Escape Velocity
WBBSE Class 11 Escape Velocity Notes
When an object is thrown vertically upwards, it moves up to a certain height and then comes down to the earth’s surface due to the gravitational pull of the earth.
At the maximum height, the velocity of the object becomes zero. The higher the projection velocity, the greater the height to which the object rises before its velocity becomes zero.
- If the velocity of projection continuously increases, a state is reached when the velocity of the object (for a finite height) does not become zero.
- This implies that it does not return to the earth’s surface. At this state, the object recedes from the earth’s surface and the force of gravity also decreases to almost a negligible value.
- Hence, it can be said that the body reaches an infinite distance from the earth. The minimum velocity of projection required for this to happen is called the escape velocity.
- It is to be noted that this concept of escape velocity is not only applicable to objects projected from the Earth’s surface but also valid for objects projected from the surface of other celestial bodies.
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Escape Velocity Formula and Derivation
Escape Velocity Definition: The minimum velocity required to project a body from the surface of the earth or a planet or a satellite such that it can escape the gravitational attraction of the earth or the planet or the satellite is called escape velocity.
Receding from the gravitational attraction of a planet or satellite implies that the projected body does not return to the planet or satellite any more.
Let the mass of the earth = M and its radius = R. The force of gravity on a body of mass m at a distance x from the
centre of the earth = \(\frac{G M m}{x^2}\)
For a small displacement dx of the body against this force of gravity, work done = \(\frac{G M m}{x^2}\)dx
Hence, the work done to displace the body from the earth’s surface to an infinite distance
= \(\int_R^{\infty} \frac{G M m}{x^2} d x=G M m\left[-\frac{1}{x}\right]_R^{\infty}=\frac{G M m}{R}\)
If the escape velocity from the earth is ve, the initial kinetic energy of the body is \(\frac{1}{2} m v_e^2\). When this energy does the work as calculated above, the body recedes from the earth’s gravitational attraction.
∴ \(\frac{1}{2} m v_e^2=\frac{G M m}{R} \text { or, } v_e^2=\frac{2 G M}{R}\)
or, \(v_e=\sqrt{\frac{2 G M}{R}}\)…(1)
The acceleration due to gravity on the Earth’s surface g= \(\frac{GM}{R^2}\) or, GM= gR²
∴ \(v_e=\sqrt{2 g R}\)…..(2)
Equation (2) gives the value of the escape velocity for a body from the earth’s surface. Note that, the escape velocity does not depend on the mass of the body. This means that for a small and a massive body, the value of the escape velocity is the same.
It is known that the acceleration due to gravity on the earth’s surface g = 9.8 m · s-2 and the radius of the earth R = 6400 km = 6.4 x 106m.
∴ \(v_e=\sqrt{2 \times 9.8 \times 6.4 \times 10^6}\)
= 11200 m · s-1 = 11.2 km· s-1
Factors Affecting Escape Velocity
Hence, if a body of any mass is projected upwards from the surface of the earth with a velocity of 11.2 km · s-1, it moves out of the gravitational field of the earth, i.e., the body does not return to the earth’s surface.
Equation (2) can also be used for calculating the escape velocity from the surface of any planet or satellite by putting the corresponding values of g and R. For example, the value of g on the surface of the moon is \(\frac{1}{6}\)th of the value on the surface of the earth and the moon’s radius is \(\frac{1}{4}\)th that of the earth.
Putting these values in equation (2), the value of the escape velocity from the moon’s surface can be evaluated and the value is about 2.4 km · s-1. This value is about \(\frac{1}{5}\)th the value of the escape velocity from the earth’s surface.
Applications of Escape Velocity in Space Exploration
Scarcity Of Light Gases In The Earth’s Atmosphere: When the Earth was formed, the Earth’s atmosphere was filled with lighter gases such as hydrogen and helium along with heavier gases like oxygen, nitrogen, etc. At present, there is practically no existence of lighter gases. The scarcity of lighter gases can be explained in the light of escape velocity.
- From the kinetic theory of gases, it is known that the rms speed of hydrogen at STP is nearly 1.6 km · s-1. During the formation stage, the earth was warmer and the rms speed of hydrogen had a value close to 5 km · s-1. Obviously, many hydrogen molecules had velocities more than or equal to the escape velocity (11km · s-1).
- Hence, gradually with the passage of time, most of the hydrogen molecules escaped the gravitational field of the Earth and went into space. The same was the fate of helium. On the other hand, heavy molecules have very low rms velocity of thermal motion.
- As the value is much lower than the value of the escape velocity, heavy gas molecules remained confined to the earth’s atmosphere. Hence, at present, the Earth’s atmosphere is primarily made up of heavier gases like oxygen, nitrogen and carbon dioxide.
- In the case of the moon or Mercury, owing to their low masses and shorter radii, the value of the escape velocity from the surface is low. As the rms speeds of both the lighter and heavier molecules are comparable to the escape velocity, these gas molecules could escape.
Hence, there is no atmosphere on the moon or on Mercury. On the other hand, the escape velocity for Jupiter or Saturn is of much higher value because of the higher mass and size. Thus, neither lighter nor heavier gas molecules could escape the gravitational attraction of these planets. So, there is an abundance of hydrogen and helium gas on these planets.
Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion
Escape Velocity Numerical Examples
Short Answer Questions on Escape Velocity
Real-Life Examples of Escape Velocity
Example 1. What is the escape velocity of a meteorite situated 1800 km above the surface of the earth? Given, the radius of the earth = 6300 km and the acceleration due to gravity on the earth’s surface = 9.8 m · s-2.
Solution:
Given, the radius of the earth = 6300 km and the acceleration due to gravity on the earth’s surface = 9.8 m · s-2.
Radius of the earth (R) = 6300 km = 63 x 105 m
Distance of the meteorite from the centre of the earth (r) = 6300 + 1800 = 8100 km = 81 x 105 km
Acceleration due to gravity on the earth’s surface, g = \(\frac{G M}{R^2}\);
acceleration due to gravity at the position of the meteorite, g’ = \(\frac{G M}{r^2}\).
∴ \(\frac{g}{g^{\prime}}=\frac{r^2}{R^2} \text { or, } g^{\prime}=\frac{g R^2}{r^2}=g\left(\frac{R}{r}\right)^2\)
Escape velocity of the meteorite, \(v_e^{\prime}=\sqrt{2 g^{\prime} r}=\sqrt{2 g\left(\frac{R^2}{r^2}\right)} r=R \sqrt{\frac{2 g}{r}}\)
= \(63 \times 10^5 \times \sqrt{\frac{2 \times 9.8}{81 \times 10^5}}\)
= \(98 \times 10^2 \mathrm{~m} \cdot \mathrm{s}^{-1}=9.8 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)
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Example 2. A man can jump up to a height of 1.5 m on the earth’s surface. What should be the radius of a planet having the same average density as that of the Earth so that man can come out of the gravitational field of that planet in one jump? The radius of the earth is 6400 km.
Solution:
Given
The radius of the earth is 6400 km.
A man can jump up to a height of 1.5 m on the earth’s surface.
The man can jump up to a height h on the earth’s surface.
Hence, his initial kinetic energy = his potential energy at height h.
Thus, \(\frac{1}{2}\)mv² = mgh [m = mass of the man, v = initial velocity]
or, v = \(\sqrt{2 g h}\)….(1)
The man can jump on the surface of any planet with this velocity v. If v equals the escape velocity from a planet, the man can go out of the gravitational pull of that planet. Let the acceleration due to gravity for a planet be g’, its radius Rf and average density ρ, then
g’ = \(\frac{4}{3} \pi G R^{\prime} \rho\)
As \(\frac{4}{3} \pi\)G and ρ are constants, g’ ∝ R’.
Hence, \(\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R} \text { or, } g^{\prime}=g \cdot \frac{R^{\prime}}{R}\)………(2) [R = radius of the earth]
Escape velocity for the planet is \(v=\sqrt{2 g^{\prime} R^{\prime}}\)…..(3)
From (1) and (3), \(\sqrt{2 g h}=\sqrt{2 g^{\prime} R^{\prime}} \text { or, } g h=g^{\prime} R^{\prime}\)
or. \(g^{\prime}=g \cdot \frac{h}{R^{\prime}}\)….(4)
From (2) and (4), \(\frac{g R^{\prime}}{R}\)=\(\frac{g h}{R^{\prime}}\)
or, \(R^{\prime}=\sqrt{h R}=\sqrt{1.5 \times\left(6400 \times 10^3\right)}\)
= \(3098 \mathrm{~m}=3.098 \mathrm{~km} .\)