Sufficiency condition for uniform circular motion under a gravitational force We say that an object follows a uniform circular motion around a center at $(0,0)$ if there exist a radius $R$ and angular velocity $\omega$ such that the motion admits parametrization
$$ \left\lbrace \begin{array}{l} x(t) = R \cos (\omega t) \\ y(t) = R \sin ( \omega t) \\ \end{array}\right.$$
or in polar coordinates
$$ \left\lbrace \begin{array}{l} r(t) = R \\ \theta(t) = \omega t \\ \end{array}\right.$$
Suppose that an object subject to only the gravitational force of the earth meets the following conditions:

*

*The initial distance from the object to the center of the earth is $r_0$.

*The initial velocity of the object is perpendicular to the radius vector.

*The initial velocity of the object has modulus $v_0$ satisfying
$$ m \frac{v_0^2}{r_0} = G \frac{M m}{r_0^2} \quad \text{or} \quad v_0 = \sqrt{\frac{GM}{r_0}} $$
where $m$ is the mass of the object, $M$ is the mass of earth and $G$ is the universal gravitational constant.

Are these conditions enough to claim that the object will follow a uniform circular motion around the earth? If so, why? If not, which condtion(s) has/have to be added and why?
 A: The best answer I found is based on the proof of Kepler's First Law. I am unsure if the force being proportional to the inverse square of the radius is a necessary requirement.
See https://faculty.etsu.edu/gardnerr/2110/notes-12E/c13s6.pdf
From equation (***) in page 8,
$$
GM \left(\frac{\vec{r}}{r} + \vec{e} \right) = \vec{\dot{r}} \times \vec{C} 
$$
Recall how the constant vector was defined: $\vec{C} = \vec{r} \times \vec{\dot{r}}$. By substituting and developing the cross product, one gets
$$
GM \left(\frac{\vec{r}}{r} + \vec{e} \right) = (\vec{\dot{r}} \cdot \vec{\dot{r}}) \vec{r} - (\vec{r} \cdot \vec{\dot{r}}) \vec{\dot{r}}
$$
or by replacing $\vec{v} = \vec{\dot{r}}$,
$$
GM \left(\frac{\vec{r}}{r} + \vec{e} \right) = (\vec{v} \cdot \vec{v}) \vec{r} - (\vec{r} \cdot \vec{v}) \vec{v}
$$
Evaluating this expression at $t=0$, we can deduce the value of the constant $\vec{e}$. Note that the assumptions imply that $\vec{v} \cdot \vec{v} = \frac{GM}{r}$ and $\vec{r} \cdot \vec{v} = 0$.
$$
GM \left(\frac{\vec{r}}{r} + \vec{e} \right) = \frac{GM}{r} \vec{r} \implies \vec{e} = \vec{0}
$$
Following the proof until (****), we obtain,
$$
r = \frac{C^2}{GM}
$$
which is constant. Hence, the object follows a circular motion.
The uniformity of the motion follows from the conservation of energy.
$$
r \text{ constant} \implies U \text{ constant} \implies K \text{ constant} \implies v \text{ constant}
$$
