Earth Satellites & Orbital Velocity

What an Earth satellite is

An Earth satellite is any object that revolves around the Earth. NCERT notes that satellite motion is very similar to the motion of planets around the Sun, so Kepler's laws of planetary motion apply equally to satellites, and their orbits are circular or elliptical. The Moon is the Earth's only natural satellite, with a near-circular orbit and a period of roughly 27.3 days. Since 1957, artificial satellites have been launched for telecommunication, geophysics and meteorology.

For NEET the workhorse case is a satellite in a circular orbit. We take the orbital radius measured from the centre of the Earth as $r = R_E + h$, where $R_E$ is the Earth's radius and $h$ is the height above the surface. Two facts drive everything that follows: the orbit is circular (so the satellite needs centripetal force pointing inward) and gravity is the only force acting (so gravity must be that centripetal force).

Orbital velocity from the force balance

Let the satellite have mass $m$ and orbital speed $v$. Circular motion of radius $r$ requires a centripetal force directed toward the centre:

$$F_{\text{centripetal}} = \frac{m v^2}{r}, \qquad r = R_E + h.$$

This force is provided entirely by the gravitational pull of the Earth (mass $M_E$):

$$F_{\text{gravitation}} = \frac{G\, m\, M_E}{r^2}.$$

Equating the two and cancelling the satellite mass $m$ from both sides:

$$\frac{m v^2}{r} = \frac{G m M_E}{r^2} \;\Longrightarrow\; v_o = \sqrt{\frac{G M_E}{r}} = \sqrt{\frac{G M_E}{R_E + h}}.$$

This is the orbital velocity: the precise speed at which gravity is fully spent on bending the path into a circle, with nothing left over. Too slow and the satellite spirals down; too fast and the orbit opens into an ellipse. The cancellation of $m$ is the single most important feature — the orbital velocity is independent of the satellite's mass.

The near-Earth value ≈ 7.9 km/s

Using $g = GM_E/R_E^{\,2}$, we can replace $GM_E$ by $gR_E^{\,2}$ and write the orbital velocity in a form NEET prefers:

$$v_o = \sqrt{\frac{G M_E}{R_E + h}} = \sqrt{\frac{g R_E^{\,2}}{R_E + h}}.$$

For a satellite very close to the surface, $h \ll R_E$, so $h$ can be neglected against $R_E$. The orbital velocity then simplifies to

$$v_o(h=0) = \sqrt{\frac{G M_E}{R_E}} = \sqrt{g R_E}.$$

Substituting $g \approx 9.8~\text{m s}^{-2}$ and $R_E \approx 6400~\text{km} = 6.4\times 10^6~\text{m}$:

$$v_o = \sqrt{9.8 \times 6.4\times 10^6} \approx 7.9\times 10^3~\text{m s}^{-1} = 7.9~\text{km s}^{-1}.$$

This is the first cosmic speed — the speed of a satellite skimming just above the surface. Notice the link to escape speed: at the same radius, $v_e = \sqrt{2GM/R} = \sqrt{2}\,v_o$, which is why escape speed near the surface (about $11.2~\text{km s}^{-1}$) is exactly $\sqrt{2}$ times the near-Earth orbital speed. See Escape Speed for the full treatment.

Time period and Kepler's law of periods

In one orbit the satellite covers the circumference $2\pi r = 2\pi(R_E+h)$ at speed $v_o$, so the time period is

$$T = \frac{2\pi (R_E + h)}{v_o} = 2\pi (R_E + h)\sqrt{\frac{R_E+h}{G M_E}} = 2\pi\sqrt{\frac{(R_E+h)^3}{G M_E}}.$$

Written compactly with $r = R_E + h$, this is

$$T = 2\pi\sqrt{\frac{r^3}{G M_E}}.$$

Squaring both sides gives

$$T^2 = \frac{4\pi^2}{G M_E}\, r^3 = k\,(R_E + h)^3, \qquad k = \frac{4\pi^2}{G M_E}.$$

This is exactly Kepler's law of periods applied to Earth satellites: $T^2 \propto r^3$. The same proportionality that governs planets around the Sun governs satellites around the Earth — only the central mass changes.

For a satellite very close to the surface, set $h \approx 0$ so $r \approx R_E$, and using $GM_E = gR_E^{\,2}$:

$$T_0 = 2\pi\sqrt{\frac{R_E^{\,3}}{G M_E}} = 2\pi\sqrt{\frac{R_E}{g}}.$$

With $g \approx 9.8~\text{m s}^{-2}$ and $R_E = 6.4\times 10^6~\text{m}$, NCERT gives $T_0 \approx 85~\text{minutes}$ — the shortest possible period of any Earth satellite.

How v and T change with height

The two formulas pull in opposite directions as the satellite climbs. Because $v_o = \sqrt{GM/r}$, increasing $r$ decreases the orbital speed. Because $T = 2\pi\sqrt{r^3/GM}$, increasing $r$ increases the period — and steeply, since $T \propto r^{3/2}$. A higher satellite is slower, and it takes longer per orbit both because it is slower and because the loop is bigger.

Geostationary vs polar satellites

Two artificial-satellite families dominate practical use, and NEET tests the contrast between them. A geostationary satellite is engineered so its period equals one rotation of the Earth — exactly 24 hours — and it orbits in the equatorial plane in the same sense as the Earth spins. With $T = 24~\text{h}$, Kepler's law of periods fixes a unique radius, which places it at an altitude of about $36{,}000~\text{km}$. Seen from the ground it appears to hang motionless over one point on the equator, which is why it is ideal for fixed communication and broadcast links.

A polar satellite orbits at much lower altitude, passing over (or near) the poles. Its period is short, and because the Earth rotates beneath it, successive orbits scan fresh strips of the surface — making it well suited to remote sensing and weather monitoring.

Worked examples