19/07/2026
VELOCITY OF A SATELLITE IN ORBIT
The velocity of a satellite in orbit, commonly known as the orbital velocity, is the minimum speed required for a satellite to remain in a stable circular orbit around a planet or any other celestial body. At this speed, the gravitational attraction between the planet and the satellite supplies the exact centripetal force required to keep the satellite moving in its circular path. If the satellite moves slower than this velocity, it will gradually fall toward the planet. If it moves much faster, it may overcome the planet's gravitational attraction and escape into space. The orbital velocity depends on the mass of the central body and the distance of the satellite from the centre of the body, but it is independent of the satellite's mass.
➡️ Orbital Velocity Formula
The orbital velocity of a satellite in a circular orbit is given by
v = √(GM/r)
where:
v = orbital velocity (m s⁻¹)
G = universal gravitational constant = 6.67 × 10⁻¹¹ N m² kg⁻²
M = mass of the Earth (or any central body)
r = distance from the centre of the Earth to the satellite
Since
r = R + h
where:
R = radius of the Earth
h = altitude of the satellite above the Earth's surface
the orbital velocity can also be written as
v = √[GM/(R + h)]
Using the relation
GM = gR²
the equation becomes
v = √[gR²/(R + h)]
For a satellite very close to the Earth's surface (h = 0),
v = √(gR)
➡️ Derivation of the Orbital Velocity Formula
The orbital velocity is obtained by equating the gravitational force acting on the satellite to the centripetal force required for circular motion.
The gravitational force acting on a satellite of mass m is
Fg = GMm/r²
The centripetal force required to keep the satellite moving in a circular orbit is
Fc = mv²/r
Since gravity provides the centripetal force,
Fg = Fc
Therefore,
GMm/r² = mv²/r
Cancelling m from both sides gives
GM/r² = v²/r
Multiplying both sides by r,
v² = GM/r
Taking the square root,
v = √(GM/r)
Or
Since
r = R + h
the final expression becomes
v = √[GM/(R + h)
Orbital Velocity Near the Earth's Surface
For a satellite orbiting just above the Earth's surface,
g = 9.81 m s⁻²
R = 6.37 × 10⁶ m
Using
v = √(gR)
Substituting the values,
v = √(9.81 × 6.37 × 10⁶)
v = √(6.24897 × 10⁷)
v = 7.90 × 10³ m s⁻¹
Therefore,
v ≈ 7.9 km s⁻¹
This value is known as the first cosmic velocity.
➡️ Factors Affecting Orbital Velocity
The orbital velocity depends primarily on the mass of the central body and the orbital radius. A more massive planet exerts a stronger gravitational pull, requiring a higher orbital velocity. Conversely, as the orbital radius or altitude increases, the gravitational attraction decreases, resulting in a lower orbital velocity. These relationships are expressed as
v ∝ √M
and
v ∝ 1/√r
➡️ Significance of Orbital Velocity
The concept of orbital velocity is fundamental to satellite technology and space science. It ensures that satellites remain in stable orbits around the Earth, making long-term communication, navigation, weather observation, and scientific research possible. Engineers use orbital velocity calculations when designing rockets, planning satellite launches, and determining orbital transfer trajectories. Accurate knowledge of orbital velocity is also essential in spacecraft docking, interplanetary missions, lunar exploration, and deep-space travel. Furthermore, the concept explains the motion of natural satellites, planets, and other celestial bodies and forms one of the foundations of celestial mechanics and astrophysics.
➡️ Problems and Solutions
Q1.
A satellite has an orbital velocity of 7.9 km s⁻¹. Determine its altitude above the Earth's surface.
Given
v = 7.9 × 10³ m s⁻¹
G = 6.67 × 10⁻¹¹ N m² kg⁻²
M = 5.97 × 10²⁴ kg
R = 6.4 × 10⁶ m
Solution
Using
v = √(GM/r)
Square both sides,
v² = GM/r
Rearranging,
r = GM/v²
Substitute,
r = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(7.9 × 10³)²
r = 6.37 × 10⁶ m
Altitude,
h = r − R
h = 6.37 × 10⁶ − 6.40 × 10⁶
h ≈ 0 m
Answer
The satellite is orbiting very close to the Earth's surface.
Q2.
A satellite is orbiting 500 km above the Earth's surface. Calculate its orbital velocity.
Given
R = 6.4 × 10⁶ m
h = 5.0 × 10⁵ m
g = 9.8 m s⁻²
Solution
Distance from Earth's centre,
r = 6.4 × 10⁶ + 5.0 × 10⁵
r = 6.9 × 10⁶ m
Using
v = √(gR²/r)
Substitute,
v = √[(9.8 × (6.4 × 10⁶)²)/(6.9 × 10⁶)]
v = 7.63 × 10³ m s⁻¹
Answer
Orbital velocity = 7.63 km s⁻¹
