For example, I don't understand why the speed of a satellite moving in an orbit of radius 'a' around the Earth must be equal to $\sqrt{GM/a}$ If I release a particle in outer space with a velocity perpendicular to the line joining the particle and Earth but magnitude not equal to $\sqrt{GM/a}$ what will happen? For al I know, the acceleration will still be perpendicular to its velocity at all instants. And, an always perpendicular acceleration is only capable of changing the direction of velocity continuously. Then, wouldn't the particle still rotate even if its speed not equal to the one given by F=$mv^2/r$?

  • 1
    $\begingroup$ Acceleration will only be perpendicular to the velocity only at the moment you released the particle (if the velocity is not equal to $\sqrt {GM/a} $). $\endgroup$
    – cobra121
    Dec 26 '16 at 12:39
  • $\begingroup$ Can you please explain why the acceleration won't be perpendicular at all other instants? $\endgroup$
    – Dove
    Dec 26 '16 at 13:37
  • $\begingroup$ For a particle having a velocity $v$ , the condition for it to move in a circle of radius $r $ is the force on it should be perpendicular to it and equal to $mv^2/r $ $\endgroup$
    – cobra121
    Dec 26 '16 at 14:04
  • $\begingroup$ Lets assume the velocity is $x\sqrt {GM/a} $ where x is not 1. Here the gravitational force is $GMm/a^2$ and the force required is $mv^2/a= x^2GMm/a^2$ which are not equal. $\endgroup$
    – cobra121
    Dec 26 '16 at 14:09
  • $\begingroup$ If $x>1$ then this force is not enough to curl the particle to move in a circle. Thus it will leave the circular orbit the moment you release it (not tangentially but some trajectory between the tangent and the circular path). In that trajectory the force will still be directed towards the earth, but the velocity will have a component in the line joining the particle and the earth. You can similarly reason for $x <1$. The force is more than what is required and the particle will curl more than required for a circle. $\endgroup$
    – cobra121
    Dec 26 '16 at 14:20

You agree that a satellite with speed $\sqrt{\frac{GM}{a}}$ will undergo a circular orbit of radius $a$.

Now suppose that you take a satellite up to that distance $a$ and give it a tangential speed greater than $\sqrt{\frac{GM}{a}}$.
The satellite will start executing a curved path of smaller curvature, larger radius, than if the speed had been $\sqrt{\frac{GM}{a}}$ because the gravitational attraction of the Earth on the satellite is not large enough for the path to be of radius $a$.
The satellite would therefore move further from the Earth gaining gravitational potential energy but at the same time losing kinetic energy i.e. moving at a slower speed.
Remember that the satellite has a mass and if no force was acting on it the satellite would travel in a straight line.
With the higher speed the gravitational attraction of the Earth cannot pull the satellite enough to make the satellite execute a circular path so it goes along a less curved, elliptical path.

What happens next depends on the speed that you give the satellite.
Below a certain value the satellite would execute an elliptical orbit about the Earth i.e. the distance between the Earth and the satellite would vary as would the speed of the satellite.

At a particular speed, the escape speed, the satellite would execute a parabolic path and escape from the Earth.
Above the escape speed the path of the satellite would be hyperbolic.


When you have a particle orbiting circularly around Earth, you can easily write the motion laws for two directions: radial direction and tangential direction. In tangential direction, you have uniform motion with constant speed. In radial direction, due to the motion of the particle, in order to have a circular motion, you should have a radial (i.e., centripetal) force with modulus: $$F=\frac{m v^2}{ R}$$ with $m$ mass of the particle, $v$ its speed and $R$ radius of the circular orbit. In this case, this force is gravity and, from the law of Gravitational attraction, you can get: $$ F = \frac{G mM}{ R^2} $$ with $G$ constant, $M$ mass of the Earth and $R$ distance between them. If you put these two expressions equal, you can get the right speed for a satellite circularly orbiting the Earth. $$ v =\sqrt{\frac{GM}{R}} $$ If you have a satellite's speed greater than the one calculated for a circular motion, you can see that in the radial direction the required centripetal force is greater than the actual Gravitational one. This means that gravity cannot keep the satellite orbiting around the Earth, you have a force term in radial direction and it will cause the satellite to go further from the Earth while it is rotating around it. On the contrary, if the speed is less than required, you still have a force in radial direction and it will cause the satellite to fall toward Earth's surface, while orbiting around it, and the orbit will become elliptical.

  • 1
    $\begingroup$ Nice answer. I hope you don't mind that I inserted a few clarifications. $\endgroup$
    – Bill N
    Dec 26 '16 at 15:01
  • $\begingroup$ @BillN Thank you for your help in providing a good answer!!! $\endgroup$
    – JackI
    Dec 26 '16 at 15:06

For example, I don't understand why the speed of a satellite moving in an orbit of radius 'a' around the Earth must be equal to $\sqrt{GM/a}.$

For the satellite to orbit at radius $a$, the Earth's gravitational field must exert a centripetal force $F_c$:


This force is the gravitational force, so:



$$\frac{mv^2}{a}=G\frac{mM}{a^2}$$ $$\implies v=\sqrt\frac{GM}{a}\tag{1}$$

If the 'launch speed' (as you defined it) is higher, the satellite will move to a different orbit, until $(1)$ is satisfied again.


So here we assume, as per the OP, that $v_0 \neq \sqrt\frac{GM}{a}$.

The 'final orbit' $r$ is found from energy conservation (we assume gravity to be the only external force). At launch the total energy $T$ is ($v_0$ is launch velocity):


In final orbit the total energy $T$ is:


From the identity, $r$ can be calculated:

$$-\frac{GM}{2r}=\frac{v_0^2}{2}-\frac{GM}{a}$$ $$\frac{GM}{r}=\frac{2GM-av_0^2}{a}$$ $$\boxed{r=\frac{aGM}{2GM-av_0^2}}$$

If $v_0 > \sqrt\frac{GM}{a}$, then $r>a$, so the satellite moves to a higher orbit. It slows down because some of its kinetic energy is converted to potential energy.

If $v_0 < \sqrt\frac{GM}{a}$, then $r<a$, so the satellite moves to a higher orbit. It will speed up because some of its potential energy is converted to kinetic energy.

  • $\begingroup$ First, you said that if the launch speed is higher, the satellite will move to a higher orbit. But, according to the equation when v is more then a should be less. Second, if I provide the satellite an initial velocity perpendicular to the gravitational force, then what would provide the satelite the force requored to move to a higher orbit. There is no radial component of initial velocity and the only force on the satelite os towards the center. Then, how will it move to a higher orbit? $\endgroup$
    – Dove
    Dec 26 '16 at 13:11
  • $\begingroup$ And,I'm not talking about launching the particle from earth with some initial velocity. I'm talking about taking it to some orbit of radius 'a', then providing it an initial velocity perpendicular to the radius but of magnitude not equal to $\sqrt{GM/a}$. Then, what will provide the satellite the force necessary to move to an orbit whose radi $\endgroup$
    – Dove
    Dec 26 '16 at 13:31
  • $\begingroup$ I've edited the post slightly. $\endgroup$
    – Gert
    Dec 26 '16 at 15:13

I will compare circular motion in the case of a rotating frame and with the case of the Schwarzshild metric. In this general relativistic context we can examine what is meant by the centripetal force in this context for a particle in a rotating frame and a particle in a circular orbit around a gravitating mass. We can compare the two and see what is modified by general relativity, and what is interpreted by centripetal acceleration.

The metric for a rotating coordinate system with $d\phi~\rightarrow~d\phi~+~\omega dt$ is $$ ds^2~=~A(\omega,~r)\left(cdt~-~\frac{\omega r^2}{c^2A(\omega,~r)}d\phi\right)^2~-~dr^2~-~\frac{r^2}{A(\omega,~r)}d\phi^2~-~dz^2, $$ $$ A(\omega,~r)~=~(1~-~\omega^2r^2/c^2) $$ gives the Christoffel symbols $$ \Gamma^r_{tt}~=~\omega^2 r,~\Gamma^\phi_{rr}~=~-\frac{\omega}{r},~\Gamma^\phi_{r\phi}~=~-\frac{1}{r},~\Gamma^r_{\phi\phi}~=~r. $$ For circular motion we can set $\Gamma^\phi_{rr}~=~\Gamma^\phi_{r\phi}~=~0$. The geodesic equation of interest is of the form $$ \frac{d^2r}{ds^2}~+~\Gamma^r_{\phi\phi}\left(\frac{d\phi}{ds}\right)^2~=~0, $$ or equivalently $$ \frac{d^2r}{ds^2}~+~\omega^2 r\left(\frac{dt}{ds}\right)^2~+~r\left(\frac{d\phi}{ds}\right)^2~=~0. $$ This is similar to centripetal accleration.

To make the connection to Newtonian physics let us transform this to acceleration in the standard coordinates of an observer. We then have $$ \frac{d^2r}{ds^2}~=~\left(\frac{d^2r}{dt^2}\right)\left(\frac{dt}{ds}\right)^2, $$ which employs $dr/dt~=~dz/dt~=~0$ for circular motion. We use the metric with $dr~=~0$ $$ ds^2~=~A(\omega,~r)\left(cdt~-~\frac{\omega r^2}{c^2A(\omega,~r)}d\phi\right)^2~-~\frac{r^2}{A(\omega,~r)}d\phi^2, $$ so the term $dt/ds$ is seen in $$ \left(\frac{ds}{dt}\right)^2~=~A(\omega,~r)\left(1~-~\frac{\omega^2r^2}{A(\omega,~r)}\right)~-~\frac{\omega^2r^2}{A(\omega,~r)}, $$ so that $dt/ds$ is a form of Lorentz gamma factor $$ \gamma(\omega)~\dot=~\frac{dt}{ds}~=~\frac{1}{\sqrt{A(\omega,~r)\left(1~-~\frac{\omega^2r^2}{A(\omega,~r)}\right)~-~\frac{\omega^2r^2}{A(\omega,~r)}}}. $$ This then gives us a gamma factor modified form of the centripetal acceleration. Since it is evident the modified gamma factor divides out on both side this is centripetal acceleration for a particle fixed to a rotating frame. The question is then whether this applies to gravitation.

For gravitation we turn to the Schwarzschild metric $$ ds^2~=~(1~-~2m/r)dt^2~-~(1~-~2m/r)^{-1}dr^2~-~r^2(d\theta^2~+~sin^2\theta d\phi^2)~with~m~=~GM/c^2, $$ where for a circular orbit we have $dr~=~d\theta=-~0$ and $\theta~=~\pi/2$ so that $$ ds^2~=~(1~-~2m/r)dt^2~-~r^2d\phi^2. $$ Dividing through by $dt^2$ and letting $\omega~=~d\phi/dt$ gives $$ ds^2~=~[(1~-~2m/r)~-~r^2\omega^2]dt^2, $$ which gives a similar gamma factor $$ \gamma_m(\omega)~=~\frac{1}{\sqrt{1~-~2m/r~-~r^2\omega^2}}. $$ The Christoffel symbol relevant for calculation is $$ \Gamma^r_{tt}~=~\frac{m(r~-~2m)}{r^3}, $$ so that $$ \frac{d^2r}{ds^2}~+~\frac{m(r~-~2m)}{r^3}\left(\frac{dt}{ds}\right)^2~=~0. $$ It is evident that the $\gamma_m(\omega)$ divides out and this leaves the dynamical equation $$ \frac{d^2r}{dt^2}~+~\frac{m(r~-~2m)}{r^3}~=~0. $$ For $r~>>~2m$ this recovers Newton's second law with gravity.

It appears that in standard coordinates centripetal acceleration is the same. What is modified by general relativity is the nature of gravitation as a force interpreted in standard coordinates.


For al I know, the acceleration will still be perpendicular to its velocity at all instants. And, an always perpendicular acceleration is only capable of changing the direction of velocity continuously.

Well, let's see. For convenience, stipulate that at $t=0$, the particle is located at $r=R, \theta=\pi/2, \phi=0$ with velocity

$$\vec v(0)=v\;\hat{\boldsymbol{y}}$$

and acceleration

$$\vec a(0)=-\frac{GM}{R^2}\;\hat{\boldsymbol{x}}$$

The velocity at the next instant is then $$\vec v(0 + dt)=-\frac{GM}{R^2}dt\;\hat{\boldsymbol{x}} + v\;\hat{\boldsymbol{y}}$$

while the acceleration is

$$\vec a(0 + dt)=-\frac{GM}{R^2}\left(\hat{\boldsymbol{x}} + \frac{v}{R}\,dt\;\hat{\boldsymbol{y}}\right)$$

and so the dot product of these velocity and acceleration vectors is

$$\vec v(0 + dt)\cdot \vec a(0 + dt)=-\frac{GM}{R^2}\left[-\frac{GM}{R^2}+\frac{v^2}{R}\right]dt$$

which is zero only if




Thus, it isn't the case that the acceleration will still be perpendicular to the velocity at all instants for arbitrary $v$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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