Physics and Mathematics
Orbital Velocity
1. Statement of the Concept
The orbital velocity of a satellite is defined as the minimum velocity that a satellite must have to stay in a stable circular orbit around a planet without any propulsion, under the influence of the planet’s gravitational force.
2. Explanation and Mathematical Derivation
Consider a satellite of mass [m] orbiting around Earth (mass [M]) in a circular orbit of radius [r] (measured from the center of Earth).
The gravitational force acting on the satellite provides the necessary centripetal force to keep it in circular motion.
[
F_g = F_c
]
Using Newton’s law of gravitation and centripetal force formula:
[
\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}
]
Simplifying, we get the expression for orbital velocity:
[
v = \sqrt{\dfrac{GM}{r}}
]
where:
- [v] = orbital velocity of the satellite
- [G] = universal gravitational constant
- [M] = mass of the Earth (or the central body)
- [r] = radius of the orbit from the Earth’s center
Thus, the orbital velocity depends on the mass of the Earth and the radius of the orbit.
If the orbit is close to the Earth’s surface, [r ≈ R_E] (Earth’s radius), so:
[
v_0 = \sqrt{\dfrac{GM}{R_E}}
]
Numerical Value for Earth
Using:
[G = 6.67 × 10^{-11} N·m²/kg²], [M = 5.97 × 10^{24} kg], [R_E = 6.37 × 10^6 m]
[v_0] [= \sqrt{\dfrac{(6.67 × 10^{-11})(5.97 × 10^{24})}{6.37 × 10^6}}] [≈ 7.9 × 10^3 m/s]
Hence, orbital velocity near Earth’s surface = 7.9 km/s.
3. Dimensions and Units
| Quantity | Symbol | Dimensional Formula | SI Unit |
|---|---|---|---|
| Orbital Velocity | [v] | [M^0 L^1 T^{-1}] | m/s |
4. Key Features
- The orbital velocity depends only on the mass of the central body (M) and radius of orbit (r), not on the mass of the satellite.
- As the altitude increases, [r] increases, hence orbital velocity decreases.
- Orbital velocity is the minimum velocity required for a stable circular orbit.
- For orbits near Earth’s surface, v = 7.9 km/s.
- It is independent of the direction of motion.
5. Important Formulas to Remember
| Concept | Formula | Description |
|---|---|---|
| Gravitational Force | [F = \dfrac{GMm}{r^2}] | Attractive force between Earth and satellite |
| Centripetal Force | [F = \dfrac{mv^2}{r}] | Force required for circular motion |
| Orbital Velocity | [v = \sqrt{\dfrac{GM}{r}}] | Velocity for circular orbit |
| Surface Orbital Velocity | [v_0 = \sqrt{\dfrac{GM}{R_E}}] | Velocity near Earth’s surface (7.9 km/s) |
6. Conceptual Questions with Solutions
1. What is meant by orbital velocity?
Orbital velocity is the velocity required by a satellite to stay in a stable circular orbit around a planet under its gravitational force.
2. Why is the orbital velocity independent of the satellite’s mass?
Because both the gravitational and centripetal forces are directly proportional to the satellite’s mass, which cancels out during derivation.
3. How does orbital velocity vary with altitude?
As altitude increases, the orbital radius [r] increases, causing [v = \sqrt{\dfrac{GM}{r}}] to decrease.
4. What happens if a satellite moves slower than its orbital velocity?
It will fall back towards the Earth due to insufficient centripetal force.
5. What happens if it moves faster than orbital velocity?
The satellite will move into a higher orbit or escape Earth’s gravity if speed exceeds escape velocity.
6. What is the orbital velocity close to Earth’s surface?
[7.9 km/s].
7. What provides the centripetal force to a satellite in orbit?
The gravitational attraction between the satellite and the planet.
8. Does orbital velocity depend on the shape of the orbit?
Yes, for elliptical orbits the velocity varies with position; for circular orbits, it remains constant.
9. Can an artificial satellite orbit any planet?
Yes, provided it achieves the required orbital velocity for that planet’s mass and radius.
10. What is the difference between orbital velocity and escape velocity?
Orbital velocity keeps a satellite in circular motion, while escape velocity allows it to leave the planet’s gravitational field.
7. FAQ / Common Misconceptions
1. Is orbital velocity the same for all planets?
No, it depends on the planet’s mass and radius.
2. Does a heavier satellite require higher orbital velocity?
No, orbital velocity is independent of the satellite’s mass.
3. Is gravity absent for a satellite in orbit?
No, gravity is essential to maintain its circular motion; it acts as the centripetal force.
4. Is orbital velocity the same as tangential velocity?
In a circular orbit, yes. The orbital velocity acts tangentially to the orbit.
5. Can a satellite have zero orbital velocity?
No, without orbital velocity, the satellite would fall toward the planet.
6. Does orbital velocity depend on the direction of motion?
No, it only depends on [G], [M], and [r].
7. Does the Moon have orbital velocity?
Yes, the Moon orbits Earth with an average orbital velocity of about [1.02 km/s].
8. Why do satellites in higher orbits move slower?
Because at greater distances, gravitational force (and thus required centripetal force) is weaker.
9. What would happen if there were no gravity?
The satellite would move off in a straight line due to inertia (Newton’s First Law).
10. Can a satellite orbit at any height?
No, the orbit must satisfy the balance between gravitational and centripetal forces.
8. Practice Questions (With Step-by-Step Solutions)
Q1. Calculate the orbital velocity 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]
[v] [= \sqrt{\dfrac{GM}{r}}] [= \sqrt{\dfrac{(6.67 × 10^{-11})(5.97 × 10^{24})}{6.97 × 10^6}}] [= 7.56 × 10^3 m/s]
✅ Orbital velocity = 7.56 km/s
Q2. At what height above Earth’s surface will a satellite have an orbital velocity of [6.9 km/s]?
Solution:
[v = \sqrt{\dfrac{GM}{r}}] [\Rightarrow r = \dfrac{GM}{v^2}]
Substituting,
[r] [= \dfrac{(6.67 × 10^{-11})(5.97 × 10^{24})}{(6.9 × 10^3)^2}] [= 8.38 × 10^6 m]
[h] [= r – R_E] [= 8.38 × 10^6 – 6.37 × 10^6] [= 2.01 × 10^6 m] [= 2010 km]
✅ Height = 2010 km
Q3. Find the ratio of orbital velocities of two satellites orbiting Earth at heights of [h_1 = 300 km] and [h_2 = 1200 km].
Solution:
[\dfrac{v_1}{v_2}] [= \sqrt{\dfrac{R_E + h_2}{R_E + h_1}}]
[\dfrac{v_1}{v_2}] [= \sqrt{\dfrac{6.37 × 10^6 + 1.2 × 10^6}{6.37 × 10^6 + 0.3 × 10^6}}] [= \sqrt{\dfrac{7.57}{6.67}}] [= 1.066]
✅ v₁ is about 6.6% greater than v₂
Q4. Derive the formula for orbital velocity using Newton’s laws.
Solution:
Gravitational force = Centripetal force
[
\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}
]
[
v = \sqrt{\dfrac{GM}{r}}
]
✅ Derived successfully.
Q5. A satellite orbits Mars (mass [6.42 × 10^{23} kg], radius [3.39 × 10^6 m]). Calculate its orbital velocity near the surface.
Solution:
[
v = \sqrt{\dfrac{GM}{R}}
]
[v] [= \sqrt{\dfrac{(6.67 × 10^{-11})(6.42 × 10^{23})}{3.39 × 10^6}}] [= 3.55 × 10^3 m/s]
✅ Orbital velocity = 3.55 km/s
Sign in with Google