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The total energy of $m_1$, orbiting around a larger mass, $m_2$, is equal to

$E =$ $-Gm_1m_2\over2r$

This is from adding kinetic energy and potential energy:

$K + U$ = $1\over2$ $mv^2$ + $-Gm_1m_2\over r$

When $v$ =$\sqrt {Gm_2\over r}$

However, since this $v$ is only the minimum required velocity for $m_1$ to orbit $m_2$, can the actual energy be higher?

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As explained in Lawrence B. Cromwell's answer the equations in your question apply circular orbits.

With that in mind the answer to your question is no.

The equation $v =\sqrt {Gm_2\over r}$ does not give the minimum speed for a circular orbit - it gives the only speed for a circular orbit at that radius. If a satellite is at a radius $r$ with a smaller or larger speed its orbit will be elliptical.

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Your question is a bit odd. The velocity of a particle may not be constant if the orbit is elliptical. The kinetic energy of a particle is $K~=~\frac{1}{2}mv^2$ for $\vec v~=~d\vec r/dt$. However, this velocity is in a plane with the variables $(r,~\theta)$, so chain rule tells us that $$ \frac{d\vec r}{dt}~=~\hat r\frac{\partial r}{\partial t}~+~\hat\theta r\frac{\partial\theta}{\partial t}~=~\hat r\dot r~+~\hat\theta r\dot\theta $$ for $\hat r$ and $\hat\theta$ unit vectors. The kinetic energy is then $$ K~=~\frac{1}{2}mv^2~=~\frac{1}{2}m\left(\dot r^2~+~r^2\dot\theta^2\right). $$ This is the kinetic energy of a particle in the polar coordinate plane.

Now consider the total energy or the Lagrangian. The potential energy for a small mass $m$ moving in a gravitation field of a large mass $M$ is $U~=~-GMm/r$. The total energy is then $E~=~K~+~U$ and the Lagrangian is $L~=~K~-~U$. I will use the Lagrangian, and the equations of motion are found by running them through the Euler-Lagrange equations $$ \frac{\partial L}{\partial r}~-~\frac{d}{dt}\left(\frac{\partial L}{\partial\dot r}\right)~=~0 $$ $$ \frac{\partial L}{\partial\theta}~-~\frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta}\right)~=~0 $$ The first of these gives the dynamical equation $$ m\ddot r~-~mr\dot\theta^2~+~\frac{GMm}{r^2} = 0 $$ and the second gives $$ mr^2\ddot\theta~=~0. $$ The angular momentum of the orbiting mass is $\vec L~=~\vec r\times \vec p$ with magnitude $mr^2\dot\theta$, so the second of these equations is equal to $d\vec r/dt~=~0$. This means angular momentum is conserved. Also the first equation becomes $$ m\ddot r~-~\frac{L^2}{mr^3}~+~\frac{GMm}{r^2} = 0. $$ If the motion is circular the radius is constant and so $\ddot r~=~0$ and this reduces to the standard formula in first year physics books with $\dot\theta~=~\omega$ which is constant.

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