How long does it take to get to Mars?
Basic equations for interplanetary travel
The basic equation for objects orbiting the sun is:
v2 µ µ
− = − (1)
2 r 2a
µ = GMsun (2)
where a is the semi-major axis of the orbit, r is the radius, G is Newton’s
gravitational constant, and v = v(r) is the speed.
Using this equation, we can calculate the speed of the Earth in its orbit
around the sun:
vE = (4)
where RE is the Earth’s orbital radius, which we assume is equal to its semi-
major axis (a circular orbit).
Using the values in Table 1, we get vE ≈ 30 km/s (≈ 67,000 MPH).
Table 1: Physical parameters
Mass of the sun 1.99 · 1030 kg
Gravitational constant 6.67 · 10−11 Nm2 /kg2
Earth orbital radius 1 A.U.
Mars orbital radius 1.52 A.U.
The Hohmann transfer orbit
The Hohmann transfer orbit1 is the lowest-cost (in terms of energy) way to move
from one circular orbit to another circular orbit. This orbital maneuver takes
an object in one orbit to another by providing a brief energy boost (usually
Figure 1: Hohmann orbital transfer between two circular orbits
signiﬁed “delta-v”, or ∆v) to change the orbital energy from a circular orbit to
an elliptical orbit. After the object has trasversed half of this elliptical orbit,
∆v is applied to make the orbit circular. Figure 1 shows a diagram of this
orbital transfer maneuver2 .
Speciﬁc solutions for Mars trip
With current technology, the Hohmann transfer, or something very similar,
oﬀers the only realistic option for a human-inhabited spacecraft to get to Mars.
Here, we calculate the energy (∆v) required to make the trip, and the time in
each part of the orbit. We assume that we start in orbit around the Sun near
the Earth (r = RE ), use a Hohmann transfer orbit, and arrive in orbit near
Mars (r = RM ).
We note here that these calculations are approximate. We assume that the
Earth and Mars orbits are circles. In fact they are not, but these approximations
are quite accurate, and the diﬀerences between these and more exact calculations
are a few percent.
Equation (4) shows the speed in a circular orbit around the Sun at Earth’s
radius. We need to know the speed in an elliptical orbit of semi-major axis equal
to the sum of the Earth and Mars orbital radii, at the Earth’s orbital distance:
v2 µ µ
− = − (5)
2 re rE + rM
v2 1 1
= µ − (6)
2 rE rE + rM
v = vE (7)
rE + rM
With this information, we now calculate the ∆v required to obtain this orbit:
∆v = vE −1 (8)
rE + rM
∆v ≈ 2.9km/s ≈ 6500MPH (9)
This means that once we have built our spacecraft in low-Earth orbit and
escaped completely from the Earth, we require an additional energy boost equal
to a change in speed of about 3000 meters/second to put the spacecraft into the
elliptical orbit which will arrive at Mars sometime later. How much later?
To answer this question, we use Kepler’s Third Law, which states that the
square of the period of an orbit divided by the cube of the semi-major axis is a
Using this, we can see that the orbit time To for the elliptical orbit for the trip
to Mars is:
rE + rM
To = TE (12)
To ≈ 1.39years (13)
where TE is the Earth’s orbital period of one year.
This is the total time for an object in the elliptical Earth-Mars orbit to
complete a full revolution of the Sun, so the time to get to Mars in this orbit is
about 0.7 years, or 255 days.
255 days! That’s a long time for a human to be exposed to the harsh
space environment (weightlessness, harmful cosmic rays from the Sun and other
Moreover, once on Mars, the travelers must wait an extended period of time
before beginning their trip homeward. How long?
The Earth moves ahead of Mars by about 170 degrees per year. When the
spacecraft ﬁrst starts from Earth to Mars, Mars needs to be about 46 degrees
ahead of Earth in its orbit. Otherwise, Mars won’t be there when the astronauts
arrive! After the arrival at Mars, Earth will be about 71 degrees ahead of Mars
in its orbit. The astronauts must wait until Earth is behind Mars in orbit by
about 71 degrees. So the total time they have to wait on Mars is:
360◦ − 2 · 71◦
∆t ≈ (14)
∆t ≈ 1.3yr (15)
That’s around 15 months! So, the total trip time is 255 days for each leg of
the journey, plus 471 days waiting on Mars, for a grand total of 981 days! That
seems like a long, long time; it’s almost three years.
Isn’t there a better way?
There is in fact a better way to go Mars and back, which takes much less time.
However, it is much more costly in terms of energy. The basic idea is to launch
a spacecraft on a parabolic or hyperbolic path to Mars (not a Hohmann orbit),
spend several days on Mars, and then race back to Earth, all in less than six
One can show that the ∆v required to put a spacecraft on a parabolic orbit
(i.e., to escape from the Sun’s gravity) is given by:
∆v = 2 − 1 vE (16)
∆v = 12.4km/s ≈ 27, 700MPH (17)
The time required to make a trip to Mars on this orbital path is about 70
days. If the astronauts spend about 12 days on the surface of Mars, they can
then begin another parabolic path back to Earth arriving about 70 days later.
The ∆v required at Mars to begin the trip back to Earth is (only!) 10.1 km/s
(≈ 22,600 MPH). The total trip time is 152 days.
This parabolic orbit is for all intents and purposes the only way to make
a trip to Mars and back in a relatively short period of time. So, the ∆v of
12.4 km/s should be considered a rough minimum in required delta-v for a
“short” trip. If the delta-v is smaller, the Earth will race ahead of Mars, and
the astronauts will have to wait a long time for Earth to come around again
before returning, as discussed above.
What if the astronauts want to stay on Mars for a longer period of time?
To stay longer on Mars requires an even larger ∆v at Earth (a hyperbolic orbit,
which is an orbit with energy greater than that required to escape from the
Sun). For example, a 30-day stay at Mars requires ∆v = 15.4 km/s, and a 45-
day stay at Mars requires ∆v =18.5 km/s. These are very large energy changes,
well outside our current technology.
The careful reader will balk at something in the above: the “minimum-speed
trip” requires a ∆v of 12.4 km/s. That’s a tremendous change in energy. By
comparison, the ∆v required to put an object in Earth orbit (arising from the
surface) is about 7.7 km/s. The escape velocity from Earth is 10.8 km/s. So, the
energy required to put a spacecraft on a parabolic path to Mars is 260 percent
larger than the energy required to put the spacecraft into orbit. Including the
∆v from low-Earth orbit to escape, it’s a whopping 410 percent larger!
Simply stated, we don’t have the technology currently to put a large space-
craft onto a parabolic orbit to Mars. For this reason, some scientists and engi-
neers want to do more research into other rocket technologies besides our current