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Escape velocity (from the earth) of the Earth-Sun system

  1. Feb 9, 2015 #1
    Escape velocity is an estimate of the launch velocity of a spacecraft (without any propulsion) to overcome a planet system's gravitational pull in order to escape to ``infinity''. In this problem we consider both the gravitational attraction of the Earth and Sun (but ignoring the effects of other bodies in the solar system). Use the following escape velocities from the Earth and from the Sun, and the orbital velocity of the earth:

    ##v_E =\sqrt{{2 G M_E}/{r_E}}## is the escape velocity from the Earth;
    ##v_S= \sqrt{{2 G M_S}/{r_{SE}}}## is the escape velocity from the gravitational field of the Sun at the orbit of the Earth, but far away from the Earth's influence;
    ##v_0## is the orbital velocity of the Earth about the Sun;

    to show that the escape velocity (relative to the Earth) from the Earth-Sun system is
    $$v = \sqrt{ v_E^2 + (v_S -v_0)^2}.$$

    Relevant equations
    Gravitational potential of a mass ##m## related to another mass ##M## at distance ##r##:
    $$U = - \frac{G M m}{r}.$$

    In the Earth-Sun two-planet system, energy equation (mechanical energy only) for a spacecraft launched from the Earth to infinity can be written as the sum of kinetic energy and the potential energy as follows:
    $$ \frac{1}{2} m v_{abs}^2 -\frac{G M_E m}{r_E} - \frac{G M_S m}{r_{SE}} = 0.$$
    This is the energy equation using the absolute velocity ##v_{abs}##, or velocity relative to the Sun (assuming the Sun is stationary). On the right-hand side of the equation above, the total energy is zero because its kinetic energy is zero at infinity (since we are looking for the minimum velocity at launch on the surface of Earth), and potential energies at infinity from the Earth and Sun are also zero. So
    $$v_{abs}^2= v_E^2 + v_S^2 \implies v_{abs} = \sqrt{v_E^2 + v_S^2}.$$
    But the escape velocity we want is relative to the Earth, we have ##v = v_{abs}- v_0## (using the simplest scenario in which the spacecraft is moving in the same direction as the Earth's orbital velocity).

    Now we have
    $$v^2 = (v_{abs} - v_0)^2 = \left( \sqrt{ v_E^2 + v_S^2} - v_0\right)^2=v_E^2 + v_S^2 + v_0^2 -
    2 v_0\sqrt{v_E^2 + v_S^2} $$
    Since ##v_E^2 \ll v_S^2##, we can further simplify the above expression by dropping ##v_E## inside the square root to get
    $$v^2 \approx v_E^2 + v_S^2 + v_0^2 - 2 v_0 v_S = v_E^2 + (v_S - v_0)^2.$$
    So we have got the solution:
    $$ v\approx \sqrt{ v_E^2 + (v_S -v_0)^2 }.$$
    This is Problem (8.61) from Douglas Giancoli's Physics for Scientists and Engineers. The problem contains a hint, yet its wording is a bit confusing (at least to me). Therefore it prompted me to write this up here. Hope it can be helpful to anybody who is curious about the form of the solution of ##v##.
    Last edited by a moderator: Feb 9, 2015
  2. jcsd
  3. Feb 9, 2015 #2

    Simon Bridge

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    Welcome to PF;
    I'm sure this will show up in the google hits for people who get confused with this problem. Well done.
  4. Feb 9, 2015 #3
    @Simon Bridge Thank you for your kind words.

    After posting this solution, I was able to see the answer given in the textbook's solution manual. The method there is very confusing. I don't have time to post that solution right now, but I will in a few days.

    When I put it up, I'll appreciate it if anyone in the PhysicsForum would make comments.
    Last edited: Feb 9, 2015
  5. Feb 9, 2015 #4


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