How Long to Mars?
Elena Voss's team is planning a cargo run to Mars orbit. Before anyone fires an engine, mission planning needs the most basic answer of all: how long is this trip going to take?
Let's work it out two ways — algebra first, then calculus.
Part 1 — Algebra: Travel Time
Mission planning gives the transfer trajectory a length of D = 480,000,000 km. The spacecraft cruises at a constant velocity of v = 24 km/s.
How long does the trip take — in seconds, hours, and days?
For constant velocity, distance, velocity, and time are related by:
$$ d = v \cdot t \quad\Longrightarrow\quad t = \frac{d}{v} $$
Plugging in the numbers:
$$ t = \frac{480\,000\,000 \text{ km}}{24 \text{ km/s}} = 20\,000\,000 \text{ s} $$
Now convert to more useful units:
- Hours: $20\,000\,000 \div 3\,600 \approx 5\,556$ hours
- Days: $20\,000\,000 \div 86\,400 \approx 231.5$ days
- Months: $231.5 \div 30.44 \approx 7.6$ months
About 7–8 months — right in line with real Mars transit times. Now mission planning knows the crew's life-support, food, and the rest of the spacecraft's systems need to last that long, with margin.
Try it yourself
Plug in any distance and cruise velocity to compute travel time.
Part 2 — Calculus: Velocity & Acceleration
Part 1 assumed the spacecraft instantly hits 24 km/s and cruises at exactly that speed the whole way — useful for a quick estimate, but not how real missions fly. Real spacecraft accelerate away from Earth and decelerate as they approach their destination.
Let's model the same trip — same total distance $D = 480\,000\,000$ km, same total time $T = 20\,000\,000$ s — but with a smooth velocity profile: slow at departure, fastest in the middle, slow again at arrival. A simple curve that does exactly this is:
$$ s(t) = D\left(3u^2 - 2u^3\right), \qquad u = \frac{t}{T} $$
This is the spacecraft's position — distance covered — at time $t$. Check the endpoints: at $t=0$, $u=0$ so $s=0$ (still at Earth); at $t=T$, $u=1$ so $s = D(3-2) = D$ (arrived).
Position vs. time
Instantaneous velocity is the derivative of position with respect to time — how fast position is changing right now, as opposed to the average velocity from Part 1. Using the chain rule (since $u = t/T$, $\frac{du}{dt} = \frac{1}{T}$):
$$ v(t) = \frac{ds}{dt} = D\left(6u - 6u^2\right)\cdot\frac{1}{T} = \frac{6D}{T}\,u(1-u) $$
Instantaneous acceleration is the derivative of velocity — equivalently, the second derivative of position:
$$ a(t) = \frac{dv}{dt} = \frac{6D}{T}\left(1 - 2u\right)\cdot\frac{1}{T} = \frac{6D}{T^2}\left(1 - 2u\right) $$
At $t=0$ ($u=0$): $v=0$, and $a = \dfrac{6D}{T^2}$ — maximum acceleration, the departure burn. At $t=T/2$ ($u=0.5$): velocity peaks at $v = 1.5\,D/T = 36$ km/s (1.5× the Part 1 average!) and acceleration is momentarily zero. At $t=T$ ($u=1$): $v=0$ again and $a = -\dfrac{6D}{T^2}$ — maximum deceleration, the arrival burn.
Plugging in numbers, $\dfrac{6D}{T^2} \approx 0.0072$ m/s$^2$ — about 7 thousandths of a gravity. That's far too gentle for a chemical rocket, but it's exactly the kind of continuous, gentle push Elena's ion thruster produces over months of cruising.
Velocity vs. time
The dashed line is the constant 24 km/s "average velocity" from Part 1 — note the curve crosses it exactly at the midpoint.
Acceleration vs. time
Positive acceleration (speeding up) for the first half of the trip, negative (slowing down) for the second half — crossing zero exactly when velocity peaks.