Information About Escape Velocity
There are several instances in our daily life when we look up to the sky and wonder if we can go out there, or maybe we wonder how the astronauts reach space. You might be thinking of how difficult it is to launch a massive object into space. How much velocity does it require to complete this phenomenon? However, all objects, be it a rocket or a baseball moving deep into space, require the same speed. Notably, this velocity or acceleration is known as escape velocity.
What is Escape Velocity?
Escape velocity in Physics is the speed that an object requires to escape from the gravitational force of the Earth. However, it must not accelerate further. It is the minimum speed with which an object must be launched so that it is able to overpower the gravitational pull of the Earth and hence is able to escape to space. This means the escape velocity should remain constant throughout the way out and must not change while the object is under the influence of the gravitational force of the Earth.
The gravitational pull of the Earth is dependent on the size and the mass of the Earth and the escape velocity is dependent on the strength of the gravitational force of the Earth. This means that the escape velocity is dependent on the size and mass of the Earth.
Also, the escape speed is dependent on several factors. It is determined by scientists that the escape rate of an enormous body like a star, or a planet is evaluated using the following escape velocity equation:
\[V_{e} = \frac{\sqrt{2GM}}{R}\]
The expression for escape velocity is derivable by taking the initial kinetic energy of a body and initial gravitational potential energy at a certain height.
Initial potential energy
\[PE_{i} = \frac{-GMm}{R_{i}}\]
Where, \[PE_{i} \]: initial gravitational potential energy in kg-\[kg-m^{2}/s^{2}\].
G: Universal Gravitational Constant = \[6.674*10 - 20 km^{3}/kg-s^{2}\]
m: mass of the attracting body (kg)
M: mass of the escaping body (kg)
Note: M is much greater than m (M >> m)
\[R_{i}\]: initial separation between the centers of the bodies (km)
Initial kinetic energy
\[KE_{i} = \frac{mv{_{e}}^{2}}{2}\]
Where \[KE_{i}\]: initial kinetic energy (\[kg-m^{2}/s^{2}\])
M: mass of the object (kg)
\[V_{e}\]: initial velocity—and thus “escape velocity” (km/s)
Hence, the sum of kinetic and potential energies equals to total initial energy:
\[ TE_{i} = KE_{i} = PE_{i}\]
\[ TE_{i} = \frac mv{_{e}}^{2}{2} - \frac{GMm}{R_{i}}\]
Final energy
Also, gravitational fields are assumed to reach infinity. Hence, a body will move to infinity if the initial speed is strong enough. As a result, it “escapes” gravitational pull.
Potential energy at infinity
\[PE \infty = \frac{-GMm}{R \infty}\]
Where, \[PE \infty\]: gravitational potential energy at infinity
\[ R \infty \]: infinite separation between the centres of the objects
As, \[ R \infty \] = ∞ (infinity), thus \[PE \infty = 0\]
Kinetic energy at infinity
\[KE \infty = \frac{mv{_{\infty}}^{2}}{2}\]
Where, \[KE_{\infty}\]: final kinetic energy
\[v_{\infty}\]: final velocity at infinity
However, at infinity, the velocity of the body is zero: \[v_{\infty}\] = 0
Therefore, \[KE_{\infty} = 0\]
Whole final energy
As, kinetic energy is acting upward and potential energy is moving downward, total energy at the initial location:
\[ TE_{\infty} = KE_{\infty} + PE_{\infty} \]
Therefore, \[ TE_{\infty} = 0 + 0 \]
Escape Velocity Derivation
According to the Law of Conservation of Energy, this total energy remains constant for a closed system. So, in this scenario, the closed system contains two bodies with a gravitational pull between each other. Moreover, no outside force or energy is acting on any of the masses.
Hence, \[TE_{i} = TE_{\infty} \]
\[ KE_{i} + PE_{i} = 0\]
By substituting values:
\[\frac{mv{_{e}}^{2}}{2} - \frac{GMm}{R_{i}} = 0\]
By adding \[\frac{GMm}{R_{i}}\] to both sides:
Therefore, \[\frac{mv{_{e}}^{2}}{2} = \frac{GMm}{R_{i}}\]
Now for \[v{_{e}}^{2}\] solution:
\[v{_{e}}^{2} = \frac{2GM}{Ri}\]
By applying square root on both sides:
\[v_{e} = \pm \sqrt{\frac{2GM}{Ri}}\]
However, in this case, the gravitational convention is away from the other body, so escape velocity expression is negative:
\[v_{e} = -\sqrt{\frac{2GM}{Ri}}\]
Note: Although with regards to this convention, the negative version is right; in maximum escape velocity books, it is considered to be positive.
Important Facts About the Escape Velocity
The escape velocity of an object from the earth is an important factor that is taken into consideration while launching satellites and other space probes, scientists very precisely calibrate the satellites and other space probes so as to ensure that at no point of time does the velocity of the satellite is higher or lower than the required escape velocity of 11.18 Km/s. Even when the satellites are launched, continuous monitoring is done to ensure the satellites are having the parameters which are necessary for a successful launch.
The launch sites are always designed at places near to the equator because at the equator the gravitational pull of the Earth is the weakest. At the two poles, the gravitational pull of the earth is very high because along with the gravitational force the magnetic forces are also very crucial in the successful launching of the satellites.
Escape velocity is always calculated when there is a very huge difference between the size and mass of the two bodies. That is, one of the bodies has a very high mass and size as compared to the body that is trying to escape. Since the escape velocity is dependent on the size and mass of the body whose gravitational pull is to escape, the higher is the difference between the weights and sizes of the two bodies, the higher would be the escape velocity. There are certain space bodies that have a very huge mass and a small size e.g black holes. In the case of black holes, the density is so high that the escape velocity is more than the speed of light and as a result, even the light can’t escape the black holes.
Do It Yourself
(i) Gas escapes from the surface of a planet as it acquires an escape speed. Determine the factors escape velocity depends on:
Mass of the planet
Mass of the element escaping
The radius of the planet
Temperature of planet
(ii) Choose the correction combination of answers from the following options:
(a) 1 and 2 are correct
(b) 1 and 3 are correct
(c) 1,2 and 3 are correct
(d) 1 and 4 are correct
FAQs on Derivation of Escape Velocity
1. What is the fundamental 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 massive body, like a planet or a star, without any further propulsion. Once an object reaches escape velocity, it will not fall back towards the body or enter into an orbit around it; instead, it will travel away indefinitely until it is effectively at an infinite distance.
2. How is the formula for escape velocity derived using the principle of conservation of energy?
The derivation of escape velocity is a key application of the law of conservation of energy. The process involves these steps:
- Consider an object of mass 'm' on the surface of a planet of mass 'M' and radius 'R'.
- The initial kinetic energy (KE) given to the object to escape is ½mvₑ², where vₑ is the escape velocity.
- The initial gravitational potential energy (PE) of the object on the surface is -GMm/R.
- For the object to just escape the gravitational field, its total energy (KE + PE) at the surface must be equal to its total energy at an infinite distance, which is zero.
- Therefore, ½mvₑ² - GMm/R = 0.
- Solving for vₑ, we get vₑ = √(2GM/R). This is the expression for escape velocity.
3. On what factors does the escape velocity from a planet depend?
The escape velocity of a planet depends only on two primary factors, as shown in its formula vₑ = √(2GM/R):
- The mass (M) of the planet: A more massive planet has a stronger gravitational pull, thus requiring a higher escape velocity.
- The radius (R) of the planet: For a planet of the same mass, a smaller radius means a more concentrated gravitational field at the surface, which also leads to a higher escape velocity.
It is also dependent on the Universal Gravitational Constant (G).
4. Why does the escape velocity of an object not depend on its own mass or the direction of launch?
This is a crucial concept that often causes confusion. During the derivation of the escape velocity formula (½mvₑ² = GMm/R), the mass of the escaping object, 'm', appears on both sides of the equation and is therefore cancelled out. This shows that the escape velocity is an intrinsic property of the planet and is the same for all objects, whether it's a small satellite or a massive spacecraft. Furthermore, since energy is a scalar quantity, the direction of launch does not affect the calculation; only the magnitude of the initial velocity matters.
5. What happens if an object is launched with a speed that is less than the escape velocity?
If an object is launched with a speed less than the escape velocity, its initial kinetic energy is insufficient to overcome the planet's gravitational potential energy. It will travel upwards, gradually slowing down due to gravity. It will reach a maximum height where its velocity momentarily becomes zero, and then it will fall back to the planet's surface. If launched at a specific angle and speed, it might also enter a stable elliptical orbit instead of falling straight back down.
6. How does escape velocity fundamentally differ from orbital velocity?
While both are critical speeds in astrophysics, they serve different purposes. The key difference lies in their objectives:
- Escape Velocity (vₑ): The speed required to escape the gravitational field entirely and travel to an infinite distance. It follows a parabolic or hyperbolic trajectory away from the planet.
- Orbital Velocity (vₒ): The speed required to maintain a stable, circular orbit around a celestial body. At this speed, the gravitational force provides the exact centripetal force needed for circular motion.
There is a direct mathematical relationship between them for a circular orbit close to the surface: vₑ = √2 × vₒ. This means the escape velocity is approximately 1.414 times the orbital velocity.
7. What is the approximate escape velocity of Earth, and why are space launch sites often located near the equator?
The escape velocity from the surface of Earth is approximately 11.2 kilometers per second (or about 40,270 km/h). Launching from sites near the equator provides a significant advantage. The Earth rotates fastest at the equator, giving any rocket a 'free' velocity boost of about 0.46 km/s in the direction of rotation. This reduces the total velocity that the rocket must generate on its own, thereby saving a considerable amount of fuel and cost.
8. How does the concept of escape velocity explain why nothing, not even light, can escape a black hole?
This is one of the most extreme applications of the escape velocity concept. A black hole is an object with an incredibly large mass (M) compressed into an extremely small radius (R). According to the formula vₑ = √(2GM/R), this combination of a huge 'M' and a tiny 'R' results in an immense escape velocity. For a black hole, this calculated escape velocity is greater than the speed of light (c ≈ 300,000 km/s). Since the laws of physics state that nothing can travel faster than light, nothing—not even light itself—has sufficient speed to escape a black hole's gravitational pull once it crosses the event horizon.
