What it says on the tin: using the most efficient launch opportunity afforded by orbital mechanics, what delta-v would you have to impart to a payload to send it from the surface of Mercury to crash on the surface of Venus?


2 Answers 2


Assuming circular, coplanar orbits, Trans Venus Insertion (TVI) is 8 km/s from Mercury's surface. That's with horizontal take off and no gravity loss. That's for the minimum energy Hohmann transfer.

Exiting the Hohmann transfer and soft landing on Venus would take 11.9 km/s if Venus had no atmosphere. But much, maybe all of that 11.9 km/s could be done by aerobraking.

I used my Hohmann spreadsheet to get these numbers.

  • $\begingroup$ What do you mean when you say "no gravity loss"? My calculations suggest 8km/s wouldn't be fast enough, but I might be misunderstanding. $\endgroup$
    – user21
    Jan 27, 2016 at 0:21
  • $\begingroup$ @barrycarter during vertical ascent, gravity subtracts from from the rocket's thrust. That's gravity loss. The trajectory you show in your answer is not Hohmann so it would take more delta V than the minimum energy ellipse. I have not examined your math. I use the Vis Viva equation, a widely used method in orbital mechanics $\endgroup$
    – HopDavid
    Jan 27, 2016 at 20:49

If you launch from the side of Mercury furthest from the Sun, parallel to the horizon in the direction of Mercury's orbit at 13 km/s when the Mercury-Sun-Venus angle is 53 degrees, your payload will reach Venus in 35 days, and its path would look something like this (each black dot = 1 day). IMPORTANT CAVEATS BELOW IMAGE.

enter image description here

This doesn't really answer your question, since I'm not sure 13 km/s is the minimal required speed. I also made several simplifying assumptions (perhaps too many), so the answer above isn't exact.

Here's the heavily commented Mathematica script I wrote, which also explains the simplifications I made.

(* cleaned up and well-commented version for SE *)


I assume that Mercury's orbit is planar, and that Venus' orbit lies in
the same plane, or at least intersects it at the appropriate time.

This allows me to use a two-dimensional equation for acceleration due
to gravity, instead of a 3 dimensional one.

The equation (below) is the acceleration imparted to an object of mass
m1 at {x1,y1} by an object of mass m2 at {x2,y2}, given that the
gravitational constant is g.

Note that the mass of the object being accelerated (m1) is actually
irrelevant; however, I include it as a parameter for symmetry




The mass, semimajor axis, period, and radius of Mercury, in kg, m (not
km), and s


mercsma = 57909050000;
mercper = 87.9691*86400
mercrad = 2439700
mercmass = 3.3011*10^23

(* solar mass, in kg *)

sunmass = 1.98855*10^30

(* gravitational constant of universe, in kg-m-s system *)

g = 6.6740*10^-11


Heliocentric, so Sun is always at origin. In theory, the positions of
the other planets (eg, Jupiter) could help boost your payload, so you
might be able to launch with a lower speed than I find below


sun[t_] = {0,0}


I assume Mercury's orbit is circular. Since the actual orbit is
elliptical, you could get a boost for your payload by launching it
when Mercury's distance from the Sun is increasing the fastest (in
other words, solar radial velocity is greatest)

I've chosen the x axis to be the line connecting the Sun to Mercury at time 0.


merc[t_] = {Cos[t*2*Pi/mercper],Sin[t*2*Pi/mercper]}*mercsma


I also ignore Venus' own gravity: you can do slightly better by noting
that Venus will pull the payload towards itself once the payload gets
close enough.

I do want to plot Venus' orbit, so I use the semi-major axis and
period values below.

Venus' starting angle (vsa below) was found by trial and error to make
sure Venus was at the right place when the payload crossed its orbit.


vensma = 108208000000
venper = 224.701*86400;
vsa = 53*Degree;
ven[t_] = {Cos[t*2*Pi/venper+vsa],Sin[t*2*Pi/venper+vsa]}*vensma


If we launch from side of Mercury furthest from the Sun, the payload's
starting position will be Mercury's position plus Mercury's radius in
the x direction

NOTE: This start position is completely arbitrary; you may get better
results by starting at different positions on Mercury's surface.


s0 = {mercsma+mercrad,0}


The initial velocity of the payload (with respect to the Sun) will be
Mercury's velocity + whatever velocity (delta v) we impart to the

Note that both the direction I choose for initial velocity (in the
same direction as Mercury's orbit) and the magnitude are
arbitrary. You may get better results by aiming the payload at a 45
degree angle or straight up or something.

NOTE: If I change 13000 to 12500 below, Mathematica will refuse to
solve the differential equation. This doesn't necessarily mean 13000
is a minimal velocity, but there is apparently some sort of important
change between 12500 m/s and 13000 m/s


v0 = merc'[0] + {0,13000}


Mathematica won't close-form integrate this problem, so we integrate
numerically, which requires a start time (0) and an end time (below).

I chose 35 days after confirming that's how long it takes the payload
to reach Venus.


timelimit = 86400*35;

nds = NDSolve[{s[0]==s0, s'[0] == v0,
 s''[t] == accel[s[t],sun[t],1,sunmass,g] + accel[s[t],merc[t],1,mercmass,g]

(* The use of [[1,1,2]] below is just Mathematica nesting weirdness *)

g= ParametricPlot[{nds[[1,1,2]][t],merc[t],ven[t]},{t,0,timelimit}, 
 Mesh -> timelimit/86400, AxesOrigin->{0,0}, PlotStyle -> {Blue,Red,Green},
 MeshStyle -> {Black}
  • 1
    $\begingroup$ For minimum energy, I think you want it to reach Mars as it reached the opposite side of the sun from where it was launched. In other words, you want an elliptical orbit with perihelion at Mercury at launch and aphelion at Mars. $\endgroup$
    – eshaya
    Jan 16, 2016 at 1:36
  • $\begingroup$ Did you mean Venus? And I'm still not sure that's correct? Wouldn't it be aphelion for Mercury and reaching Venus at perihelion for minimal orbital distance traveled? $\endgroup$
    – user21
    Jan 16, 2016 at 2:26
  • 1
    $\begingroup$ I meant aphelion at Venus. Mercury is closer to Sun so it has to be the perihelion. This is minimum Energy orbit that connects the two points. $\endgroup$
    – eshaya
    Jan 18, 2016 at 1:21
  • $\begingroup$ Oh, you mean the payload should assume an elliptical orbit where Mercury is its perihelion and Venus is its aphelion? I'm still not sure that's true, since it ignore Mercury's own gravity. $\endgroup$
    – user21
    Jan 18, 2016 at 1:23
  • $\begingroup$ @barrycarter Outside of the planetary spheres of influence, transfer orbit is modeled as an ellipse with a focus at the sun center. Within planetary sphere of influence orbit is modeled as a hyperbola with a focus at planet center. Speed of a hyperbola is $sqrt(Vesc^2+Vinf^2)$. This is the patched conics approximation. $\endgroup$
    – HopDavid
    Jan 27, 2016 at 20:40

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