How can I find a estimation of the delta-V needed to lift-off from a planet and go into a low orbit around it?

  • $\begingroup$ Your question about the effect of Mars' atmosphere would be best asked as a separate question. $\endgroup$
    – Joe
    Commented Jun 13, 2014 at 17:19
  • 1
    $\begingroup$ For dV off Earth, you might want to take a gander at clowder.net/hop/railroad/deltaveemap.html $\endgroup$
    – Everyone
    Commented Jun 13, 2014 at 18:29

4 Answers 4


For any spherical body with a density $\rho$ and radius $R$ and no atmosphere, we can calculate this easily. Let's assume you launch to a very low orbit that just skims the surface of the sphere. You can add 10% or 20% later for a planet like Earth with its atmosphere. Venus would be a lot harder! (so I've asked separately Launch to orbit delta-v penalty from Venus compared to Earth?)

$$M = \frac{4}{3} \pi R^3 \rho.$$

From the vis-viva equation orbital velocity (roughly the delta-v you need to enter a very low orbit around a sphere with no atmosphere launching from the surface) is

$$v= \sqrt{\frac{GM}{R}} = 2R\sqrt{\frac{\pi G \rho}{3}},$$

and the period is

$$T = \frac{2 \pi R}{v} = \pi \sqrt{\frac{3}{G\pi \rho}}.$$

Surprisingly, while the velocity increases linearly with radius (for a given density) the period is independent of size. Earth's high average density will give a period of about 90 minutes but the low density of a bowling ball or a comet will have a period closer to 4 hours.

The gravitational constant G is 6.674E-11 m^3/kg s^2 and some densities and examples are ((almost) all from Wikipedia):

object or substance        ρ (kg/m^3)    R(m)        v(m/s)          T(min)
-------------------        ----------  ---------     ------          ------
    Earth                    5514     6,371,000      7,910             84
    Moon                     3344     1,737,000      1,679            108
    2-Pallas                 3000       256,000        234            114
    1-Ceres                  2160       469,730        365            135
    Pluto                    1854     1,188,000        855            145
    162173 Ryugu             1330           432          0.26         172
    bowling ball*           ~1000             0.22       0.00012      198
    Haley's comet             600         5,500          2.2          255


As a first approximation, you can calculate the orbital speed:

$v_o \approx \sqrt{\frac{GM}{r}}$

where Vo is the orbital speed, G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit.
That gives you the speed you need to achieve. You can't say $v_o = \Delta V$, because $\Delta V$ contains several other factors:

  • potential energy due to the height of the orbit
  • speed gained or lost by launching along or against the planet's rotation (you need to calculate the rotation speed of your launch site). On Earth's equator, this is some 1600 km/h.
  • energy losses due to e.g. drag
  • 1
    $\begingroup$ Such a rocket will not reach orbit. 1) You have to climb to orbital height. 2) You will lose a certain amount of energy countering gravity. $\endgroup$ Commented Jun 13, 2014 at 23:37
  • $\begingroup$ So how could we take into account those effects? $\endgroup$
    – Pierpaolo
    Commented Jun 17, 2014 at 11:18
  • $\begingroup$ Isn't a Hohmann transfer a better first approximation $\endgroup$
    – user20636
    Commented Sep 27, 2019 at 10:38

I too am going to start with the orbital speed as a first approximation, but we can do slightly better than that.

$$\Delta v = v_{orbit} = \sqrt{\frac{\mu}{r}}$$

If you are using this approximation alone, you would want to use the objects radius for $r$ and not the radius of the low orbit, as that gives a slightly higher cost that better accounts for the other losses involved.

To account for the difference in altitude, the perfect case would involve an elliptic transfer orbit touching the surface and the target orbital altitude. This is applicable for objects with low mass and no atmosphere, such as asteroids:

$$\Delta v = \sqrt{\mu\left(\frac{2}{r_{surface}} - \frac{2}{r_{surface} + r_{orbit}}\right)} + \sqrt{\frac{\mu}{r_{orbit}}} - \sqrt{\mu\left(\frac{2}{r_{orbit}} - \frac{2}{r_{surface} + r_{orbit}}\right)}$$

For longer continuous burns, a trajectory that looks more like a low thrust transfer is more applicable, and slightly simpler:

$$\Delta v = 2\sqrt{\frac{\mu}{r_{surface}}}-\sqrt{\frac{\mu}{r_{orbit}}}$$

Other factors

The Earth rotates (465 m/s at the equator), and many other objects do too. The rotational velocity can generally be directly subtracted from your $\Delta v$ cost, in the case of a prograde launch. It scales by cos(latitude).

Gravity losses. The rocket needs to counteract gravity to remain in the air. A first approximation would be:

$$\Delta v_{gl} =g\cdot T_{burn}$$

but this is generally too much. A more reasonable model is:

$$\Delta v_{gl} =\frac{2}{3} \cdot T_{burn}\left(\frac{\sqrt{(g\cdot T_{burn})^2/2 + \Delta v^2}-\Delta v}{g}\right)$$

Drag. No simple approximation. Assume 1-3 km/s for Earth.

If you want more accurate results than these, you would need numerical simulations of your spacecraft, or you would have to look up values for such simulations for existing spacecraft.

  • $\begingroup$ Nice answer, but may I suggest, in order to round it off, to add numerical values to each of the $\Delta v$s? This would give an idea about how important the discussed effects are compared to the simple $\sqrt{\mu/r}$ estimate. $\endgroup$ Commented Sep 27, 2019 at 7:58

Or for a quick lookup:

The Solar System: a subway map


(NB. The values look like they include atmospheric drag, so for atmospheric bodies the incoming delta-v is a bit lower, or in the case of Venus a lot)

  • $\begingroup$ @uhoh I thought I was only embedding. I've added a link to where I found it. Unfortunately I can't find the owner other than what's in the image already. $\endgroup$
    – JanKanis
    Commented Jan 8, 2021 at 7:10
  • $\begingroup$ okay if the source is also imgur, then I guess that that's the best that can be done. Thanks! $\endgroup$
    – uhoh
    Commented Jan 8, 2021 at 12:49

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