I'm having a lot trouble with this problem. I need the minimum escape velocity for the (earth sun) solar system. I can't simply integrate the potential with respect to radius since this changes with respect to time. I have worked out the velocity needed to purely escape the sun's influence ignoring earth which is $\sqrt{2\frac{GM_s}{R}}-v_e$ where $M_s$ is the mass of the sun and $v_e$ is the orbital velocity of earth and $R$ is sun earth radius. This is approximately 12.3km/s which can also be seen here


What's troubling me is that must account for the changing potential of the earth with respect to the sun, we can't simply add the escape velocity of the earth can we? I have worked out that the velocity of the earth in the direction of launch is $v_l - v_ecos(\omega t) =v_l-v_e$ approximately where $v_l$ is velocity of launch. How do I go on from here? Please feel free to leave a comment if the question is unclear


1 Answer 1


The easy way to calculate the escape velocity is to use the fact that for an object to just escape a gravitational well its total energy must be zero. That is, the negative gravitational potential energy and the positive kinetic energy sum to zero. So if we take the Earth as an example we have the gravitational potential energy:

$$ V = -\frac{GM_em}{r_e} $$

and the kinetic energy:

$$ T = \tfrac{1}{2}mv^2 $$

and requiring these sum to zero gives:

$$ T + V = \tfrac{1}{2}mv^2 -\frac{GM_em}{r_e} = 0 $$

and we get the usual equation for the escape velocity:

$$ v = \sqrt{\frac{2GM_e}{r_e}} $$

The point of all this is that potential energies are simply additive. For an object on the Earth's surface we add together the potential energy due to the Sun's gravity and the potential energy due to the Earth's gravity to give a total potential energy:

$$ V_\text{tot} = -\frac{GM_\text{Sun}m}{r_\text{Sun}} - \frac{GM_\text{Earth}m}{r_\text{Earth}} $$

Simply require that the kinetic energy of the object be equal to the total potential and you have calculated the escape velocity from the Earth's surface to out of the Solar System.

You are correct to note that the Earth is already moving due to it's orbital velocity. The escape velocity relative to the Earth will of course depend on which direction you launch in. For high accuracy don't forget that the surface of the Earth is rotating, so include that velocity as well.

  • $\begingroup$ It definitely makes it easier when you realise that the loss in kinetic energy just depends on the position in the field $\endgroup$
    – user86425
    Jul 25, 2017 at 14:09

Your Answer

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