15/04/2026
ORBITAL VELOCITY OF SATELLITE
Orbital motion refers to the movement of a satellite around a planet under the influence of gravity. When a satellite is launched into space, it must possess a certain velocity to remain in orbit. This velocity ensures that the gravitational pull of the planet provides the necessary centripetal force to keep the satellite moving in a circular path rather than falling back to the surface. This required velocity is known as orbital velocity.
Definition of Orbital Velocity
Orbital velocity is defined as the minimum velocity a satellite must have to remain in a stable circular orbit around a planet without falling towards it or escaping into space.
➡️ Derivation of the Orbital Velocity Formula
Assumptions and Parameters
Consider a satellite of mass m revolving around a planet of mass M in a circular orbit of radius r. Let G represent the universal gravitational constant and v the orbital velocity.
Gravitational Force
The gravitational force acting between the planet and the satellite is given by:
Fg = (G × M × m) / r²
Centripetal Force
The centripetal force required to keep the satellite in circular motion is:
Fc = (m × v²) / r
Equating Forces
Since gravitational force provides the centripetal force, we equate:
(G × M × m) / r² = (m × v²) / r
Simplification
Cancel m from both sides and simplify:
(G × M) / r² = v² / r
Multiplying both sides by r gives:
(G × M) / r = v²
Final Expression
Taking square root:
v = √(G × M / r)
This is the formula for orbital velocity of a satellite.
➡️ Orbital Velocity Near the Surface of the Earth
Relation with Acceleration Due to Gravity
We know that:
g = (G × M) / R²
where R is the radius of the Earth.
Rewriting:
G × M = g × R²
Substitution
Substitute into the orbital velocity formula:
v = √(g × R² / R)
Simplified Formula
v = √(g × R)
This gives the orbital velocity for a satellite close to the Earth’s surface.
➡️ Dimensional Analysis
From formula:
v = √(G × M / r)
Dimensions:
G = [M⁻¹ L³ T⁻²]
M = [M]
r = [L]
Substitute:
v = √( (M⁻¹ L³ T⁻² × M) / L )
= √( L² T⁻² )
= L T⁻¹
[v] = L T⁻¹
➡️ Mathematical Problems and Solutions
Q1.
A satellite is orbiting at a height of 400 km above Earth’s surface. Calculate its orbital velocity.
Given:
R = 6.4 × 10⁶ m
h = 400 km = 4 × 10⁵ m
g = 9.8 m/s²
Solution
First find total radius:
r = R + h
r = 6.4 × 10⁶ + 4 × 10⁵
r = 6.8 × 10⁶ m
We use:
v = √(G × M / r)
But G × M = g × R²
So:
v = √(g × R² / r)
Substitute values:
v = √(9.8 × (6.4 × 10⁶)² / (6.8 × 10⁶))
Simplify:
v ≈ 7.67 × 10³ m/s
Answer:
v ≈ 7.7 km/s
Q2.
A satellite has an orbital velocity of 5 km/s. Find its orbital radius.
Given:
G × M = 3.986 × 10¹⁴ m³/s²
Solution
Convert velocity:
v = 5 km/s = 5000 m/s
Use:
v = √(G × M / r)
Square both sides:
v² = G × M / r
Rearrange:
r = G × M / v²
Substitute:
r = (3.986 × 10¹⁴) / (5000)²
r = (3.986 × 10¹⁴) / (2.5 × 10⁷)
r ≈ 1.594 × 10⁷ m
Answer:
r ≈ 1.59 × 10⁷ m
Q3.
Calculate the orbital velocity of a satellite close to the Earth’s surface.
Given:
g = 9.8 m/s²
R = 6.4 × 10⁶ m
Solution
We use the formula:
v = √(g × R)
Substitute the values:
v = √(9.8 × 6.4 × 10⁶)
v = √(62.72 × 10⁶)
v ≈ 7.92 × 10³ m/s
Answer:
v ≈ 7.9 km/s
