Mars has a $g$ of about 38% of Earth's, but an escape velocity that is 45% of ours.

Not a large difference, but there must be a reason.


2 Answers 2


Suppose a planet has mass $M$ and radius $R$. Its surface gravity $g$ is the free fall acceleration that an object experiences at its surface. Thus we can find $g$ by Newton's 2nd law:

$$mg = \frac{GMm}{R^2} \implies g = \frac{GM}{R^2}.$$

The planet's escape velocity is the minimum velocity that an object at its surface requires to go an infinite distance away from the planet. The gravitational potential energy of an object at distance $r$ from the center of the planet is $-\frac{GMm}{r}$, so at an infinite distance away, the GPE of the object is $0$. Thus, we need total energy at least $0$ to escape to infinity.

$$\frac{1}{2}mv^2 - \frac{GMm}{R} \geq 0 \implies v \geq \sqrt{\frac{2GM}{R}} \implies v_{e} = \sqrt{\frac{2GM}{R}}.$$

Because these formulas depend on $M$ and $R$ in different ways, the surface gravity and escape velocity are not proportional.

  • $\begingroup$ You can also from the result being dependent of M and R, say the escape velocity depends somehow on the planet (or other body) mass density.... $\endgroup$ Sep 7, 2020 at 6:21

Not a large difference, but there must be a reason.

Yep there sure is!

tl;dr: You can't compare them directly. You need one more independent piece of information about the planet before you can connect the velocity you need to escape to infinity with surface gravity.


Since "Because the equation says..." is often needlessly unsatisfying in classical mechanics, let's first look at it this way.

The surface gravitational acceleration $g$ tells us something about the combination of the planet's mass and how close the surface is to the center. We know it's proportional to mass $m$ and inversely proportional to the radius squared ($r^2$) so we'll feel the same $g$ standing on a planet that's 1% of Earth's mass and 10% of Earth's radius.

But, the escape velocity also depends on how that acceleration drops off with distance.

On the mini-Earth mentioned above, the gravitational acceleration will fall off much faster with distance. By the time you climb one Earth radius, the acceleration (depending on $1/r^2$) has dropped by a factor of 100 because you started from 0.1 only Earth radius.

But by the time you climb one more Earth radius from Earth, the acceleration has only dropped off a factor of 4.

The discussion above only shows that you can't use the acceleration measured at the surface only to judge how quickly it falls off with distance or how much initial velocity you'll need to escape to infinity.


Since "Because..." is often needlessly unsatisfying in classical mechanics without the equations to back it up, now let's look at it this way.

$$g = \frac{GM}{r^2}$$

$$v_{esc} = \sqrt{\frac{2GM}{r}}$$

from here. You can get the escape velocity using calculus, integrating acceleration as the rocket climbs, but there's an easier way just using kinetic and potential energy.

When we stand on Earth we have negative potential energy. When we jump we have a little positive kinetic energy, but the sum with potential is still negative. Having escape velocity means that our kinetic energy $K$ plus our potential energy $U$ are equal in magnitude and opposite in sign, so sum to zero.

Again with the math:

$$U = - \frac{GM}{r}$$

$$K = \frac{1}{2} v^2$$

These are called "reduced" energy because we leave out the mass $m$ of the person, or the rocket. Since we're looking for the sum to be zero, it doesn't matter. The escape velocity doesn't depend on the objects mass as long as we ignore drag in the atmosphere and the object is not also planet-sized.

If we rearrange $U + K = 0$ to $K = -U$ we get

$$2GM/r = v_{esc}^2$$

and thus the equation for escape velocity

$$v_{esc} = \sqrt{\frac{2GM}{r}}$$

If we rearrange the equation for surface gravity we have

$$r = \frac{\sqrt{GM}}{\sqrt{g}}$$

putting that back into escape velocity:

$$v_{esc} = (4 GM g)^{1/4}$$

So the escape velocity depends both on the surface gravity $g$ and the planet's mass. You could get rid of the mass but then the planet's radius pops up.

$$ v_{esc} = \sqrt{2 g r}$$

Either way, you need one more independent piece of information about the planet before you can connect escape velocity with surface gravity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.