STEM Puzzles

For those who enjoyed high school or college physics, here is the math I did to keep an element of realism in this fiction story.

In this story, the starship Eden uses a theoretical Bussard ramjet for propulsion. The ramjet uses an electromagnetic scoop to collect rarified hydrogen from interstellar space, which is then used as fuel. This theoretical ramjet is very efficient but has very low thrust. The tricky part of the Bussard ramjet is decelerating, as the ship has to turn around — pointing the ram in the wrong direction! If ever built, a Bussard ramjet would be ideal for long journeys but could not make them quickly. Here’s the math the author used to calculate the journey for this story, written in the voice of those terrible textbooks, which teachers used to afflict us all.

A spaceship leaves Alpha Centauri Ba (Tarth) for Sol c (Earth). It accelerates continuously for half the journey then decelerates the rest of the way, beginning and ending the journey with no (zero) velocity and experiencing maximum velocity at the midpoint. As the fuel is rarified hydrogen collected from interstellar space, the mass of the ship does not change, thus the acceleration remains constant for the whole journey. The journey is 160122 Earth days (a little more than 438 Earth years).

A. What is Eden’s acceleration on the first half of the journey and its deceleration on the second half of the journey? (Hint: both are the same.)

The relevant equation is x = ½ • a • t2; position is equal to one-half the acceleration times the square of the time. For this problem, position (distance) and time are known but acceleration is not. The equation is re-worked to anormal = [ 2 • x ] ÷ t2, acceleration is twice the distance divided by the square of the time.

anormal = the acceleration and deceleration of the Eden (in meters per sec ond-squared)

x = half the distance from Alpha Centauri to Sol, which is 4.37 light years divided bytwo (in meters)

t = half the duration of the journey, which is 160122 days divided by two (in seconds)

anormal = 2 • x ÷ t2 = ( 2 • 2.0672e16 m ) ÷ ( 6.9173e9 s )2 = 8.6405e-4 m/s2

In one second, the ship will increase (first half of the journey) or decrease (second half of the journey) its velocity by about one millimeter per second.

B. How far will Eve drift in one minute while she’s in Eden’s zero-g dome?

For this problem the equation is the same as above, x = ½ • a • t2. The equation does not need to be reordered. The value of a (acceleration) was calculated above, 8.6403e-4 m/sec2. One minute is 60 sec. This is all the information needed to solve the equation.

x = the distance Eve will drift in one minute (in meters)

a = the deceleration of the Eden, from problem A (in meters per second-squared)

t = the time Eve drifts, one minute (in seconds)

x = ½ • a • t2 = ( 0.5 • 8.6405e-4 m/s2 ) • ( 60 s )2 = 1.5553e0 m

Because time is squared, in two minutes Eve drifts almost 6 meters.

C. What is Eden’s maximum velocity, which occurs at the journey’s half-way point?

The velocity equation is simpler than the position equation; it is v = a • t, velocity is acceleration times time. The acceleration was calculated in A, above, and is 8.6405e-4 m/sec2. The time is half the journey, which is 80061 days, which is 6917270400 sec. This is all the information needed to solve the equation.

vmid = the velocity of the Eden at the midpoint of the journey (in meters per second)

a = the acceleration of the Eden, from problem A (in meters per second-squared)

t = half the duration of the journey, which is 160122 days divided by two (in seconds)

vmid = a • t = ( 8.6405e-4 m/s2 ) • ( 6.9173e9 s ) = 5.9768e6 m/s

Converting this to the speed of light, it is 0.0199 c, which is about 2% the speed of light, more than 13 million miles per hour, more than 21 million kilometers per hour! That’s fast enough to do one lap around Earth’s equator in just under 7 seconds. At this speed, there will be minor relativistic effects (time dilation); they are minor enough to disregard for a work of fiction but not for real mission planning!

D. This is a more challenging problem. If Adam hadn’t been an unruly teenager and had not taken a saw to the black box, their acceleration would have dropped to 38.12%, extending their journey by 51 years, 9 months, and 5 days. This is the answer to the problem. See if you can solve it yourself. Here are some needed variables: The Bussard ramjet dropped from 100% thrust to 38.12% thrust on day 155959. Here’s how to solve it:

Calculate the velocity of the Eden on the day the Bussard ramjet failed: vfailure = vmid + anormal • tmid2fail

vfailure = velocity of Eden on day 155959 (in meters per second)

vmid = maximum speed at the midpoint, from problem C (in meters per second)

anormal = the normal 100% acceleration times -1, because it is deceleration, not acceleration, from problem A (in meters per second-squared)

tmid2fail = time from the midpoint to the failure, day 155959 minus day 80061 (in seconds)

vfailure = vmid + anormal • tmid2fail = ( 5.9768e6 m/s ) + ( –1.0 • 8.6405e-4 m/s2 ) • ( 6.5576e9 s ) = 3.1078e5 m/s

Calculate the distance the Eden traveled from the midpoint to the day the Bussard ramjet failed: xfailure = vmid • tmid2fail + ½ • anormal • tmid2fail2

xfailure = the distance the Eden traveled from the midpoint to the point of the failure (in meters)

vmid = the velocity of the Eden at the midpoint, from problem C (in meters per second)

anormal = the acceleration at 100%, make it negative because it is decelerating, from problem A (in meters per second-squared)

tmid2fail = the time the Eden traveled from the midpoint to the failure point, day 155959 minus day 80061 (in seconds)

xfailure = vmid • tmid2fail + ½ • anormal • tmid2fail 2 = ( 5.9768e6 m/s ) • ( 6.5576e9 s ) + ( 0.5 • –8.6405e-4 m/s2 ) • ( 6.5576e9 s )2 = 2.0616e16 m

From the point of failure, calculate the time it takes the Eden to stop; use its velocity at the point of failure and its reduced deceleration: tfullstop = vfail ÷ afail

tfullstop = time to stop from the point of failure (in seconds)

vfail = velocity at the point of failure, calculated immediately above (in meters per second)

afail = the reduced acceleration, problem A times 38.12% (in meters per second-squared)

tfullstop = vfail ÷ afail = ( 3.1078e5 m/s ) ÷ ( 3.1316e-4 m/s2 ) = 9.9239e8 s

Calculate the distance the Eden traveled from the failure point to full stop, using its velocity at that point, the reduced acceleration, and the amount of time it takes to stop: xfullstop = vfailure • tfullstop + ½ • afail • tfullstop 2

xfullstop = the distance the Eden traveled from the point of the failure to full stop (in meters)

vfailure = the velocity of Eden on day 155959, calculated above (in meters per second)

afail = the acceleration at 38.12%, make it negative because it is decelerating (in meters per second-squared)

tfullstop = the time the Eden traveled from the point of failure to full stop, calculated above (in seconds)

xfullstop = vfailure • tfullstop + ½ • afail • tfullstop 2 = ( 3.1078e5 m/s ) • ( 1.3488e9 s ) + 0.5 • ( -3.1316e-4 m/s2 ) • ( 1.3488e9 s )2 = 1.3432e14 m

Calculate the time to travel back to Earth from the stop point. Just as the trip from Tarth to Earth, the ship will accelerate for half the journey then decelerate for the other half, reaching its maximum speed in the middle. Thus, we calculate the the time to reach the midpoint then double it. First we need to know the distance that Eden overshot Earth. A method to determine this is to add the distance to the Tarth-Midpoint, the add the distance from the midpoint to the failure, then add the distance from the failure to the full stop and then subtract the distance between Tarth and Earth: xovershoot = xmidpoint + xfailure + xfullstop - xtarth-earth

xovershoot = the distance the Eden traveled past Earth until it reached a full stop (in meters)

xmidpoint = half the distance from Tarth to Earth, 4.37 lightyears divided by 2 (in meters)

xfailure = the distance the Eden traveled from the midpoint to the point of the failure, see below (in meters)

xfullstop = the distance the Eden traveled from the failure point to full stop, see below (in meters)

xtarth-earth = the distance between Tarth and Earth, 4.37 lightyears (in meters)

xovershoot = xmidpoint + xfailure + xfullstop – xtarth-earth = ( 2.0672e16 m ) + ( 2.0616e16 m ) + ( 1.3432e14 m ) – ( 4.1343e16 m ) = 7.8428e13 m

(For clarity, note that only 5 significant digits are shown; you’ll need more in your calculation.)

Now calculate the time to accelerate to the halfway point (between the full stop and Earth). Double that to get both the acceleration and deceleration times: tovershoot = 2 • √ { [ 2 • ( xovershoot ÷ 2) ] ÷ afail }

tovershoot = time it take the Eden to travel from the full stop point to Earth (in seconds)

xovershoot = the distance the Eden traveled past Earth until it reached a full stop (in meters)

v = the acceleration at 69.93% (in meters per second-squared)

tovershoot = 2 • √ { [ 2 • ( xovershoot ÷ 2) ] ÷ afail } = 2 • √ { [ 2 • ( ( 7.8428e13 m ) ÷ 2) ] ÷ ( 3.1316e-4 m/sec2 ) } = 1.0009e9 sec

Finally, calculate the tie difference between the time of the planned voyage and the time it would have taken if they hadn’t fixed the Bussard ramjet: tdifference = tfailure + tfullstop + tovershoot – tplanned

tdifference = The time difference between the original planned voyage and if the voyage had proceeded with the failure.

tfailure = The time at which the Eden encountered its failure, 155959 days.

tfullstop = The time required for Eden to come to a full stop after the failure.

tovershoot = The time required for Eden to return to Earth after the full stop.

tplanned = The original planned duration of the mission, 160122 days.

tdifference = tfailure + tfullstop + tovershoot – tplanned = (1.3475e10 sec ) + ( 9.9239e8 sec ) + ( 1.0009e9 sec ) – ( 1.3835e10 sec ) = 1.6336e9 sec

(For clarity, note that only 5 significant digits are shown; you’ll need more in your calculation.)

In converting 1.6336e9 seconds to years, months, and days, you’ll come up with approximately 51 years, 9 months, and 5 days. Small errors (less than a few weeks) might be due to the number of significant digits you used and whether you rounded your years (365 days per year), rounded your leap years (365.25 days per year), or didn’t round your leap years (365.2425 days per year), and whether you considered leap seconds (which are not constant).

Disclaimer: My proofreader didn’t really know what to do with this section. If you find errors, please let me know (via https://www.markwilx.com). Future editions of the story will credit the first to report any errors. The beauty of writing science fiction is that you can tweak the variables. The beauty of writing fantasy is that you can tweak the equations.