Physics and Mathematics
Time Period of a Satellite
1. Statement of the Concept
The time period of a satellite is the time taken by the satellite to complete one full revolution around the planet in its orbit.
It is denoted by [T] and depends on the radius of the orbit (r) and the mass of the planet (M) around which it revolves.
2. Explanation and Mathematical Derivation
Consider a satellite of mass [m] revolving around the Earth in a circular orbit of radius [r] with orbital velocity [v].
From Newton’s law of gravitation and centripetal force balance:
[
\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}
]
Simplifying:
[
v = \sqrt{\dfrac{GM}{r}}
]
Now, the time period (T) of revolution is the time taken to complete one orbit of circumference [2\pi r]:
[T = \dfrac{\text{Distance}}{\text{Velocity}}] [= \dfrac{2\pi r}{v}]
Substitute [v = \sqrt{\dfrac{GM}{r}}]:
[T = 2\pi r \sqrt{\dfrac{r}{GM}}] [= 2\pi \sqrt{\dfrac{r^3}{GM}}]
Hence,
[
\boxed{T = 2\pi \sqrt{\dfrac{r^3}{GM}}}
]
This relation shows that the square of the period (T²) is directly proportional to the cube of the orbital radius (r³):
[
T^2 \propto r^3
]
This is known as Kepler’s Third Law of Planetary Motion.
3. Dimensions and Units
| Quantity | Symbol | Dimensional Formula | SI Unit |
|---|---|---|---|
| Time Period | [T] | [M^0 L^0 T^1] | seconds (s) |
4. Key Features
- The time period of a satellite depends only on [r] and [M].
- It is independent of the satellite’s mass [m].
- [T^2 \propto r^3] — a direct consequence of Kepler’s Third Law.
- For satellites closer to Earth, the time period is smaller.
- For higher orbits, the time period increases rapidly.
- Geostationary satellites have a time period equal to 24 hours.
5. Important Formulas to Remember
| Concept | Formula | Description |
|---|---|---|
| Orbital Velocity | [v = \sqrt{\dfrac{GM}{r}}] | Required velocity for stable circular orbit |
| Time Period | [T = 2\pi \sqrt{\dfrac{r^3}{GM}}] | Time for one complete revolution |
| Kepler’s Law | [T^2 = \dfrac{4\pi^2}{GM}r^3] | Relation between T² and r³ |
| Geostationary Orbit | [T = 24\text{ h}] | Satellite period equals Earth’s rotation period |
6. Conceptual Questions with Solutions
1. What is meant by the time period of a satellite?
It is the time taken by a satellite to complete one full revolution around the Earth.
2. On what factors does the time period of a satellite depend?
It depends on the orbital radius [r] and the planet’s mass [M].
3. Is the time period dependent on the satellite’s mass?
No, it is completely independent of the satellite’s mass.
4. What is the relation between the time period and orbital radius?
[T^2 \propto r^3], as derived from [T = 2\pi \sqrt{\dfrac{r^3}{GM}}].
5. Why do higher satellites have longer time periods?
Because their orbital radius [r] is larger, and [T] increases with [r^{3/2}].
6. What is the time period of a geostationary satellite?
It is 24 hours, equal to Earth’s rotation period.
7. What provides the centripetal force for a satellite in orbit?
The gravitational attraction of the Earth provides the required centripetal force.
8. If the radius of the orbit doubles, how will the time period change?
[T \propto r^{3/2}], so if [r] doubles, [T] becomes [2^{3/2} = 2.828] times greater.
9. What is the period of a satellite near Earth’s surface?
About **84.4 minutes** or **1.4 hours**.
10. What happens to the time period if the planet’s mass increases?
[T \propto \dfrac{1}{\sqrt{M}}], so the time period decreases.
7. FAQ / Common Misconceptions
1. Does the time period depend on the satellite’s weight?
No, it depends only on the planet’s mass and the orbital radius.
2. Do all satellites have the same time period?
No, each satellite’s time period depends on its orbital radius.
3. Is the orbital time period affected by air resistance?
In space (where air resistance is negligible), it is not affected.
4. Does the Earth’s rotation affect the satellite’s time period?
No, except for geostationary satellites that are synchronized with Earth’s rotation.
5. Why do communication satellites have a 24-hour period?
To remain fixed relative to the same point on Earth’s surface.
6. Can two satellites have the same time period at different altitudes?
No, because [T^2 \propto r^3]; time period changes with orbital radius.
7. What happens if a satellite’s speed changes?
If its speed increases, its orbit (and hence time period) also changes.
8. Does a polar satellite have the same time period as a geostationary one?
No, polar satellites have much shorter time periods.
9. Is Kepler’s Third Law valid for artificial satellites?
Yes, it applies to all bodies in orbit, natural or artificial.
10. Is the time period measured from the satellite or Earth?
It is measured with respect to the planet’s frame (Earth’s frame for Earth satellites).
8. Practice Questions (With Step-by-Step Solutions)
Q1. Calculate the time period of a satellite revolving at an altitude of [600 km] above Earth’s surface.
Given: [R_E = 6.37 × 10^6 m], [G = 6.67 × 10^{-11}], [M = 5.97 × 10^{24} kg].
Solution:
[
r = R_E + h = 6.37 × 10^6 + 0.6 × 10^6 = 6.97 × 10^6 m
]
[
T = 2\pi \sqrt{\dfrac{r^3}{GM}} = 2\pi \sqrt{\dfrac{(6.97 × 10^6)^3}{(6.67 × 10^{-11})(5.97 × 10^{24})}} = 5.85 × 10^3 s
]
✅ T = 1.62 hours
Q2. Find the time period of a satellite very close to Earth’s surface.
Solution:
[
T = 2\pi \sqrt{\dfrac{R_E^3}{GM}} = 2\pi \sqrt{\dfrac{(6.37 × 10^6)^3}{(6.67 × 10^{-11})(5.97 × 10^{24})}} = 5.07 × 10^3 s
]
✅ T = 84.4 minutes
Q3. If the radius of orbit becomes 4 times, how does the time period change?
Solution:
[
T \propto r^{3/2} \Rightarrow \dfrac{T_2}{T_1} = (4)^{3/2} = 8
]
✅ The new time period becomes 8 times greater.
Q4. A satellite revolves around Mars (mass [6.42 × 10^{23} kg], radius [3.39 × 10^6 m]). Find its time period near the surface.
Solution:
[
T = 2\pi \sqrt{\dfrac{R^3}{GM}}
]
[
T = 2\pi \sqrt{\dfrac{(3.39 × 10^6)^3}{(6.67 × 10^{-11})(6.42 × 10^{23})}} = 7.6 × 10^3 s = 2.1 h
]
✅ T = 2.1 hours
Q5. Show that [T^2 \propto r^3] for a satellite in circular motion.
Solution:
From [v = \sqrt{\dfrac{GM}{r}}] and [T = \dfrac{2\pi r}{v}],
[
T = 2\pi r \sqrt{\dfrac{r}{GM}} \Rightarrow T^2 = \dfrac{4\pi^2}{GM}r^3
]
✅ Hence proved that [T^2 \propto r^3].
Sign in with Google