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Escape Velocity and Orbital Velocity

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Relation between Escape Velocity and Orbital Velocity

Have you ever wondered how rockets take off? While going off to space, rockets need a push, some sort of force that lets them leave the earth and go into space. This is a huge kickstart that they have to be given, otherwise the result, that is them leaving the earth’s surface, would not happen. 


But, what is the reason behind the need for such a huge kickstart? Well, it all comes down to the gravitational field that is there on the surface of the earth. In case you didn't know, it is a very strong force for which an all the more strong “push” is required for the rockets to send them into space. 


Escape Velocity is the minimum Velocity that is required for a body or an object to leave or “Escape” the gravitational field of the surface of the earth. Furthermore, it is always the goal for Escape Velocity to have an object leave the face of the earth without a chance for it to ever fall back due to the gravitational pull of our planet.

Orbital Velocity

Do you know what happens if an aeroplane stops? Well, on such an occasion, it will be pulled towards the earth’s surface due to the force of gravity. There are satellites, amongst other objects in the space sent by us, which require the velocity that can withstand as well as defy the gravitational pull of our planet. 


Any object requires to sustain a certain speed that can easily let them have a good alignment with the celestial body’s rotational velocity. The speed is also important to be enough to sustain the kind of gravitational pull that comes from the earth. 

Definition of Orbital Velocity

When an object needs to enter into an orbit of any other celestial bodies, it is a must that they obtain a speed that can help them counteract the gravitational force.  


The lowest velocity an object must have to escape the gravitational force of a planet or an object. The relationship between the Escape Velocity and the Orbital Velocity is defined by Ve = 2 Vo where Ve is the Escape Velocity and Vo is the Orbital Velocity. And the Escape Velocity is root-two times the Orbit Velocity.


Escape Velocity, as it relates to rocket Science and space travel, is the Velocity required for an object (such as a rocket) to Escape the gravitational Orbit of a celestial body (such as a planet or a star).


We have studied in kinematics that the range of the projectile depends on the initial Velocity of the projectile. ⇒ Rmax ∝ u2 ⇒ Rmax = \[\frac {u_2}{2g}\], which means that the particle flies away from the gravitational impact of the earth at a certain initial Velocity provided to the particle. 


This minimum amount of Velocity for which the particle Escapes the gravitational sphere of influence of the planet is known as the Velocity of Escape (ve). When an Escape Velocity is given to a body, it theoretically goes to infinity.


As gravitational force is a conservative force, the law on energy conservation is fine. Applying the law on the conservation of energy for a particle with the necessary minimum Velocity to infinity


Ui + Ki  = Uf + Kf


At infinity, the particles undergo no interaction, so the final potential energy, and we know from motion in the 1D chapter that the final Velocity of the body is zero after reaching its maximum height so that we can deduce the final kinetic energy of the particle.


Then ,


Ui + Ki  = 0 and   we   know   that, Ui ​= \[\frac {-GMm}{R}\] , Ki​= \[\frac {1}{2mv_{e}^2}\]  


We get,  


\[\frac {1}{2mv_{e}^2}\] + \[(\frac {-GMm}{R})\] = 0 ⇒ \[\frac {1}{2mv_{e}^2}\] = \[(\frac {GMm}{R})\]


That implies,


ve​= \[\sqrt {\frac {2GM}{R}}\] ​​ ……………(1)  


It is obvious from the above formula that the Escape Velocity does not depend on the test mass (m). If the source mass is earth, the Escape Velocity has a value of 11.2 km / s. When v = ve the body leaves the gravitational field or control of the planets, when 0 ≤v < ve the body either falls down to Earth or proceeds to Orbit the earth within the sphere of influence of the earth.


Orbital Velocity is the Velocity that the body will sustain in order to Orbit another body. Escape Velocity is the speed at which an object leaves the Orbit. Escape Velocity will be a square-root of 2 times the Orbital Velocity in order to exit the Orbit.


(l) If the Velocity is equal, the body must remain in constant Orbit, not in elevation.


(ll) less than the Orbit, the Orbit will decay and the object will crash.


(ll) rather than Orbital, and the body will have an ascending Orbit, which will fly out into space.


The speed at which the test mass travels around the source mass is known as Orbital Velocity


(vo) when the test mass Orbits around the source mass in a circular path of radius 'r' having the center of the source mass as the center of the circular path, the centripetal force is provided by the gravitational force as it is always the attracting force having its direction towards the center of a source mass.


⇒ \[\frac {mv_o^2}{r}\] = \[\frac {GMm}{r^2}\]


⇒ \[\frac {v_o^2}{r}\] = \[\frac {GM}{r^2}\]

 

⇒ vo ​= \[\sqrt {\frac {GM}{R}}\] 


If the test mass is small distances from the source mass  r ≈ R (radius of the source mass)

Then,


vo​ = \[\sqrt {\frac {GM}{R}}\] …………..(2)


The above formula indicates that the Orbital Velocity is independent of the test mass (the mass which is Orbiting).

Relation Escape Velocity And Orbital Velocity Formula

In astrophysics, the relationship between Escape Velocity and Orbit Velocity can be mathematically described as -


Vo=\[\frac {V_e}{\sqrt{2}}\]


Or


Ve= \[\sqrt{2}V_o\]


Where,

  • Ve is the Escape Speed Measurement using km / s.

  • Vo is the calculation of angular Velocity using km / s.

We know that Escape Velocity= \[\sqrt {2} \times \] Orbital Velocity that means that the Escape Velocity is directly proportional to the Orbital Velocity. This means for any big body-

  • When the angular Velocity increases, the Escape Velocity will also increase and aim-versely.

  • If the angular Velocity decreases, the Escape Velocity decreases as well as the vise-versa.

Escape Velocity and Orbital Velocity

The relationship between Escape Velocity and Orbital Velocity equations is very important for understanding the definition. For any kind of massive body or planet.

  • Escape Velocity is given by – Ve = \[\sqrt {2gR}\] ———-(1)  

  • Orbital Velocity is given by – Vo = \[\sqrt {gR}\] ———–(2)


Where,


g is the acceleration due to gravity.


R is the radius of the planet.


From equation (1) we can write that-


Ve= \[\sqrt {2}\]  \[\sqrt {gR}\]


Substituting Vo = \[\sqrt {gR}\] we get-


Ve=\[\sqrt {2} V_o\]


The above equation can be rearranged for Orbital Velocity as-


Vo = \[\frac {V_e}{\sqrt{2}}\]

FAQs on Escape Velocity and Orbital Velocity

1. What is the definition of escape velocity in Physics?

Escape velocity is the minimum speed an object must have to completely break free from the gravitational pull of a celestial body, like a planet or star, without any further propulsion. For an object to escape, its kinetic energy must overcome the planet's gravitational potential energy. The formula for escape velocity (vₑ) is vₑ = √(2GM/R), where G is the gravitational constant, M is the mass of the celestial body, and R is its radius.

2. What is meant by orbital velocity?

Orbital velocity is the precise speed at which an object must travel to maintain a stable, circular orbit around a celestial body. At this velocity, the gravitational force pulling the object towards the central body provides the exact centripetal force required for circular motion. The formula for orbital velocity (vₒ) for an orbit close to the body's surface is vₒ = √(GM/R), where M is the mass and R is the radius of the central body.

3. What is the fundamental difference between escape velocity and orbital velocity?

The key difference lies in the object's final trajectory. Orbital velocity is just enough to keep an object in a continuous path *around* a celestial body, balancing gravity. In contrast, escape velocity is the higher speed needed to overcome gravity *entirely* and travel away from the celestial body indefinitely, never to return. Essentially, one is for staying, and the other is for leaving.

4. How are escape velocity and orbital velocity mathematically related?

Escape velocity and orbital velocity are directly related. The escape velocity is precisely the square root of 2 (approximately 1.414) times the orbital velocity for the same celestial body at the same radius. This relationship is expressed by the formula: vₑ = √2 × vₒ. This shows that an object needs about 41.4% more speed to escape gravity than to simply orbit the body.

5. Why doesn't the escape velocity of an object depend on its own mass?

This is a common point of confusion. The escape velocity does not depend on the mass of the escaping object because both the kinetic energy needed (½mv²) and the gravitational potential energy to overcome (-GMm/R) are proportional to the object's mass (m). When we set these energies equal to derive the velocity, the 'm' on both sides of the equation cancels out, leaving the formula vₑ = √(2GM/R), which only depends on the planet's mass (M) and radius (R).

6. What happens if a satellite is launched with a velocity that is between its orbital and escape velocities?

If a satellite is launched with a speed greater than the required orbital velocity but less than the escape velocity, it will not fall back to Earth, nor will it escape Earth's gravity. Instead of a circular path, it will enter a stable elliptical orbit. The central celestial body (e.g., Earth) will be at one of the two foci of this ellipse.

7. What are the approximate orbital and escape velocities for an object near the Earth's surface?

For an object orbiting close to the surface of the Earth, the required velocities are:

  • The orbital velocity (vₒ) is approximately 7.9 km/s (or about 28,440 km/h).
  • The escape velocity (vₑ) is approximately 11.2 km/s (or about 40,320 km/h).
You can see the relationship vₑ ≈ √2 × vₒ holds true (11.2 ≈ 1.414 × 7.9).

8. How is the formula for orbital velocity derived from fundamental principles?

The formula for orbital velocity is derived by equating the gravitational force with the centripetal force required for circular motion. Here are the steps:

  • The gravitational force attracting the object is Fg = GMm/R².
  • The centripetal force needed to keep it in a circular orbit is Fc = mvₒ²/R.
  • For a stable orbit, these two forces must be equal: GMm/R² = mvₒ²/R.
  • By cancelling 'm' from both sides and one 'R', we can rearrange the equation to solve for vₒ, which gives the final formula: vₒ = √(GM/R).