I am trying to gain some intuition about Delta-V requirements.

My understanding is if the Delta-V budget was doubled, the difficulty of the mission is more than double because of the exponential nature of having to carry more fuel.

My question is - how are Delta-V requirements influenced by the planet's gravity?

For example, if Earth's gravity were to double, would Delta-V (to an orbit of similar height) double as well? Or is this a non-linear effect?


What do you mean by "Earth's gravity"? Do you mean the mass of the Earth? Do you mean the acceleration of gravity at the surface of the Earth?

The acceleration of gravity is $g={GM\over r^2}$, where $M$ is the mass, and $r$ is the radius of the surface. $G$ is Newton's gravitational constant. The escape velocity from Earth is $v_e=\sqrt{2GM\over r}$.

So the answer to your question depends on what you mean by "gravity", and how you plan to increase the "gravity". Do you plan to increase the mass of the Earth but not the radius? Do you plan to increase the mass of the Earth and keep its density constant?

You would also need to know where you're going. $\Delta V$ to where?

Yes, the rocket equation is exponential in $\Delta V$. For a single stage, ${m_i\over m_f}=e^{\Delta V\over v_e}$, where $m_i$ is the initial mass, $m_f$ is the final mass (the difference being the mass of the propellant expended), and $v_e$ is the exhaust velocity of the engine.

  • $\begingroup$ Thanks. Let's take the case of doubling mass without doubling the radius of Earth. So plugging that in to escape velocity, that will mean you need sqrt(2) ~ 1.41 times more velocity to escape earth? Is that the same as 1.41 more times delta-v? $\endgroup$ – Andrew Nov 24 '13 at 3:27
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    $\begingroup$ Yes. Going into orbit and escaping are proportional. There are pieces that are not proportional, in particular atmospheric drag loss and the additional velocity beyond escape go to a particular destination. $\endgroup$ – Mark Adler Nov 24 '13 at 3:29

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