Find orbital velocity and time period of satellite at height h above Earth

Question

Derive expressions for the orbital velocity and time period of a satellite revolving at height $h$ above Earth’s surface. Let Earth’s radius be $R$ and mass $M$ , and gravitational constant be $G$ .

Solution — Step by Step

A satellite at height $h$ above Earth’s surface revolves at a distance:

r = R + h

from Earth’s centre. This $r$ is the orbital radius — the distance from Earth’s centre (not from the surface).

For circular orbital motion, the gravitational force provides the centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r}

The satellite mass $m$ cancels:

\frac{GM}{r^2} = \frac{v^2}{r}

v^2 = \frac{GM}{r} = \frac{GM}{R+h}

\boxed{v_0 = \sqrt{\frac{GM}{R+h}}}

We know $g = \frac{GM}{R^2}$ , so $GM = gR^2$ .

Substituting:

v_0 = \sqrt{\frac{gR^2}{R+h}} = R\sqrt{\frac{g}{R+h}}

For a satellite close to Earth’s surface ( $h \ll R$ , so $R + h \approx R$ ):

v_0 \approx \sqrt{gR} \approx \sqrt{10 \times 6.4 \times 10^6} \approx 8 \text{ km/s}

Time period = circumference ÷ orbital speed:

T = \frac{2\pi r}{v_0} = \frac{2\pi(R+h)}{\sqrt{\frac{GM}{R+h}}}

T = 2\pi(R+h) \sqrt{\frac{R+h}{GM}}

\boxed{T = 2\pi\sqrt{\frac{(R+h)^3}{GM}}}

This is Kepler’s Third Law for satellite orbits: $T^2 \propto r^3$ .

Why This Works

The key insight is that for circular orbital motion, gravity provides the entire centripetal force. There’s no need for any engine — the satellite “falls” continuously but keeps missing Earth because it’s moving so fast horizontally.

Notice that orbital velocity depends only on $r$ (the orbital radius), not on the mass of the satellite. A 1 kg probe and the International Space Station at the same height orbit at exactly the same speed. This is a direct consequence of the equivalence principle — gravity accelerates all masses equally.

Common Mistake

The most common error is using $r = h$ (height) instead of $r = R + h$ (orbital radius from Earth’s centre). The gravitational force law uses distance from Earth’s centre, not from the surface. If a satellite is 400 km above Earth (as ISS is), the orbital radius is $6400 + 400 = 6800$ km, not 400 km. Using $r = h$ drastically overestimates velocity and underestimates period.

For JEE: relate these formulas using $GM = gR^2$ . Then $v_0 = \sqrt{gR^2/(R+h)}$ and $T = \frac{2\pi}{R}\sqrt{\frac{(R+h)^3}{g}}$ . JEE often asks you to find $v_0$ when $g$ and $R$ are given rather than $G$ and $M$ — the second form is more convenient.

Question

Solution — Step by Step

Why This Works

Common Mistake

Try These Next