Time Period of Satellite — Revision Notes

⚡ 30-Second Revision

Definition: — Time for one complete orbit.
Formula: — $T = 2\pi \sqrt{\frac{r^3}{GM}}$
Orbital Radius: — $r = R_E + h$
Dependencies:

- $T^2 \propto r^3$ (Kepler's Third Law) - $T \propto 1/\sqrt{M}$

Independence: — $T$ is independent of satellite's mass ($m$).
Geostationary Satellite: — $T = 24,\text{hours}$, fixed position relative to Earth's surface.

2-Minute Revision

The time period of a satellite is the time it takes to complete one full revolution around its central body. It's derived by equating the gravitational force ($G M m / r^2$) to the centripetal force ($m v^2 / r$), and then substituting orbital velocity ($v = 2\pi r / T$).

The resulting formula is $T = 2\pi \sqrt{\frac{r^3}{GM}}$. Key takeaways: the time period depends on the mass of the central body ($M$) and the orbital radius ($r$), but *not* on the satellite's own mass ($m$).

According to Kepler's Third Law, $T^2$ is directly proportional to $r^3$. Remember that $r$ is the distance from the center of the central body, so if height above surface $h$ is given, $r = R_E + h$.

A geostationary satellite has a time period of 24 hours, matching Earth's rotation, making it appear stationary from the ground.

5-Minute Revision

To thoroughly revise the time period of a satellite, start with its fundamental definition: the time required for one complete orbit. The derivation of its formula is crucial. We begin by balancing the gravitational force ($F_g = G M m / r^2$) with the centripetal force ($F_c = m v^2 / r$) needed for circular motion.

Equating them, $G M m / r^2 = m v^2 / r$, which simplifies to $v^2 = G M / r$. Next, recall that orbital velocity $v$ is the circumference divided by the time period, so $v = 2\pi r / T$. Substituting this into the previous equation gives $(2\pi r / T)^2 = G M / r$, leading to $4\pi^2 r^2 / T^2 = G M / r$.

Rearranging for $T^2$ yields $T^2 = 4\pi^2 r^3 / (G M)$, and finally, $T = 2\pi \sqrt{\frac{r^3}{GM}}$.

Key Points to Remember:

1
Independence from Satellite Mass: — Notice that the satellite's mass ($m$) cancels out during the derivation. This means $T$ is independent of $m$, a common NEET trap.
2
Dependencies: — $T$ depends on the mass of the central body ($M$) and the orbital radius ($r$). Specifically, $T^2 \propto r^3$ (Kepler's Third Law) and $T \propto 1/\sqrt{M}$.
3
Orbital Radius vs. Height: — $r$ is the distance from the center of the central body. If given height $h$ above the surface, $r = R_E + h$.
4
Geostationary Satellites: — These are special cases with $T = 24,\text{hours}$, orbiting at a specific radius (approx. $42,164,\text{km}$ from Earth's center) above the equator, appearing stationary.

Worked Mini-Example: If a satellite orbits at a radius of $R$ and has a period $T$, what is the period of a satellite orbiting at $9R$? Using $T^2 \propto r^3$, we have $\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$. So, $\frac{T_2^2}{T^2} = \frac{(9R)^3}{R^3} = 9^3 = 729$. Thus, $T_2^2 = 729T^2$, and $T_2 = \sqrt{729}T = 27T$. This demonstrates the strong dependence of $T$ on $r$.

Prelims Revision Notes

1
Definition: — Time taken for a satellite to complete one revolution around its central body.
2
Fundamental Formula: — $T = 2\pi \sqrt{\frac{r^3}{GM}}$

* $T$: Time period (seconds) * $r$: Orbital radius (meters) = $R_{\text{central body}} + h_{\text{height above surface}}$ * $G$: Universal Gravitational Constant ($6.674 \times 10^{-11},\text{N m}^2/\text{kg}^2$) * $M$: Mass of the central body (kilograms)

1
Independence from Satellite Mass: — The time period $T$ is independent of the satellite's own mass ($m$). This is a critical conceptual point for NEET.
2
Kepler's Third Law: — $T^2 \propto r^3$. This proportionality is vital for comparative problems. If $r$ increases, $T$ increases significantly.

* Example: If $r$ becomes $2r$, $T$ becomes $2^{3/2}T = 2\sqrt{2}T \approx 2.828T$.

1
Dependency on Central Body Mass: — $T \propto 1/\sqrt{M}$. If the central body's mass increases, the time period decreases.
2
Relationship with Orbital Velocity: — $v = \frac{2\pi r}{T}$. Therefore, $T = \frac{2\pi r}{v}$.
3
Geostationary Satellite:

* Time period $T = 24,\text{hours}$. * Orbits above the equator. * Appears stationary from Earth's surface. * Orbital radius $r \approx 42,164,\text{km}$ from Earth's center (or $h \approx 35,786,\text{km}$ above surface).

1
Units: — Always use SI units for calculations (meters, kilograms, seconds).

Vyyuha Quick Recall

Three Radii Get Massive Time: $T^2 \propto r^3 / GM$. (Helps remember $T^2$ is proportional to $r^3$ and inversely to $GM$). Also, Mass of Satellite Not Important (MSNI) for period.