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Suppose we have an object orbiting the Earth with a circular orbit of radius $R$. We then change the velocity by an angle $\alpha$, without changing the speed. The textbook then asks what orbit will the object describe.

My guess is an ellipse, but of course I want to show it. The initial angular momentum is $L_0 = mRv$ and the final one is $L = mRv\cos(\alpha) = L_0\cos(\alpha)$.

I want to study the initial and final energies in order to deduce the path of the orbit.

The initial energy of the particle is : $$E_0 = T_0 + U_{eff}^0 = \frac{1}{2}mv_0^2 + \frac{L_0^2}{2mR^2} + \frac{K}{R^2}$$ And the final one is : $$E = T + U_{eff} = T_0 + \frac{L_0^2\cos^2(\alpha)}{2mr^2} + \frac{K}{r^2}$$

Now the textbook says that the mechanical energy of the particle does not change, since $E = T_0 + \frac{K}{r} = T + \frac{K}{r}$, but isn't the mechanical energy of the particle $E = T + {U_{eff}}$?

In that case, the mechanical energy of the particle does change, so the textbook would be wrong. If not, can someone explain it to me?

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2 Answers 2

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You are correct with your guess of ellipse. Perhaps the argument is most directly shown using vector addition. For pure circular motion, the centripetal acceleration will always be perpendicular ($90^\circ$) to the tangential acceleration of the object. If we change the velocity direction by angle $\alpha$, there can no longer be circular motion, as we have lost orthogonality at that point.

The only other path that can be physically taken if the magnitude of velocity is kept the same, is that of an ellipse. But let's dive deeper...

According to Wikipedia, the mechanical energy of an object in orbit is given by the following, where $U$ is gravitational potential, and $K$ is kinetic energy. $$E_{mech} = U + K = -G\frac{Mm}{r}+\frac{1}{2}mv^2$$

Which for circular motion (where $G\frac{Mm}{r^2}=\frac{mv^2}{r}$) would reduce to $$E_{mech}=-G\frac{Mm}{2r}$$

But when we look at the first form, where $G\frac{Mm}{r^2}\neq\frac{mv^2}{r}$, we are lead to the vis-viva equation, otherwise known as the orbital energy invariance law, where assuming that $M\gg m$, we obtain $$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$

This expression directly results from conservation of mechanical energy and is derived from principles of elliptic orbits, specifically the eccentricity of the ellipse $\mathcal{E}$.

Specific total energy does not change through the orbit ($U+K$ is constant), and total angular momentum can be stated as $$L=mb\sqrt{\frac{GM}{a}}$$ Where $a$ and $b$ are the elliptical semimajor and semiminor axes respectively.

So in closing, for the scenario you've given, we must use the gravitational potential for our mechanical energy - where which we cannot just slap on a $\cos(\alpha)$ to in order to obtain a correct expression for $U$.

If the object changes with angle $\alpha$, it will be forced in to a situation where that angle will then define an elliptical eccentricity $\mathcal{E}$, which describes our orbit.

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Your textbook's definition of mechanical energy is defined in an inertial frame that is at rest with respect to the Earth. Your definition with an effective potential energy ($U_{eff}$) is defined in rotating reference frame that keeps the orbiting object motionless. Either one can be useful on different situations. The effective potential energy is useful for small perturbations from a circular orbit since the angular velocity is constant. In this problem, since the angle of deflection is not specified to be small, it's probably better to use the potential energy from the inertial reference frame. The object will have an elliptical orbit that changes speed as it gets closer and farther away from the Earth.

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