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?

  • $\begingroup$ I think these conditions are enough to claim that it will be a uniform circular motion because initial velocity is perpendicular to radius vector and since the Force vector will always be towards the center, it will remain perpendicular to the radius vector $\endgroup$ Commented Sep 22, 2021 at 21:05
  • $\begingroup$ @QuantumOscillator The conditions are enough but it is not so clear why velocity will remain perpendicular to the radius vector. Bear in mind that, if you modify the starting velocity the slightest bit, the orbit will be elliptical and not circular. The initial position will have perpendicular velocity and radius because the object will be in the apogee or perigee. $\endgroup$
    – mendus
    Commented Oct 13, 2021 at 9:39
  • $\begingroup$ The given condition does not apply at the apogee or perigee of an elliptical orbit. $\endgroup$
    – R.W. Bird
    Commented Oct 13, 2021 at 14:36
  • $\begingroup$ @R.W.Bird The first two conditions do. The last one doesn't, unless the ellipse is actually a circle and all points can be considered both apogee and perigee. I believe this is what I mention in my previous comment. $\endgroup$
    – mendus
    Commented Oct 14, 2021 at 8:13

1 Answer 1


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. $$ r \text{ constant} \implies U \text{ constant} \implies K \text{ constant} \implies v \text{ constant} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.