Class 11 > Unit # 04: Rotational & Circular Motion > Orbital Velocity & Time Period of Satellite

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Unit No. 4 Rotational and Circular Motion - Orbital Velocity and Satellite Time Period

Define Orbital Velocity. Derive expressions for the Orbital speed and time period of satellite.

Orbital Velocity

"Orbital velocity is the speed required for an object to achieve and maintain a stable orbit around a heavenly body, such as a planet, moon, or star."

Figure: Satellite in circular orbit around a central planet.

Expression for Orbital Speed

Suppose a satellite of mass $m$ is orbiting a central heavenly body of mass $M$ at a radius $r$ from its center. When the satellite moves in a circular orbit, the necessary centripetal force acting upon it is provided by the gravitational attraction between the satellite and the central body.

1. Centripetal Force:

$F_c = \frac{mv^2}{r} \quad \text{--- (i)}$

2. Gravitational Force:

$F_g = \frac{G Mm}{r^2} \quad \text{--- (ii)}$

Equating the centripetal force to the gravitational force ($F_c = F_g$):

$\frac{mv^2}{r} = \frac{G Mm}{r^2}$

Canceling the mass of the satellite ($m$) on both sides and multiplying both sides by $r$:

$v^2 = \frac{G M}{r}$

Taking the square root on both sides yields the formula for orbital speed:

$v = \sqrt{\frac{GM}{r}}$

Time Period and Orbital Radius

The time period ($T$) of a satellite is the total time taken to complete one full revolution around the central body along the circumference of its orbit ($S = 2\pi r$).

Using the linear speed formula ($v = \frac{\text{Distance}}{\text{Time}}$):

$v = \frac{2\pi r}{T} \implies T = \frac{2\pi r}{v}$

Substituting the expression for orbital speed $v = \sqrt{\frac{GM}{r}}$ into the time period equation:

$T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}}$

Simplifying the complex fraction by bringing $r$ inside the square root ($r = \sqrt{r^2}$):

$T = 2\pi r \cdot \sqrt{\frac{r}{GM}} = 2\pi \sqrt{\frac{r^2 \cdot r}{GM}}$

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

Conclusion & Kepler's Third Law:
Squaring both sides of the time period equation:

$T^2 = \frac{4\pi^2}{GM} r^3$

Since $\frac{4\pi^2}{GM}$ is a constant for a given central body, this equation shows that $T^2 \propto r^3$. This mathematical relationship demonstrates that the square of the orbital period is directly proportional to the cube of the orbital radius, which is precisely Kepler's Third Law of Planetary Motion.

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