Class 9 > Unit # 06:Gravitation > Orbital Velocity and Time Period of Satellites

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Gravitation - Orbital Velocity & Time Period of Satellite

Q. Derive an expression for Orbital Velocity of Satellite and prove that it is independent of mass of satellite.

Orbital Velocity

"The velocity required to keep a satellite in its circular orbit around a planet or the Earth is called Orbital Velocity ($v_o$)."

Figure: Satellite of mass $m$ revolving around the Earth of mass $M$ at orbital radius $r$.

Derivation

Consider the motion of a satellite revolving around the Earth:

• $m$ = Mass of the satellite
• $M$ = Mass of the Earth
• $R$ = Radius of the Earth
• $h$ = Altitude (height) of the satellite above the Earth's surface
• $r = R + h$ = Radius of the orbit

The necessary centripetal force required for circular motion is provided by the gravitational force of attraction between the Earth and the satellite:

$F_c = F_g$    --- (i)

We know the standard formulas for centripetal force and gravitational force:

$F_c = \frac{m v_o^2}{r}$    and    $F_g = G \frac{M m}{r^2}$

Substituting these values into equation (i):

$\frac{m v_o^2}{r} = G \frac{M m}{r^2}$

Multiplying both sides by $r$ and dividing by $m$:

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

Taking the square root on both sides gives the general expression for orbital velocity:

$v_o = \sqrt{\frac{G M}{r}}$    --- (ii)

Independence of Satellite Mass: Notice that the mass of the satellite '$m$' cancelled out during the derivation. This proves mathematically that the orbital velocity of a satellite is completely independent of its own mass.

Since $r = R + h$, we can also express orbital velocity in terms of Earth's radius and altitude:

$v_o = \sqrt{\frac{G M}{R + h}}$

Furthermore, since $g = \frac{G M}{R^2}$ (implying $G M = g R^2$), substituting this into the equation yields an alternative useful form:

$v_o = \sqrt{\frac{g R^2}{R + h}}$

Q. Derive an expression for Time Period of Satellite.

Time Period of Satellite

"The total time required for a satellite to complete one full revolution around the Earth in its orbit is called its Time Period ($T$)."

The time period can be calculated from the orbital circumference and orbital speed:

$T = \frac{\text{Circumference of Orbit}}{\text{Orbital Velocity}} = \frac{2\pi r}{v_o}$    --- (i)

Substituting the expression for orbital velocity ($v_o = \sqrt{\frac{GM}{r}}$) into equation (i):

$T = \frac{2\pi r}{\sqrt{\frac{G M}{r}}}$

Rearranging the terms by bringing $r$ inside the square root ($r = \sqrt{r^2}$):

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

Since the orbital radius is $r = R + h$, substituting this gives:

$T = 2\pi \sqrt{\frac{(R + h)^3}{G M}}$

Conclusion: This equation gives the final expression for the time period of a satellite orbiting around the Earth, proving that Newton's law of universal gravitation successfully describes and predicts satellite dynamics in space.

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