Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am very confused about orbital velocity for satellite launched from earth. It is states that: $$ v=\sqrt{\frac{GM}{r}} $$ Some says that it is equal to 8000 m/s and some say 3.1 km/s. Please remove my doubt.

share|improve this question
add comment

2 Answers 2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.

share|improve this answer
add comment

The mass $M$ in the equation for orbital velocity would be equal to the mass of the Earth (assuming that the mass of the satellite can be neglected) and thus constant. The gravitational constant, $G$, like the name suggests is constant as well. But the radius of the (circular) orbit, $r$, can vary. So by using different altitudes at which the satellite would orbit Earth would yield different orbital velocities.

The product of the gravitational constant and the mass of a celestial body (like Earth) is often called the gravitational parameter and can be measured more accurately than the each of them separately. The gravitational parameter of Earth is equal to: $$ \mu_{\oplus}=3.986004418\times 10^{14}\ m^3s^{−2} $$

If you would like to know which velocity would be needed to have an satellite in LEO (low Earth orbit) you could use a radius slightly bigger than Earths equatorial radius, $r_{\oplus}=6.3781\times10^6\ m$, say at $300\ km$ above the surface. Plugging these numbers into the equation yields an orbital velocity of: $$ v_{LEO}=\sqrt{\frac{\mu_{\oplus}}{r_{\oplus}+h}}=7725.8\ ms^{-1} $$ This is close to your first value. If you would look at the equation you can see that the orbital velocity will increase the decreasing the radius. So to get closer to $8000\ ms^{-1}$ you would have to decrease the altitude even more. However I suspect that it is rounded up, since the altitude required to have that orbital velocity will put you underneath Earths surface.

And like Nathaniel said, $3.1\ km\ s^{-1}$ is the orbital velocity of a geostationary orbit, which has a radius of roughly $42164\ km$. A quick check yields an orbital velocity of $3.0747\ km\ s^{-1}$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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