Escape Velocity I पलायन वेग I

Escape Velocity I पलायन वेग I #1stgrade By Dr. S.K. Mehta Sir
*Escape Velocity (पलायन वेग)* is the minimum speed an object must reach to break free from the gravitational pull of a celestial body, such as a planet or moon, without further propulsion.
Derivation of Escape Velocity
To derive the escape velocity, we start with the concept of gravitational potential energy and kinetic energy.
1. *Gravitational Potential Energy (U):*
The gravitational potential energy of an object of mass \( m \) at a distance \( r \) from the center of a celestial body with mass \( M \) is given by:
\[
U = -\frac{GMm}{r}
\]
where \( G \) is the gravitational constant.
2. *Kinetic Energy (K):*
The kinetic energy of an object with mass \( m \) moving with velocity \( v \) is:
\[
K = \frac{1}{2} mv^2
\]
3. *Total Energy (E):*
To escape the gravitational field, the total energy of the object must be zero or positive. The total energy \( E \) is the sum of the kinetic and potential energies:
\[
E = K + U = \frac{1}{2} mv^2 - \frac{GMm}{r}
\]
4. *Condition for Escape:*
For the object to just escape, the total energy \( E \) must be zero. Therefore:
\[
\frac{1}{2} mv^2 - \frac{GMm}{r} = 0
\]
5. *Solve for Escape Velocity (v):*
Rearranging the equation to solve for \( v \), we get:
\[
\frac{1}{2} mv^2 = \frac{GMm}{r}
\]
\[
v^2 = \frac{2GM}{r}
\]
\[
v = \sqrt{\frac{2GM}{r}}
\]
Final Formula
So, the escape velocity \( v \) from the surface of a celestial body is:
\[
v = \sqrt{\frac{2GM}{r}}
\]
where:
- \( G \) is the gravitational constant \((6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})\),
- \( M \) is the mass of the celestial body,
- \( r \) is the radius of the celestial body from which the object is escaping.
Example Calculation
For Earth:
- Mass \( M \approx 5.972 \times 10^{24} \) kg,
- Radius \( r \approx 6.371 \times 10^6 \) m,
The escape velocity can be calculated as:
\[
v = \sqrt{\frac{2 \times 6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}}
\]
\[
v \approx 11.2 \text{ km/s}
\]
This is approximately the velocity required for a rocket to escape Earth’s gravitational pull without any further propulsion.
#EscapeVelocity
#PalaayanVeg
#SpaceScience
#Physics
#GravitationalPhysics
#Astrophysics
#RocketScience
#CelestialMechanics
#SpaceExploration
#ScientificKnowledge
#Astronomy
#PhysicsEquations
#SpaceTravel
#PhysicsConcepts
#Gravity