Flip and Burn
The problem
In The Expanse, a ship in a hurry doesn’t coast. It burns the whole way: accelerate toward the destination for the first half of the trip, flip end over end at the midpoint, and decelerate for the second half so it arrives at rest. That is the flip-and-burn. It is ordinary constant-acceleration kinematics, and every number in it can be checked.
Take a ship leaving Mars for Jupiter. On the planets’ average orbits, the distance between them swings from about 550 million km, when both are on the same side of the Sun, to about 1 billion km, when they are on opposite sides. Take a distance toward the near end of that range and fix these figures:
- Distance to cover: d = 600 million km, in a straight line
- Constant thrust acceleration: a = 0.3 g, with g = 9.80665 m/s²
- Accelerate for the first half of the distance, flip, and decelerate for the second half, arriving at rest. The flip itself takes no time.
- Treat both planets as stationary and ignore the Sun’s gravity. At these speeds the planets’ own orbital motion is a small correction.
Find:
- The travel time, in days.
- The peak speed, reached at the flip, in km/s.
Notes
- Keep the distance in millions of km and the acceleration in km/s². The factor of a million then comes out of the square root as exactly 1000, and nothing awkward has to be typed in.
- 0.3 g in km/s² is 0.3 × 0.00980665.
- The trip is two identical halves, each covering d/2 from rest. Solve one half, then use the symmetry.
- Nothing here needs addition or subtraction, which is what makes a slide rule a fair contender.
Hint
Click for a hint
One half of the trip is a start from rest at constant acceleration over a distance d/2:
d/2 = ½ · a · t₁² → t₁ = √(d / a)
v = a · t₁
The full trip is two such halves, so the total time is 2·t₁. The peak speed can also be written without the time at all, as v = √(a · d).
Solution
Click to reveal the solution
Acceleration in km/s²:
a = 0.3 × 9.80665 / 1000 = 0.002941995 km/s²
Time for the first half, covering d/2 from rest:
d/2 = ½ · a · t₁² → t₁ = √(d / a)
The whole trip is two equal halves:
t = 2 · √(d / a) = 2 · √(6×10⁸ / 0.002941995) s = 903 201 s
Dividing by 86 400 s per day gives the travel time.
Answer (time): 10.45 days.
Peak speed at the flip:
v = a · t₁ = √(a · d) = √(0.002941995 × 6×10⁸) = 1328.6 km/s
Answer (peak speed): 1329 km/s, a little under half a percent of the speed of light. That is fast, but nowhere near relativistic, so the Newtonian treatment above is sound.
Notice that nothing in this problem is added or subtracted, only multiplied, divided and rooted.
Calculator solutions
HP-42S
No memory registers needed.
Travel time in days:
600 ENTER
0.3 ÷
0.00980665 ÷
√x
2000 ×
86400 ÷
Result: 10.4537
Peak speed in km/s:
0.3 ENTER
0.00980665 ×
600 ×
√x
1000 ×
Result: 1328.6072
HP-15C
Same keystrokes. The 15C shows the thousands separator.
Travel time in days:
600 ENTER
0.3 ÷
0.00980665 ÷
√x
2000 ×
86400 ÷
Display: 10.4537
Peak speed in km/s:
0.3 ENTER
0.00980665 ×
600 ×
√x
1000 ×
Display: 1,328.6072
HP-41C
Same keystrokes.
Travel time in days:
600 ENTER
0.3 ÷
0.00980665 ÷
√x
2000 ×
86400 ÷
Display: 10.4537
Peak speed in km/s:
0.3 ENTER
0.00980665 ×
600 ×
√x
1000 ×
Display: 1,328.6072
HP-48GX
Plain stack entry, no PI, so no symbolic traps.
Travel time in days:
600 ENTER
0.3 ÷
0.00980665 ÷
√x
2000 ×
86400 ÷
Display: 10.453721198
Peak speed in km/s:
0.3 ENTER
0.00980665 ×
600 ×
√x
1000 ×
Display: 1328.60716542
HP-25
The machine powers up in FIX 2. Press f FIX 4 first to compare against the other machines.
Travel time in days:
600 ENTER
0.3 ÷
0.00980665 ÷
√x
2000 ×
86400 ÷
Display: 10.4537
Peak speed in km/s:
0.3 ENTER
0.00980665 ×
600 ×
√x
1000 ×
Display: 1328.6072
Elektronika MK-61
Square root is F then the √ key. Enter every number starting with a digit: 0.3 is typed
0 . 3.
Key 0.00980665 as 9.80665 ВП 3 /-/, that is 9.80665 × 10⁻³. Typed out in full, the
MK-61 runs out of digit positions: its 8-digit entry counts the leading zeros, so it keeps
0.0098066 and silently drops the final 5, which moves both answers in the sixth or seventh figure.
Travel time in days:
600 В↑
0.3 ÷
9.80665 ВП 3 /-/ ÷
F √
2000 ×
86400 ÷
Display: 10.453721
Peak speed in km/s:
0.3 В↑
9.80665 ВП 3 /-/ ×
600 ×
F √
1000 ×
Display: 1328.6071
The MK-61 works to 8 digits, so its last digit can differ slightly from the other machines’.
TI-84
2000*√(600/(.3*.00980665))/86400
√(.3*.00980665*600)*1000
Results: 10.4537212 and 1328.607165
Casio fx-991
Natural display mode:
2000×√(600÷(0.3×0.00980665))÷86400
1000×√(0.3×0.00980665×600)
Results: 10.4537212 and 1328.607165
Slide Rule
Everything is multiplication, division and a square root, so the rule handles all of it. Track the powers of ten by hand and expect three significant figures.
- Acceleration. 0.3 × 9.81 on C and D gives 2.94, so a ≈ 0.00294 km/s².
- Distance over acceleration. Divide 600 by 0.00294: set 6 on D, bring 2.94 on C over it, read ≈ 2.04 at the index, so d/a ≈ 2.04 × 10⁵.
- Root. Put 2.04 on the A scale (the 10⁵ means use the right-hand half) and read the root on D: ≈ 4.52, so √(d/a) ≈ 452 s.
- Time. The trip is twice that, in seconds, divided by 86 400 for days. The constants combine to one division by 43.2 (86 400 ÷ 2000): set 4.52 on D, bring 4.32 on C over it, and read ≈ 1.046 at the index, which is 10.46, so ≈ 10.5 days.
- Peak speed. Multiply 0.00294 × 600 to get 1.764. Put 1.764 on A, read ≈ 1.33 on D, and scale by 1000.
Hand-checked against the verified results above: both agree to three significant figures.
Bonus 1: program it
For programmable calculators only.
The main problem fixed the distance at 600 million km and the acceleration at 0.3 g. A trip planner takes both as inputs. Change the distance or the burn, and the answer changes.
The task. Write a program with two inputs:
- d, the distance in million km
- a, the acceleration in g
It returns the travel time in days and the peak speed in km/s. Keep the time on the display when the program stops, and leave the speed in a register for the user to recall.
Test your program on these three cases:
| Case | Distance d (million km) | Acceleration (g) |
|---|---|---|
| A | 550, about the closest approach | 0.3 |
| B | 1000, about the farthest | 0.3 |
| C | 600, the main problem’s distance | 1.0, a full g |
Click for a hint
Convert the acceleration first: a = g × 0.00980665 km/s². With d in million km, the factor of a million comes out of each square root as 1000:
peak speed = 1000 · √(a · d) km/s
time = 2 · 1000 · √(d / a) / 86400 days = √(d / a) / 43.2
Keep a in a register, since both formulas use it.
Click to reveal the solution
| Case | Travel time (days) | Peak speed (km/s) |
|---|---|---|
| A | 10.01 | 1272 |
| B | 13.50 | 1715 |
| C | 5.726 | 2426 |
The time grows as the square root of the distance and shrinks as the square root of the acceleration. Case C is the main problem’s trip at a full g instead of 0.3 g: that is 3.33 times the acceleration, so the time falls by a factor of √3.33 = 1.83, not by 3.33.
Programs
Programs follow for the HP-42S, HP-15C, HP-41C, Elektronika MK-61 and HP-25. Open the one for your calculator, key it in once, then for each case store d and a and run it. The text to the right of a listing explains it; do not key it in.
HP-42S
This program uses 19 lines. Press PRGM, key in the lines, press PRGM again.
01 LBL 01 start
02 RCL 02 g, the acceleration in g
03 0.00980665 1 g in km/s²
04 × a, in km/s²
05 STO 04 save a
06-08 RCL 01 RCL 04 × d · a
09-11 √x 1000 × peak speed 1000·√(d·a), in km/s
12 STO 03 save the speed in register 3
13-15 RCL 01 RCL 04 ÷ d / a
16-18 √x 43.2 ÷ √(d/a) / 43.2, the travel time in days
19 RTN stop, result in X
To run it, store the inputs and start it. For case A:
550 STO 01
0.3 STO 02
XEQ 01
The time in days is in X, 10.0087. For the speed, press RCL 03: 1272.0445, in km/s.
HP-15C
This program uses 34 of the machine’s steps. g P/R toggles program mode; LBL 1 is f LBL 1.
01 f LBL 1 start
02 RCL 2 g, the acceleration in g
03-12 0.00980665 1 g in km/s²
13 × a, in km/s²
14 STO 4 save a
15-17 RCL 1 RCL 4 × d · a
18-23 √x 1000 × peak speed 1000·√(d·a), in km/s
24 STO 3 save the speed in register 3
25-27 RCL 1 RCL 4 ÷ d / a
28-33 √x 43.2 ÷ √(d/a) / 43.2, the travel time in days
34 R/S stop, result showing
To run it, store the inputs and start it. For case A:
550 STO 1
0.3 STO 2
GSB 1
The time in days, 10.0087, shows when the program stops. For the speed, press RCL 3: 1,272.0445, in km/s.
HP-41C
This program is 19 lines long. PRGM toggles program mode; LBL is the gold shift of STO.
01 LBL 01 start
02 RCL 02 g, the acceleration in g
03 0.00980665 1 g in km/s²
04 × a, in km/s²
05 STO 04 save a
06-08 RCL 01 RCL 04 × d · a
09-11 √x 1000 × peak speed 1000·√(d·a), in km/s
12 STO 03 save the speed in register 3
13-15 RCL 01 RCL 04 ÷ d / a
16-18 √x 43.2 ÷ √(d/a) / 43.2, the travel time in days
19 R/S stop, result showing
To run it, store the inputs and start it. For case A:
550 STO 01
0.3 STO 02
XEQ 01
The time in days, 10.0087, shows when the program stops. For the speed, press RCL 03: 1,272.0445, in km/s.
Elektronika MK-61
This program uses 38 of the machine’s 105 steps. F ПРГ enters program mode and F АВТ leaves it. The listing puts В↑ before each typed number. The MK-61 would lift the stack by itself there, so it is not strictly needed, but it does no harm. Type every number starting with a digit. Addresses start at 00.
00 П→x 2 g, the acceleration in g
01-11 В↑ 9 . 8 0 6 6 5 ВП 3 /-/ 1 g in km/s²
12 × a, in km/s²
13 x→П 4 save a
14-16 П→x 1 П→x 4 × d · a
17-23 F √ В↑ 1 0 0 0 × peak speed 1000·√(d·a), in km/s
24 x→П 3 save the speed in register 3
25-27 П→x 1 П→x 4 ÷ d / a
28-34 F √ В↑ 4 3 . 2 ÷ √(d/a) / 43.2, the travel time in days
35-36 x→П 7 П→x 7 store and re-read X so that the stopped display is clean
37 С/П stop
To run it, store the inputs and start it. For case A:
5 5 0 x→П 1
0 . 3 x→П 2
БП 0 0
С/П
The time in days, 10.008675, shows when it stops. For the speed, press П→x 3: 1272.0445, in km/s.
The constant 0.00980665 is keyed as 9.80665 ВП 3 /-/. Typed out in full, the MK-61’s 8-digit entry counts the leading zeros and silently drops the final 5. The MK-61 works to 8 digits, so its last digit can differ slightly from the other machines’.
HP-25
The HP-25 has room for 49 program steps, and this program uses 33. With the switch at PRGM, press f PRGM to clear program memory, then key in the steps. A typed number takes one step per key.
01 RCL 2 g, the acceleration in g
02-11 0.00980665 1 g in km/s²
12 × a, in km/s²
13 STO 4 save a
14-16 RCL 1 RCL 4 × d · a
17-22 f √x 1000 × peak speed 1000·√(d·a), in km/s
23 STO 3 save the speed in register 3
24-26 RCL 1 RCL 4 ÷ d / a
27-32 f √x 43.2 ÷ √(d/a) / 43.2, the travel time in days
33 R/S stop, result showing
Set the switch to RUN, press f FIX 4 to see four decimals, store the inputs, then GTO 0 0 and R/S. For case A:
550 STO 1
0.3 STO 2
GTO 0 0
R/S
The time in days, 10.0087, shows when the program stops. For the speed, press RCL 3: 1272.0445, in km/s.
Bonus 2: where Newton runs out
For programmable calculators only.
Run the same flip-and-burn between the stars and the Newtonian formulas stop being trustworthy: nothing in √(a · d) keeps the peak speed below the speed of light. This bonus replaces them with the relativistic version, and it needs a program because the formulas now have several steps and a logarithm.
The setup. The ship keeps a constant acceleration as its own crew feels it, flips at the midpoint, and arrives at rest. Work in units where c = 1 light-year per year. In those units one g is 1.0323 ly/yr² (using the Julian year). Two clocks matter: the ship’s own, and the one at home, in Earth’s frame.
The task. Write a program with two inputs:
- d, the distance in light-years
- a, the acceleration in g
It returns the ship’s own travel time in years, and leaves the Earth-frame time in a register.
Test it on three cases:
| Case | Distance d (light-years) | Acceleration (g) |
|---|---|---|
| A | 4.37, Alpha Centauri | 1 |
| B | 4.37, Alpha Centauri | 0.3 |
| C | 25 | 1 |
Click for a hint
Write a for the acceleration in ly/yr² (that is, g × 1.0323) and let
x = 1 + a · d / 2
Then, with c = 1:
Earth-frame time t = 2 · √(x² − 1) / a
ship's own time τ = (2 / a) · ln( x + √(x² − 1) )
No hyperbolic keys are needed: the logarithm form works on every calculator here. A neat way to program it is to work with half of a, and with h = a · d / 2, so that x = 1 + h and x² − 1 = h · (h + 2).
Click to reveal the solution
| Case | Ship’s own time (years) | Earth-frame time (years) |
|---|---|---|
| A | 3.582 | 6.002 |
| B | 7.143 | 8.691 |
| C | 6.440 | 26.87 |
In every case the home clock runs longer than the ship’s, and longer than the d years that light would need. Case C is the stark one: the crew ages about 6.4 years while 26.9 pass at home.
Now compare the Newtonian formula, peak speed = √(a · d), in the same units: it gives 2.12, 1.16 and 5.08 times the speed of light for cases A, B and C. All three exceed c, which is what shows the Newtonian treatment has failed here. It was fine for Mars to Jupiter, where the peak speed was under half a percent of c.
The two time formulas both come from the hyperbolic-motion solution for constant proper acceleration. They were cross-checked here by stepping the ship’s motion forward in small slices of its own time, which gave the same six figures.
Programs
The same five calculators. The structure is the same as in Bonus 1, with one extra register used for the intermediate value √(x² − 1).
HP-42S
This program uses 28 lines. Press PRGM, key in the lines, press PRGM again.
01 LBL 01 start
02-04 RCL 02 0.51615 × c = a/2 in ly/yr² (g times half of 1.0323)
05 STO 04 save c
06-08 RCL 01 × STO 05 h = c · d = a·d/2, saved
09-13 RCL 05 RCL 05 2 + × h · (h + 2)
14-15 √x STO 06 y = √(h² + 2h), which is √(x² − 1) with x = 1 + h; saved
16-19 RCL 06 RCL 04 ÷ STO 03 y / c, the Earth-frame time; saved in register 3
20-24 RCL 06 RCL 05 + 1 + x + y = 1 + h + y
25-27 LN RCL 04 ÷ ln(x + y) / c, the ship's own time
28 RTN stop, result in X
To run it, store the inputs and start it. For case A:
4.37 STO 01
1 STO 02
XEQ 01
The ship’s own time in years is in X, 3.5824. For the Earth-frame time, press RCL 03: 6.0025.
HP-15C
This program uses 34 of the machine’s steps. g P/R toggles program mode; LBL 1 is f LBL 1.
01 f LBL 1 start
02-10 RCL 2 0.51615 × c = a/2 in ly/yr² (g times half of 1.0323)
11 STO 4 save c
12-14 RCL 1 × STO 5 h = c · d = a·d/2, saved
15-19 RCL 5 RCL 5 2 + × h · (h + 2)
20-21 √x STO 6 y = √(h² + 2h), which is √(x² − 1) with x = 1 + h; saved
22-25 RCL 6 RCL 4 ÷ STO 3 y / c, the Earth-frame time; saved in register 3
26-30 RCL 6 RCL 5 + 1 + x + y = 1 + h + y
31-33 g LN RCL 4 ÷ ln(x + y) / c, the ship's own time
34 R/S stop, result showing
To run it, store the inputs and start it. For case A:
4.37 STO 1
1 STO 2
GSB 1
The ship’s own time in years, 3.5824, shows when the program stops. For the Earth-frame time, press RCL 3: 6.0025.
HP-41C
This program is 28 lines long. PRGM toggles program mode; LBL is the gold shift of STO.
01 LBL 01 start
02-04 RCL 02 0.51615 × c = a/2 in ly/yr² (g times half of 1.0323)
05 STO 04 save c
06-08 RCL 01 × STO 05 h = c · d = a·d/2, saved
09-13 RCL 05 RCL 05 2 + × h · (h + 2)
14-15 √x STO 06 y = √(h² + 2h), which is √(x² − 1) with x = 1 + h; saved
16-19 RCL 06 RCL 04 ÷ STO 03 y / c, the Earth-frame time; saved in register 3
20-24 RCL 06 RCL 05 + 1 + x + y = 1 + h + y
25-27 LN RCL 04 ÷ ln(x + y) / c, the ship's own time
28 R/S stop, result showing
To run it, store the inputs and start it. For case A:
4.37 STO 01
1 STO 02
XEQ 01
The ship’s own time in years, 3.5824, shows when the program stops. For the Earth-frame time, press RCL 03: 6.0025.
Elektronika MK-61
This program uses 36 of the machine’s 105 steps. F ПРГ enters program mode and F АВТ leaves it. The listing puts В↑ before each typed number. The MK-61 would lift the stack by itself there, so it is not strictly needed, but it does no harm. Type every number starting with a digit. Addresses start at 00.
00-09 П→x 2 В↑ 0 . 5 1 6 1 5 × c = a/2 in ly/yr² (g times half of 1.0323)
10 x→П 4 save c
11-13 П→x 1 × x→П 5 h = c · d = a·d/2, saved
14-19 П→x 5 П→x 5 В↑ 2 + × h · (h + 2)
20-21 F √ x→П 6 y = √(h² + 2h), which is √(x² − 1) with x = 1 + h; saved
22-25 П→x 6 П→x 4 ÷ x→П 3 y / c, the Earth-frame time; saved in register 3
26-31 П→x 6 П→x 5 + В↑ 1 + x + y = 1 + h + y
32-34 F ln П→x 4 ÷ ln(x + y) / c, the ship's own time
35 С/П stop
To run it, store the inputs and start it. For case A:
4 . 3 7 x→П 1
1 x→П 2
БП 0 0
С/П
The ship’s own time in years, 3.5823838, shows when it stops. For the Earth-frame time, press П→x 3: 6.0024961.
The MK-61 works to 8 digits, so its last digit or two can differ slightly from the other machines’.
HP-25
The HP-25 has room for 49 program steps, and this program uses 33. With the switch at PRGM, press f PRGM to clear program memory, then key in the steps. A typed number takes one step per key.
01-09 RCL 2 0.51615 × c = a/2 in ly/yr² (g times half of 1.0323)
10 STO 4 save c
11-13 RCL 1 × STO 5 h = c · d = a·d/2, saved
14-18 RCL 5 RCL 5 2 + × h · (h + 2)
19-20 f √x STO 6 y = √(h² + 2h), which is √(x² − 1) with x = 1 + h; saved
21-24 RCL 6 RCL 4 ÷ STO 3 y / c, the Earth-frame time; saved in register 3
25-29 RCL 6 RCL 5 + 1 + x + y = 1 + h + y
30-32 f ln RCL 4 ÷ ln(x + y) / c, the ship's own time
33 R/S stop, result showing
Set the switch to RUN, press f FIX 4 to see four decimals, store the inputs, then GTO 0 0 and R/S. For case A:
4.37 STO 1
1 STO 2
GTO 0 0
R/S
The ship’s own time in years, 3.5824, shows when the program stops. For the Earth-frame time, press RCL 3: 6.0025.
Solvable on: HP-42S, HP-15C, HP-41C, HP-48GX, HP-25, Elektronika MK-61, TI-84, Casio fx-991, Slide Rule