 10km/s with 10km/s exhaust velocity (nuclear battery powered thrusters)

I'm rusty, but it seems to me, for a craft velocity of x, you need an exhaust velocity of at least 2x.

 

[Ryan_m_b]

I'm rusty, but it seems to me, for a craft velocity of x, you need an exhaust velocity of at least 2x.

Using this online calculator a delta-V of 10km/s with a 10km/s exhaust velocity requires 1.77kg of fuel per 1kg of vessel. If you want to get up to 10km/s and slow down again that becomes 6.7kg of fuel per 1kg of vessel.

 

[DaveC426913]

requires 1.77kg of fuel per 1kg of vessel.

Deos that take into account the fuel weight too?
i.e. the fueled ship is 2.7kg for each 1kg of payload?

 

[Janus]

Deos that take into account the fuel weight too?
i.e. the fueled ship is 2.7kg for each 1kg of payload?

The basic formula is

MR = e^{\frac{\Delta v}{v_e}}

Where MR is the mass ratio of fully fueled ship to unfueled ship and Ve is the exhaust velocity.
Thus in this example the mass ratio is 2.7

 

[Ryan_m_b]

Deos that take into account the fuel weight too?
i.e. the fueled ship is 2.7kg for each 1kg of payload?

The fuelled ship would have 1.7kg of fuel per 1kg payload. So a one tonne vessel would mass a total of 2.7 tonnes when fully fueled.

 

[GTOM]

Using this online calculator a delta-V of 10km/s with a 10km/s exhaust velocity requires 1.77kg of fuel per 1kg of vessel. If you want to get up to 10km/s and slow down again that becomes 6.7kg of fuel per 1kg of vessel.

I think the mass ratio shouldn't be much higher than e*e with external fuel tanks for spacecraft , so double delta-v as exhaust velocity.

Otherwise what about the manuevers i described, could they work?

A: interplanetary ship make significant course change and still land on destinated, or nearby celestial ( in asteroid belt ) with sustainable 4-2 miliG overall delta-v 100-200 km/s?

Bn Mercury take ballistic course sharp turn orbital speed decelerate with overall delta-v around 10 km/s

 

[Ryan_m_b]

I think the mass ratio shouldn't be much higher than e*e with external fuel tanks for spacecraft , so double delta-v as exhaust velocity.

What do you mean by "external fuel tanks"? That the craft are staged? If not I have no idea what you're getting at here.

Otherwise what about the manuevers i described, could they work?

I assume you've calculated the journey times based on a Brachistochrone transfer correctly. With regards to the 100MW figure that's the theoretical maximum. In reality it's going to take more. A 100 tonne craft accelerating at ~40mm/s with an exhaust velocity of 50km/s would have a propellent flow of 80g/s. The kinetic energy of 80g at 50km/s is 100Mj. No engine is 100% efficient at converting energy to propulsion so it's going to take more than that.

A: interplanetary ship make significant course change and still land on destinated, or nearby celestial ( in asteroid belt ) with sustainable 4-2 miliG overall delta-v 100-200 km/s?

The Delta-V seems high enough but 4mG isn't enough thrust to take off from any of the planets or Pluto. Something like Ceres or a small moon then yeah.

Bn Mercury take ballistic course sharp turn orbital speed decelerate with overall delta-v around 10 km/s

I honestly don't get what you're talking about with this "maneuver". Do you mean the vessels accelerate to orbital speed, or above, and then decelerate to a sub-orbital trajectory to land on their target? With only 4mG of thrust they won't even get off the ground. Also if you're running a nuclear reactor with hundreds of megajoule output you're going to need some solid radiators to not cook the vessel. The dayside of Mercury is nearly 500°C. How are you going to account for this problem?

 

[GTOM]

What do you mean by "external fuel tanks"? That the craft are staged? If not I have no idea what you're getting at here.

External fuel tank is a bit similar to staging, but a staged rocket drops a burnedout rocket, in this case, the spacecraft only drops an empty fuel tank, the rocket and power core remains.

I assume you've calculated the journey times based on a Brachistochrone transfer correctly. With regards to the 100MW figure that's the theoretical maximum. In reality it's going to take more. A 100 tonne craft accelerating at ~40mm/s with an exhaust velocity of 50km/s would have a propellent flow of 80g/s. The kinetic energy of 80g at 50km/s is 100Mj. No engine is 100% efficient at converting energy to propulsion so it's going to take more than that.

The Delta-V seems high enough but 4mG isn't enough thrust to take off from any of the planets or Pluto. Something like Ceres or a small moon then yeah.

Interplanetary fusion ships only land on small asteroids or orbital docks.

I honestly don't get what you're talking about with this "maneuver". Do you mean the vessels accelerate to orbital speed, or above, and then decelerate to a sub-orbital trajectory to land on their target? With only 4mG of thrust they won't even get off the ground. Also if you're running a nuclear reactor with hundreds of megajoule output you're going to need some solid radiators to not cook the vessel. The dayside of Mercury is nearly 500°C. How are you going to account for this problem?

There is a large difference between interplanetary and orbital spacecraft . While i imagine the later to have some super tritium lithium nuclear battery (not much dangerous decaying isotope in exhaust) but basically, the difference is similar to ion thrusted and chem fuel spacecraft . Orbital spacecraft can maintain multiple g acceleration but delta-V only around 10km/s. Unfit for fast interplanetary travel in a realm of fusion ships.

I don't know how it is possible to deal with that heat radiate from surface, but they hope that no surface batteries can withstand it, that is why they want to attack enemy base from sunny side.
Current script : take off from night side. Reach sunny side in a ballistic trajectory, than take a sharp turn.
(I know, with low orbital speed it is next to impossible to take sharp turn. So I wonder how much fuel could be saved by ballistic way? )
Accelerate to orbital speed, attack north pole base, than land on night side.
Or probably reach an orbital dock requires less delta-V. On Mercury, orbital bases can have a polar orbit, since the only city is under north pole.

 

[Ryan_m_b]

External fuel tank is a bit similar to staging, but a staged rocket drops a burnedout rocket, in this case, the spacecraft only drops an empty fuel tank, the rocket and power core remains.

I'm pretty sure that still counts as staging. The space shuttle was a two-stage + booster system, one of those stages was jettisoning the external fuel tank. In any case do you have any working out for the masses of the fuel tanks and the amount of fuel they hold? If you're looking for people to check the ma

Current script : take off from night side. Reach sunny side in a ballistic trajectory, than take a sharp turn.
(I know, with low orbital speed it is next to impossible to take sharp turn. So I wonder how much fuel could be saved by ballistic way? )
Accelerate to orbital speed, attack north pole base, than land on night side.

I guess you may save fuel if the sub-orbital velocity is quite low so that the total change in velocity of going there and back is less than achieving orbit and slowing down to land once you've completed one circuit. I'm not familiar with what equations you'd need for working out suborbital flight. Remember that Mercury has no atmosphere so you could orbit exceedingly close. Using this calculator orbiting at just 5km above the surface would require a speed of ~3km/s. Double that if you intend to circle round the planet and land where you started. Your four suborbital maneuvers (launch, decelerate, accelerate in reverse direction, decelerate to land) would have to total less than 6km/s of velocity change to save any fuel.

 

[Janus]

I also wonder about a situation at Mercury. The fighters take off at a base on the night side (not very far from North Pole), and want to attack defence towers of the North Pole, coming from the sunny side.
I can think about two ways, go around the entire planet with low orbital speed, but how about the shorter way, take a ballistic course from the night side to bright side, than stop and take orbital speed toward North? Would it be very energy consuming? I think the fighters delta-V could be about 10km/s with 10km/s exhaust velocity (nuclear battery powered thrusters)

It really depends on the exact parameters of the problem, Where exactly is the base located with respect to the day/night terminator? How deep into the daylight side do you need the attack to come from? (In other words are you planning to do a "Red Baron" out of the Sun?)

Here's an diagram that might help.
The red line are the possible orbital/ballistic trajectory lines from the base (located where these lines cross).
The blue line is the day/night terminator and the yellow line an orbit the "comes out of the Sun" relative to the North pole (where the blue and yellow lines cross)

From the location of the base in the image, in order for the fighter to approach the North pole along the yellow line from the daylight side in two legs, the first leg would have to be along one of the red lines until it crosses the yellow line and then follow the yellow line in. Note that some trajectories of the first leg would pass close by the North Pole (one actually crosses the yellow line at the North pole. If this is to be stealth mission, you are going avoid those.

The the further the base is from the terminator, the fewer trajectories you have to choose from that will not pass close to the North pole. If the base were on the yellow line, then you have only one choice, fly completely around the planet. There is only one ballistic path line that intersects the yellow line, and that's the one that follows the yellow line.

The other option would be to break it up into three legs, One that crosses the terminator line at some point far enough away from the North pole, then changes to a new course that crosses the yellow line and finally one that follows the yellow line.

However, these types of course change can be costly in terms of delta V. A 90 degree change in an orbit of 100 km above the surface of Mercury requires ~4km/sec delta v. Just one such change will eat up your available delta V. It will take a minimum of 3.06 km/sec to get your fighter into such an orbit, and unless this is a suicide mission, at least the same to soft land the fighter. 3.06+3.06+4= 10.12 km/sec which exceeds your available delta v.

 

[chasrob]

Could you clear up a couple things about your calculation in this post? Say I choose the handier prograde orbit-

From my figures, it works out that you would have to change the trajectory of the asteroid by an angle of 57.8 degrees and give it a velocity of 37 km/sec.
If this final trajectory is prograde, then this means you need to apply a delta v of 31.33 km/sec.

Does this mean 57.8 deg from the bodies’ direction of travel in its orbit, as in this drawing-

So supe needs to get the asteroid going along that trajectory at 31.33 km/s, correct? That will get it to Earth in 90 days?

When it arrives at the Earth it will be moving at ~49.4 km/sec, so in order to to put it into orbit, you will need to kill most of the difference between that and the Earth's orbital velocity.

The earth’s velocity is 30 so he needs to kill ~19km/s?

 

[Janus]

Could you clear up a couple things about your calculation in this post? Say I choose the handier prograde orbit-

Does this mean 57.8 deg from the bodies’ direction of travel in its orbit, as in this drawing-

So supe needs to get the asteroid going along that trajectory at 31.33 km/s, correct? That will get it to Earth in 90 days?

Yes 57.8 degrees from its initial velocity vector. 31.33 km/sec is the delta v he will have to supply, the final velocity will closer to 37 km/sec.

The earth’s velocity is 30 so he needs to kill ~19km/s?[/QUOTE]

Not quite. Let's say you wanted to insert it into that 1759 km altitude orbit with an orbital velocity of 7km/sec. To do this you would need to kill ~14.2 km/sec. (Basically it works out like this: You start with the relative velocity difference between asteroid and Earth, and then add in the velocity it would gain falling into a altitude of 1759 km above the Earth. Then subtract the 7km/sec orbital velocity you want to leave it with.)

 

[chasrob]

Yes 57.8 degrees from its initial velocity vector. 31.33 km/sec is the delta v he will have to supply, the final velocity will closer to 37 km/sec.

Right. His “push”, plus the asteroid has a vector in that direction equals the 37 he needs.

Not quite. Let's say you wanted to insert it into that 1759 km altitude orbit with an orbital velocity of 7km/sec. To do this you would need to kill ~14.2 km/sec. (Basically it works out like this: You start with the relative velocity difference between asteroid and Earth, and then add in the velocity it would gain falling into a altitude of 1759 km above the Earth. Then subtract the 7km/sec orbital velocity you want to leave it with.)

Just a thought... could it matter, for a couple of km/s, if his fly—by was to the right or left-considering the Earth's rotation?

 

[Janus]

Right. His “push”, plus the asteroid has a vector in that direction equals the 37 he needs.

Just a thought... could it matter, for a couple of km/s, if his fly—by was to the right or left-considering the Earth's rotation?

No. The Earth's rotation has no bearing on the orbital insertion.

 

[GTOM]

It really depends on the exact parameters of the problem, Where exactly is the base located with respect to the day/night terminator? How deep into the daylight side do you need the attack to come from? (In other words are you planning to do a "Red Baron" out of the Sun?)

Here's an diagram that might help.
The red line are the possible orbital/ballistic trajectory lines from the base (located where these lines cross).
The blue line is the day/night terminator and the yellow line an orbit the "comes out of the Sun" relative to the North pole (where the blue and yellow lines cross)
View attachment 112752
From the location of the base in the image, in order for the fighter to approach the North pole along the yellow line from the daylight side in two legs, the first leg would have to be along one of the red lines until it crosses the yellow line and then follow the yellow line in. Note that some trajectories of the first leg would pass close by the North Pole (one actually crosses the yellow line at the North pole. If this is to be stealth mission, you are going avoid those.

The the further the base is from the terminator, the fewer trajectories you have to choose from that will not pass close to the North pole. If the base were on the yellow line, then you have only one choice, fly completely around the planet. There is only one ballistic path line that intersects the yellow line, and that's the one that follows the yellow line.

The other option would be to break it up into three legs, One that crosses the terminator line at some point far enough away from the North pole, then changes to a new course that crosses the yellow line and finally one that follows the yellow line.

However, these types of course change can be costly in terms of delta V. A 90 degree change in an orbit of 100 km above the surface of Mercury requires ~4km/sec delta v. Just one such change will eat up your available delta V. It will take a minimum of 3.06 km/sec to get your fighter into such an orbit, and unless this is a suicide mission, at least the same to soft land the fighter. 3.06+3.06+4= 10.12 km/sec which exceeds your available delta v.

Yes Red Baron approach would be the best.

Now I think i rather skip that course change thing, it isn't that much time to go around entire planet.

However much delta-V could be spared if they land on 300 km high orbiting dock instead of surface after reach orbital speed?
The orbiting dock has a polar orbit. Let's have a 10% inclination difference between desired yellow line and the dock's present orbit.

 

[Janus]

Yes Red Baron approach would be the best.

Now I think i rather skip that course change thing, it isn't that much time to go around entire planet.

However much delta-V could be spared if they land on 300 km high orbiting dock instead of surface after reach orbital speed?
The orbiting dock has a polar orbit. Let's have a 10% inclination difference between desired yellow line and the dock's present orbit.

To attain a 300 km high circular orbit requires ~3.7 km/sec delta v. It would take the same to de-orbit and land. Matching orbit with the dock would take ~0.5 km/sec, so there would be a significant difference. (4.2 km/sec compared to 7.4 km/sec)

 

[chasrob]

Ok, the superguy got himself in a bit of trouble. He is within a couple million klicks of Earth the flyby of earth. His cohorts radio him that he’s screwed up the angle of departure and is going to miss by a million kms-

I ‘ve heard that orbital motions can be counter-intuitive. Can he make a simple thrusting shove as above, orthogonal to his vector, and close the gap that way, since this is a hyperbolic orbit he’s on? Or will the body move in some other unwanted direction?

 

[chasrob]

I just realized that the pic is rather confusing! It's supposed to be a perspective from close behind the asteroid viewing toward the distant earth.
Perhaps this would be clearer-

EDIT: I wonder if I could find out if I got Kerbal?

 

[Janus]

I just realized that the pic is rather confusing! It's supposed to be a perspective from close behind the asteroid viewing toward the distant earth.
Perhaps this would be clearer-

EDIT: I wonder if I could find out if I got Kerbal?

From your diagram I'm going to assume that the Sun is to the left. In this case the asteroid will miss 1,000,000 km further out. The problem with just shoving the Asteroid in is that as the asteroid moves in towards the Sun, it will gain gravitational potential energy. This will result in a velocity gain of ~4.3 km/sec . This will have it arriving at the intercept with the Earth's orbit sooner (earlier than the Earth reaches that point). You would have to compensate the Shove direction and magnitude to compensate. More of a problem is the fact that if your trajectory was that far off from the start, that means that the asteroid was following a different orbital path. The proper trajectory was chosen because it got to Earth orbit distance at the same time as the Earth was there. If you alter the orbit, you also alter the timing. An orbit that misses by 1000,000 further out would also miss by a factor of time. (In other words, when it arrives at its closest approach to Earth orbit, the Earth won't be there.

If you need the 1,000,000 km miss for your plot, I'd suggest an orbital inclination error instead, The asteroid will arrive at Earth orbit distance at the same time as the Earth, but above its orbital plane by 1,000,000 km. This is a much easier problem to correct as you don't have the GPE or orbital timing issues.
In this case, A miss by 1,000,000 km that needs to be corrected when the asteroid is 2,000,000 km short of the Earth's position will take a delta v of ~18 km/sec in order to correct the inclination for a proper interception. Though honestly, I would think that 2,000,000 km, is awfully close for the asteroid to have gotten before the cohorts noticed the discrepancy( by this time the asteroid would be ~26 degrees from where it should be as seen from the Earth.)

 

[chasrob]

Why did I consider the miss? Delving deeper into my plot, I see that superguy has a major problem. Such precision in orbital maneuvers would not be possible as I see him. He can get the velocity down to +/- 1km/s but the vector angle? His cohorts on Earth number a half dozen trusted friends, the handiest of which has a minor in astronomy. She plots his courses and has an automated 16” Dobsonian with a camera attachment. Which I suppose could be used to track the planetoid (plus use Stellarium?) and alert him of trouble.

The point is superfella is a couple hundred million kms from home and ~15-20 minute wait on radio, has no obvious way to determine the body’s own directional vector; I can’t see how he can even visualize a new vector accurate to a tenth of a degree. So he’s going to screw the pooch there by 5 +/- deg I reckon. And the velocity too, as above.

If you need the 1,000,000 km miss for your plot, I'd suggest an orbital inclination error instead, The asteroid will arrive at Earth orbit distance at the same time as the Earth, but above its orbital plane by 1,000,000 km. This is a much easier problem to correct as you don't have the GPE or orbital timing issues.

Interesting. Wouldn’t the asteroid have its own inclination that would probably have to be taken into account too?

Though honestly, I would think that 2,000,000 km, is awfully close for the asteroid to have gotten before the cohorts noticed the discrepancy( by this time the asteroid would be ~26 degrees from where it should be as seen from the Earth.)

True, even though they’re using the rather crude amateur equipment I mentioned above.

If he launched the asteroid from the belt at ~53-54 deg and a velocity of 38 km/s could his miss be that 1 million km? Or even more of a whopper of a miss?

Maybe if its that or more, he could just do what Superman would do, just grab the chunk and fly, piloting the last 1 million klicks.

 

[Janus]

Why did I consider the miss? Delving deeper into my plot, I see that superguy has a major problem. Such precision in orbital maneuvers would not be possible as I see him. He can get the velocity down to +/- 1km/s but the vector angle? His cohorts on Earth number a half dozen trusted friends, the handiest of which has a minor in astronomy. She plots his courses and has an automated 16” Dobsonian with a camera attachment. Which I suppose could be used to track the planetoid (plus use Stellarium?) and alert him of trouble.

The point is superfella is a couple hundred million kms from home and ~15-20 minute wait on radio, has no obvious way to determine the body’s own directional vector; I can’t see how he can even visualize a new vector accurate to a tenth of a degree. So he’s going to screw the pooch there by 5 +/- deg I reckon. And the velocity too, as above.
Interesting. Wouldn’t the asteroid have its own inclination that would probably have to be taken into account too?
True, even though they’re using the rather crude amateur equipment I mentioned above.

If he launched the asteroid from the belt at ~53-54 deg and a velocity of 38 km/s could his miss be that 1 million km? Or even more of a whopper of a miss?

Maybe if its that or more, he could just do what Superman would do, just grab the chunk and fly, piloting the last 1 million klicks.

Some quick rough calculations show that 38 km/sec at 54 deg deflection results in a perihelion that is a better part of 1 AU further out from the Sun than Earth orbit is. Put another way, your asteroid wouldn't even get inside of Mars' orbit. Your superfella is going to either going to need to start with very precise calculations or being making constant corrections along the way. And the only way he is going to know what those corrections need to be is if someone is tracking the asteroid in the whole way.

Inclination correction is generally taken care of with what is known as a "broken plane" maneuver, in which you wait until you reach the point in your trajectory where it is crossing the orbital plane of the target body to make the correction. This is an easier method than trying to include the correction at the onset.

 

[chasrob]

Some quick rough calculations show that 38 km/sec at 54 deg deflection results in a perihelion that is a better part of 1 AU further out from the Sun than Earth orbit is. Put another way, your asteroid wouldn't even get inside of Mars' orbit. Your superfella is going to either going to need to start with very precise calculations or being making constant corrections along the way. And the only way he is going to know what those corrections need to be is if someone is tracking the asteroid in the whole way.

Inclination correction is generally taken care of with what is known as a "broken plane" maneuver, in which you wait until you reach the point in your trajectory where it is crossing the orbital plane of the target body to make the correction. This is an easier method than trying to include the correction at the onset.

I was afraid of that :), hoping that a small velocity error would minimize the mistake, until I finally realized a 1km/s error is actually a huge screw-up on McGuyver-man’s part.

If he instead made the same deflection but the body ended up with a 37.05 km/sec velocity, would the perihelion be closer to Earth (a few million klicks)?

 

[Janus]

I was afraid of that :), hoping that a small velocity error would minimize the mistake, until I finally realized a 1km/s error is actually a huge screw-up on McGuyver-man’s part.

If he instead made the same deflection but the body ended up with a 37.05 km/sec velocity, would the perihelion be closer to Earth (a few million klicks)?

Sorry I took so long to get back to you, but this took a bit finer calculation then my earlier one did. In order to determine by how much the asteroid would miss, you need to consider a number of factors, 1) the new perihelion distance, 2) how the perihelion position shifts with respect to the original target position and 3) how this effects the timing of the orbit. (remember we calculated for what it would take for the asteroid to arrive at a given point at the same time as the Earth was there.)
Also, the 37 km/sec value was a rounded number, so I had to go back and refine it first.
The end result is that the perigee will shift by ~1,000,000 km clock-wise (the orbit directions are CCW), it will be ~1.4 million km further out, and it will arrive there ~4 hrs later than originally planned. In other words, when it arrives at perigee, it will be ~1.5 million km "behind" the Earth (The Earth will have moved past the original rendezvous in that 4 hrs) . and 1.4 million km further from the Sun.
It will take most of 1 day for it to catch up to the Earth, so its nearest approach will be a bit over 1 day later than originally planned. It will move out from the Sun a bit more as it will be past perihelion, but this won't account for much and the asteroid will miss the Earth by just a bit over 1.4 million km. This seems to be right in range you wanted.

 

[chasrob]

Sorry I took so long to get back to you, but this took a bit finer calculation then my earlier one did. In order to determine by how much the asteroid would miss, you need to consider a number of factors, 1) the new perihelion distance, 2) how the perihelion position shifts with respect to the original target position and 3) how this effects the timing of the orbit. (remember we calculated for what it would take for the asteroid to arrive at a given point at the same time as the Earth was there.)
Also, the 37 km/sec value was a rounded number, so I had to go back and refine it first.
The end result is that the perigee will shift by ~1,000,000 km clock-wise (the orbit directions are CCW), it will be ~1.4 million km further out, and it will arrive there ~4 hrs later than originally planned. In other words, when it arrives at perigee, it will be ~1.5 million km "behind" the Earth (The Earth will have moved past the original rendezvous in that 4 hrs) . and 1.4 million km further from the Sun.
It will take most of 1 day for it to catch up to the Earth, so its nearest approach will be a bit over 1 day later than originally planned. It will move out from the Sun a bit more as it will be past perihelion, but this won't account for much and the asteroid will miss the Earth by just a bit over 1.4 million km. This seems to be right in range you wanted.

Yes, thanks to your fine calculations, the plot goes as I planned. Protag is transformed. Hears about the greater than $100 trillion value of these asteroids. Gets greedy and plots, along with his small team of cohorts, to get the body into Earth orbit. Superguy is told to launch according to your parameters, so he can just let the planetoid come to him in 3 months while he’s doing various feats of derring-do. Unfortunately, he fouls up, although unintentionally, and is eventually informed of that fact when said cohorts comparatively crude equipment reveals he’s screwed the pooch. They decide to sit tight and let it reach perigee, and since its only 1-2 Mkm distant, he can guide constantly and ride it into orbit within a few hours.

Now I only have to dig up some magic somewhere that allows him to apply pressure to the asteroid during maneuvers that equals hundreds of thousands of times that in the center of the Sun. Enough to fuse the iron for an extra kick?

 

[GTOM]

I speculated about 100MW fusion drives (for a 100ton ship).
Then i guess, a space destroyers cannons could expend 10 MJ x-ray beams in every second. ( at least for a while, focusing lenses are more vulnerable than the thrusters, i expect, that the focusing stuff of laser cannon can be damaged by a scale smaller cannon)
A fleet of 15 ships, destroyers, cruisers, battleship could expend around 400 MJ in every sec.
As far as i know, martian atmosphere swallows 10-20% of x-rays at most.

Is it a bit plausible, that a reinforced government building (on Mars, capital at equator, so they could built a space elevator near to it) could withstand such a laser siege for about a minute?
I read the WTC bomb was around 600kg, and it wasnt enough for collapse.

 

[chasrob]

Concerning the superhero story above; I am at a loss for how the super could perform a maneuver which requires him to slow/accelerate the body at a precise vector, considering the tools he could use. If he wants to do that discreetly, how would he, McGuyver-man, sitting on an asteroid a couple hundred million klicks from nowhere, determine its motion close enough for even a rough maneuver?Would a potato-shaped Ni-Fe asteroid, undisturbed for millennia, settle into an orbit with its long axis identical to its velocity vector? It couldn’t be that easy, could it? He could then position a protractor and...

 

[chasrob]

I have a protagonist who is super powered and doesn't need to breathe. Ergo, he needs no spacesuit when he ventures into outer space.

He is helping construct a moon base. In one instance Superguy absent-mindedly steps outside his habitat into the vacuum in his everyday clothes. His shirt is polyester-cotton and he's wearing denim jeans, and say, he's wearing cotton socks and leather shoes.

Would this outer wear stand up long in the rigors of space? Particularly the wild temperature ranges in shade and sunlight would be too much on the cloth; even spandex, eh?