I know that if we project a body with velocity equal to escape velocity the body will fly to infinity but after reaching there it would have no energy as such, so it is the minimum velocity in fact to send a body to infinity.

However if the velocity of the body is greater than escape velocity then it will have some kinetic energy upon reaching infinity.I am interested in finding out the trajectory of the body. My book says the path to be a hyperbola without any explanation.

Can anyone put forward a trajectory of such a body?

  • $\begingroup$ A simple answer would be that as the orbital speed increases for an object with constant closest approach distance, the eccentricity of the orbit increases until the eccentricity of the orbit reaches 1, at which point the orbit becomes a parabolic escape trajectory. Increasing the "orbital" speed now continues to smoothly increase the eccentricity of the escape trajectory, and since the conic section with eccentricity greater than 1 is the hyperbola, the escape trajectory will be a hyperbola. $\endgroup$ – GenericHumanoid Dec 17 '16 at 0:01

The answer has to do with relationship between energy and eccentricity. $ e= \sqrt{1+ \frac{2\epsilon L^2}{G(m_1m_2)^2} }$ where $\epsilon $ is the specific orbital energy. The formula is valid for a gravitational force. For Hyperbolic orbits we need $e$ to be be greater than 1 and this means that $\epsilon $ i.e energy is positive. For parabolic orbits we need $e=1$ which means $\epsilon$ must equal 0.

Now when one shoots a particle with greater than escape velocity, it has a residual energy which is positive. This is what creates the hyperbolic orbit.

  • $\begingroup$ Thanks for the answer.can you please suggest a derivation of the formula you cited. $\endgroup$ – Pink Dec 22 '16 at 23:12
  • $\begingroup$ @navinstudent This is a result of the solution of the two body problem. Here is a link web.mit.edu/8.01t/www/materials/modules/guide17.pdf $\endgroup$ – Amara Dec 22 '16 at 23:38

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