## how does velocity affects the shape of orbit

suppose a mass m reduces its orbital velocity by 2 at some point in its circular orbit, but leaving its direction of travel unchanged at that moment, how does the orbit change? I know the new orbit is elliptical but i have no idea how to get to that conclusion.
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 Recognitions: Gold Member Science Advisor http://orbitsimulator.com/orbiter/orbit.gif The orbit becomes very elliptical with an eccentricity of 0.75 when circular velocity is halved. In the above link the picture shows a purple planet and a green planet orbiting a solar-mass red star. The purple planet is in a circular orbit 0.1 AU from the star. Its orbital speed is 94 km / s. The green planet's apihelion is also 0.1 AU. But at apihelion, its orbital velocity is exactly half (47 km / s) of the purple planet.
 Recognitions: Gold Member Homework Help Science Advisor Another way: $$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$$ $$\epsilon$$ is the specific energy per unit of mass. It determines the size of your orbit. It always winds up being negative. The closer to zero, the greater the specific energy (In fact, if it reaches zero, you have a parabolic escape orbit - greater than zero, a hyperbolic escape orbit). $$\mu$$is the geocentric gravitational constant (or the gravitational constant for the object you're orbiting). $$v$$ is velocity. $$r$$ is the radius of your current position. If the velocity decreases, the size of the orbit decreases. Your current position hasn't changed (assuming you instantaneously changed the velocity). Due to conservation of energy, you have to return to the same point that you made your maneuver at. If you started with a circular orbit, that means your new point is apogee. Any change in velocity made at perigee, apogee, or at any point in a circular orbit affects the point 180 degrees opposite the most, but does not affect the point you made the maneuver at (you have to return to the same point you made your maneuver at). The formula for determing the semi-major axis of your orbit is: $$a=-\frac{\mu}{2\epsilon}$$ Notice that all terms on the right are constants, except for $$\epsilon$$ The closer $$\epsilon$$ gets to zero, the bigger the orbit. The radius of apogee is found by: $$r_a=a(1+e)$$ with $$e$$ being your eccentricity. Wherever you were at in the circular orbit when you decreased the velocity is the apogee point. You have to pass through the maneuver point again. Decreasing the velocity decreased the size of the semi-major axis (a). So e has to increase in order to keep the radius of apogee constant.