Orbital Velocity of Satellite-Formula, Derivation, Solved Examples

Define Orbital Velocity

Orbital velocity is defined as the minimum velocity a body must maintain to stay in orbit.

Orbital velocity is the minimum velocity required for a satellite to stay in orbit around a celestial body, such as Earth, without falling back due to gravity. Understanding orbital velocity is crucial for space exploration, satellite communications, and astrophysics.

For an artificial or natural satellite, orbital velocity can be interpreted as the velocity necessary to maintain it in its orbit.

Derivation of Orbital Velocity

To derive the orbital velocity, we concern ourselves with the following two concepts:

• Gravitational Force
• Centripetal Force

It is important to know the gravitational force because it is the force that allows orbiting to exist. A central body exerts a gravitational force on the orbiting body to keep it in its orbit. Centripetal force is also important, as this is the force responsible for circular motion.

For the derivation, let us consider a satellite of mass m revolving around the Earth in a circular orbit of radius r at a height h from the surface of the Earth. Suppose M and R are the mass and radius of the Earth respectively, then r = R + h.

To revolve the satellite, a centripetal force of

\(\begin{array}{l}\frac{mv_{o}^2}{r}\end{array} \)

is needed which is provided by the gravitational force

\(\begin{array}{l}G\frac{Mm}{r^2}\end{array} \)

between the satellite and the Earth.

Therefore, equating both the equations, we get

\(\begin{array}{l}\frac{mv_{0}^2}{r}=G\frac{Mm}{r^2}\end{array} \)

\(\begin{array}{l}v_{0}^2=\frac{GM}{r}=\frac{GM}{R+h}\end{array} \)

Simplifying the above equation further, we get

\(\begin{array}{l}v_{0}=\sqrt{\frac{GM}{R+h}}\end{array} \)

……(eqn 1)

Hence orbital velocity depends on the mass of the central body (say earth) and the radius of the orbit (r).

But \(\begin{array}{l}GM = gR^2\end{array} \) , where g is the acceleration due to gravity.

Therefore,

\(\begin{array}{l}v_0=\sqrt{\frac{gR^{2}}{R+h}}\end{array} \)

Simplifying the above equation, we get

\(\begin{array}{l}v_0=R\sqrt{\frac{g}{R+h}}\end{array} \)

Let g’ be the acceleration due to gravity at a height h from the surface

\(\begin{array}{l}g’=\frac{GM}{(R+h)^2}\end{array} \)

Simplifying further, we get

\(\begin{array}{l}\frac{GM}{(R+h)}=g'(R+h)=g’r\end{array} \)

……(eqn 2)

Substituting (2) in (1), we get

\(\begin{array}{l}v_0=\sqrt{g’r}=\sqrt{g'(R+h)}\end{array} \)

This derives the formula for orbital velocity of an object or satellite revolving around a planet.

Factors Affecting Orbital Velocity

• Altitude of the Satellite (r = R+h) : The farther a satellite is from the planet (h), the lower its orbital velocity.
• Mass of the Celestial Body (M): More massive planets require higher orbital velocities.
• Gravitational Constant: A fundamental constant affecting all celestial orbits.

Important Points on Orbital Velocity

1. Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of the orbit i.e., satellites of deferent masses have same orbital velocity, if they are in the same orbit.
2. Orbital velocity depends on the mass of central body and radius of orbit.
3. For a given planet, greater the radius of orbit, lesser will be the orbital velocity of the satellite v ∝ 1/√r .
4. Orbital velocity of the satellite when it revolves very close to the surface of the planet, V_o = √(gR) = √9.8 x 6.4 x 10⁶ = 7.9 km/s ≈ 8 km/s

Solved Examples on Orbital Velocity

Question 1:

Calculate the orbital velocity of the earth so that the satellite revolves around the earth if the radius of earth R = 6.5 × 10⁶ m, the mass of earth M = 5.9722×10²⁴ kg and Gravitational constant G = 6.67408 × 10^-11 m³ kg^-1 s^-2

Solution:

Given:

R = 6.5 × 10⁶ m

M = 5.9722×10²⁴ kg

G = 6.67408 × 10^-11 m³ kg^-1 s^-2

The Orbital velocity formula is given by

V_o = √(GM/R)

       = √6.67408 × 10^-11 ×5.9722×10²⁴/6.5 × 10⁶

       = √36.68 x 10¹³/6.5 x 10⁶

       = 7.5 x 10⁹km/s

Example 2:

A satellite launch is made for the study of Jupiter. Determine its velocity so that its orbit around the Jupiter.

Given: Radius of Jupiter R = 70.5 × 106 m,

      Mass of Jupiter M = 1.5 × 10²⁷ Kg,

      Gravitational constant G = 6.67408 × 10^-11 m³ kg^-1 s^-2

Solution:

When the given parameters are substituted in the orbital velocity formula, we get

V_o = √(GM/R)

    = √6.67408 × 10^-11 × 1.5 × 10²⁷/70.5×10⁶

    = √10.0095 x 10¹⁶  / 70.5 x 10⁶

    = √0.141 x 10¹⁰

     = 3.754 x  10⁴m/s.

Solved Problems on Orbital Velocity

Problem 1. Find the orbital velocity of an object revolving around the planet Earth if the radius of Earth is 6.5 × 10⁶ m, the mass of Earth is 5.9722 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10^-11 m³ kg^-1 s^-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 6.5 × 10⁶

M = 5.9722 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(5.9722 × 10²⁴)/(6.5 × 10⁶)

= 29.8 km/s

Problem 2. Find the orbital velocity of an object revolving around the planet Mercury if the radius of Mercury is 2439.7 km, the mass of Mercury is 0.33 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10^-11 m³ kg^-1 s^-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 2439.7

M = 0.33 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(0.33 × 10²⁴)/(2439.7)

= 47.4 km/s

Problem 3. Find the orbital velocity of an object revolving around the planet Venus if the radius of Venus is 6051.8 km, the mass of Venus is 4.87 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10^-11 m³ kg^-1 s^-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 6051.8

M = 4.87 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(4.87 × 10²⁴)/(6051.8)

= 35 km/s

Problem 4. Find the orbital velocity of an object revolving around the planet Mars if the radius of Mars is 3389 km, the mass of Mars is 0.642 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10^-11 m³ kg^-1 s^-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 3389

M = 0.642 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(0.642 × 10²⁴)/(3389)

= 24.1 km/s

Problem 5. Find the orbital velocity of an object revolving around the planet Jupiter if the radius of Jupiter is 69911 km, the mass of Jupiter is 1898 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10-11 m3 kg-1 s-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 69911

M = 1898 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(1898 × 10²⁴)/(69911)

= 13.1 km/s

Problem 6. Find the orbital velocity of an object revolving around the planet Saturn if the radius of Saturn is 58232 km, the mass of Saturn is 568 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10-11 m3 kg-1 s-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 58232

M = 568 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(568 × 10²⁴)/(58232)

= 9.7 km/s

Problem 7. Find the orbital velocity of an object revolving around the planet Uranus if the radius of Uranus is 25362 km, the mass of Uranus is 86.8 × 10²⁴ kg and the gravitational constant G is 6.67408 × 10-11 m3 kg-1 s-2.

Solution:

We have,

G = 6.67408 × 10^-11

R = 25362

M = 86.8 × 10²⁴

Using the formula we have,

V = √(GM/R)

= (6.67408 × 10^-11)(86.8 × 10²⁴)/(25362)

= 6.8 km/s

Practical Applications

• Space Stations: The International Space Station orbits at around 7.66 km/s.
• Geostationary Satellites: These satellites orbit at 3.07 km/s at 35,786 km above Earth.
• Lunar Missions: The Moon orbits Earth at 1.02 km/s.

Multiple Choice Questions (MCQs)

Q1: What is the orbital velocity of a satellite at a height equal to Earth’s radius?

A) 7.9 km/s
B) 11.2 km/s
C) 8.3 km/s
D) 11.9 km/s
Answer: A)
Explanation: The orbital velocity at the surface of the Earth is 7.9 km/s, and it decreases with altitude.

Q2: Which factor does NOT affect the orbital velocity of a satellite?

A) Mass of the satellite
B) Mass of the Earth
C) Radius of orbit
D) Gravitational constant
Answer: A) Mass of the satellite
Explanation: The orbital velocity is independent of the satellite’s mass; it depends on the central body’s mass and orbit radius.

Q3: What happens if a satellite travels at a velocity lower than the required orbital velocity?

A) It will continue in the same orbit
B) It will escape Earth’s gravity
C) It will fall back to Earth
D) It will move to a higher orbit
Answer: C) It will fall back to Earth
Explanation: A satellite needs to maintain orbital velocity to counteract gravity. If the velocity decreases, the satellite will descend into the atmosphere and burn up.

Frequently Asked Questions – FAQs

Q1: How does orbital velocity change with altitude?

A: Orbital velocity decreases as altitude increases because the gravitational pull weakens at higher distances.

Q2: What happens if a satellite moves slower than its orbital velocity?

A: It will eventually spiral down and fall back to Earth due to gravitational pull.

Q3: Why is the orbital velocity of satellites in low Earth orbit higher than those in geostationary orbit?

A: The closer the satellite is to Earth, the stronger the gravitational pull, requiring a higher velocity to maintain orbit.

Q4: Can a satellite have an elliptical orbit?

A: Yes, satellites can follow elliptical orbits as described by Kepler’s laws. In such orbits, the velocity changes depending on the distance from the Earth.

Q5: Why do astronauts experience weightlessness in orbit?

A: Astronauts are in free fall along with the spacecraft, creating a sensation of weightlessness.

Test Your Knowledge

Conclusion

Orbital velocity is a fundamental concept in astrophysics and space technology. It determines how satellites, space stations, and even planets maintain their paths in space. Understanding this helps in satellite deployment, space exploration, and communication technologies.

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