Orbital Velocity Derivation

The orbital velocity, also referred to as orbital speed is the speed at which an astronomical body or object orbits the barycenter.

• Barycenter is the point around which the objects orbit.
• Examples of such objects include planets, moons, artificial satellites, spacecraft, and stars.
• If one body is significantly more massive than the other bodies in the system combined, then the velocity relative to the center of mass of the most massive body is also known as the orbital velocity.

The formula for orbital velocity for a satellite orbiting at height h from the surface of the earth is given by

\(V_0 = \sqrt{\frac{GM_e}{R_e+h}}\)

The formula for orbital velocity for a satellite orbiting near the surface of the earth is given by

\(V_0=\sqrt{\frac{GM_e}{R_e}}\)

The relationship between escape velocity and orbital velocity

V_e = \(\sqrt{2V_0}\)

Key Terms: Radius of the Earth, Orbit, Orbital velocity, Escape velocity, Gravitational constant, Acceleration due to gravity, Satellite.

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Orbital Velocity

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The velocity at which a body appears to rotate around another body is known as its orbital velocity. 

• An Earth's orbit is a term used to describe the regular circular motion of the object around the Earth.
• Orbital velocity depends on the distance of the separation between an object and the center of the earth. 
• Any planet's orbital velocity may be calculated using its known mass M and radius R. 
• Metres per second (m/s) is the unit of measurement of orbital velocity.
• The orbital velocity for a satellite revolving near the surface of the earth is equal to 7.92 km/s or 7.92 x 10³ m/s.

Read More: Differences Between Acceleration and Velocity

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Orbital Velocity Formula

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For a satellite orbiting at height h from the surface of the earth, the formula for orbital velocity is given by

\(V_0 = \sqrt{\frac{GM_e}{R_e+h}}\)

Where

• G is gravitational constant
• M_e is the mass of the earth
• R_e is the radius of the earth

For a satellite orbiting near the surface of the earth, the formula for orbital velocity is given by

\(V_0=\sqrt{\frac{GM_e}{R_e}}\)

In terms of acceleration due to gravity, the orbital velocity is given by

V₀ = \(\sqrt{gR_e}\)

In terms of escape velocity (V_e), the orbital velocity is given by

\(V_0= \frac{V_e}{\sqrt{2}}\)

Read More: Central Force

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Orbital Velocity Derivation

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Let us consider a satellite of mass m revolving with v in a circular orbit of the earth at height h from the surface of the earth.

Orbital velocity

The gravitational force between the earth and the satellite provides the necessary centripetal force to revolve the satellite around the orbit of the earth. i.e. 

F_{Gravitational} = F_Centripetal ….(i)

The gravitational force acting on the satellite due to earth at height h is given by

F_{Gravitational} = GM_em / (R_e + h)²

Where

• G is the gravitational constant
• M_e is the mass of the earth
• R_e is the radius of the earth

Also, the centripetal force acting on the satellite is given by

F_Centripetal = mv² / (R_e + h)

Where v is the orbital velocity of the satellite at height h.

Using equation (i), we get

GM_em / (R_e + h)² = mv² / (R_e + h)

⇒ v = √(GM_e / R_e + h)

If h = 0 i.e. if a body is thrown from the surface of the earth, then the orbital speed is given by

V₀ = √(GM_e / R_e)

If M be the mass of any planets or massive bodies and R be its radius, then orbital velocity is given by

V₀ = √(GM / R)

Multiplying numerator and denominator by R, we get

V₀ = √(GMR / R²)

But GM/R² = g, acceleration due to gravity

Therefore

V₀ = √(gR)

For Earth

• Acceleration due to gravity, g = 9.8 m/s²
• Radius of the earth, R = 64 x 10⁵ m

Therefore orbital velocity is given by

V₀ = √(9.8 x 64 x 10⁵) = 7.92 x 10³ m/s

Hence, the orbital velocity of an Earth’s satellite orbiting near its surface is about 7.92 x 10³ m/s or 7.92 km/s

Read More: Centripetal Acceleration

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Orbital Velocity in terms of Escape Velocity

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The minimum velocity with which a body should be projected from the surface of a planet so as to reach infinity (i.e. it will never come again).

The formula of escape velocity is given by

\(V_e = \sqrt{2gR_e}\)

Also, the formula of orbital velocity is given by

\(V_0 = \sqrt{gR_e}\)

From the above two equations, the relation between orbital velocity and escape velocity is given by

\(V_e= \sqrt{2}V_0\)

Read More: Difference between Force and Pressure

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Solved Examples

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Ques. What will be the orbital speed from a planet of mass 10²⁷ kg and a radius of 10⁹ m?

Ans. Given

• The mass of the planet, M = 10²⁷ kg
• The radius of the planet, R = 10⁹ m

The orbital velocity is given by

V₀ = √(GM/R)

Where G is the Gravitational constant and its value is 6.67 x 10^-11 Nm² kg^-2

On substituting the values, we get

⇒ V₀ = √(6.67 x 10^-11 x 10²⁷ / 10⁹) = 8.16 x 10³ m/s

Ques. What will be the orbital speed of a satellite near the surface of a planet if its escape velocity is 12.4 x 10³ m/s?

Ans. The relationship between the orbital velocity and the escape velocity is given by

V_e = √2 V₀

Given, the escape velocity, V_e = 12.4 x 10³ m/s

Therefore, V₀ = V_e / √2

⇒ V₀ = 12.4 x 10³ / √2 = 12.4 x 10³ / 1.414

⇒ V₀ = 8.77 x 10³ m/s

Also Read:

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Things to Remember

• The orbital velocity is the velocity at which an object orbits the barycenter.
• The SI unit of orbital velocity is meters per second (m/s).
• The orbital velocity for a satellite revolving near the surface of the earth is equal to 7.92 km/s or 7.92 x 10³ m/s.
• The speed at which an object must be moving in order to escape from the gravitational field of a planet or moon and leave it without extra energy is known as the escape velocity.
• If a body is thrown from the surface of the earth, then the orbital speed is given by

\(V_0=\sqrt{\frac{GM_e}{R_e}}\)

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