Question

Define the term of escape velocity. Derive expression for escape velocity of a body from the surface of earth.

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Answer 1

Escape Velocity :

It is the minimum velocity with which a body must be projected from the surface of the earth so that it escapes from the gravitational influence of the planet.

Derivation :

 Consider a body of mass "m" which is at rest initially on the surface of the earth.

Let "M" be the mass of the earth and "R" is the radius of the earth.

The work done to take the body from the surface of the earth (x = R) to the infinity (x = ∞) is

   $\sf W=\int\limits^\infty _R {dW} \\\\\\ \sf W=\int\limits^\infty _R {\dfrac{GMm}{x^2}} \, dx \\\\\\ \sf W=GMm \int\limits^\infty _R {x^{-2}dx} \\\\\\ \sf W=GMm \bigg[\dfrac{x^{-2+1}}{-2+1} \bigg] ^\infty _R \\\\\\ \sf W=GMm \bigg[\dfrac{x^{-1}}{-1} \bigg] ^\infty _R \\\\\\ \sf W=-GMm\bigg[\dfrac{1}{x} \bigg] ^\infty _R \\\\\\ \sf W=-GMm \bigg[\dfrac{1}{\infty}-\dfrac{1}{R}\bigg] \\\\\\ \sf W=-GMm\bigg[0-\dfrac{1}{R}\bigg] \\\\\\ \sf W=-GMm\bigg(\dfrac{-1}{R}\bigg) \\\\\\ \boxed{\sf W=\dfrac{GMm}{R}}$

Let the minimum velocity required by the body to escape from the earth be "vₑ"

Kinetic Energy of the body =  $\sf \dfrac{1}{2}mv_e ^2$

The work done against the gravity to take the body from the surface to infinity must be equal tot he Kinetic Energy of the body in order to escape from the surface of the earth.

   KE = W

 $\sf \dfrac{1}{2}mv_e ^2=\dfrac{GMm}{R} \\\\ \sf \dfrac{v_e ^2}{2}=\dfrac{GM}{R} \\\\ \sf v_e ^2=\dfrac{2GM}{R} \\\\ \boxed{\sf v_e=\sqrt{\dfrac{2GM}{R}}}$

We know,

acceleration due to gravity,  $\sf g=\dfrac{GM}{R^2}$

 ⇒ GM = gR²

 

Substitute GM = gR² in the expression of escape velocity.

  $\sf v_e=\sqrt{\dfrac{2GM}{R}} \\\\ \sf v_e=\sqrt{\dfrac{2gR^2}{R}} \\\\ \underline{\boxed{ \sf v_e=\sqrt{2gR}}}$

➙ The escape velocity does not depend on the mass of the body.

➙ The escape velocity depends on the mass and radius of the Planet.

Escape velocity of a body on Earth :

➙ vₑ = √(2gR)

where

g = 9.8 m/s²

R ≈ 6400 km = 6400 × 1000 m = 6.4 × 10⁶ m

Substitute the values,

 $\sf v_e=\sqrt{2 \times 9.8 \ m/s^2 \times 6.4 \times 10^6 m} \\\\ \sf v_e=\sqrt{19.6 \times 6.4 \times 10^6 \ m^2/s^2} \\\\ \sf v_e=\sqrt{14^2 \times 8^2 \times 10^4} \ m/s \\\\ \sf v_e=(14 \times 8 \times 10^2) \ m/s \\\\ \sf v_e=112 \times 10^2 \ m/s \\\\ \sf v_e=\dfrac{11200}{1000} \ km/s \\\\ \sf v_e=11.2 \ km/s$

∴ The escape velocity of a body on the surface of the earth = 11.2 km/s

 

Answer 2

Answer:

Escape velocity: -The minimum speed with which a body must be projected. in order that it will escape from the earth gravitational filed is called as. escape velocity. Expression of it:-consider a body of mass m on the surface of earth. whose mass and radius are respectively represented by M and R.