Question

Define escape velocity. Obtain an expression for escape velocity of a body from the surface of the earth. Write
the factor on
which it depends.​

Answer 1

Answer:

the escape velocity of a body is the velocity requered  for a body to overcome the gravitational force of earth and escape from it's feild of attraction.

expression :-

V= root 2GM/R

escape velocity depends on the mass of earth (M) and radius of earth(R)

these are the two factors which escape velocity depends.

Explanation:

the total energy of a body = potential energy + kinetic energy

let total energy = T.E

so  T.E = P.E + KE

for a body which is in rest on the surface of earth it has only P.E

                                                                           P.E = -GMm/R  

                                                                            K.E = 1/2mV2

so T.E = -GMm/R

whenever a body need to escape from gravity its T.E must be zero

which means K.E = P.E

so   -GMm/R + 1/2mV2 = 0

so    1/2mV2 = -GMm/R

so     V power 2 = 2GM/R

so      V  = root 2GM/R

Answer 2

Escape Velocity-:

The minimum velocity required for an object located on earth's surface so that it just escapes the earth's gravitational field.

Or

The minimum velocity required to project a body vertically upward from the surface of the earth so that it never returns to the surface of earth is called escape velocity. _____________________________

Expression for the escape velocity of a body from the surface of the Earth-:

Let's consider a body of mass m thrown upward with a velocity of vₑ from the surface of the earth.

Initial K.E. of the body,

$K_i=\frac{1}{2}m\:v_e^2$

Initial gravitational potential,

$U_i=-\frac{GMm}{R}$

Here, M = mass of the earth

R = radius of the earth

Total initial energy,$E_i=K_i+U_i$

$\:\:\:\:\:= \frac{1}{2}m\:(v_e)^2- \frac{GMm}{R}$

Now, let's suppose that the body reaches a height h and finally its velocity becomes $v_f$

Therefore, K.E. of the body at height h,

$K_f = \frac{1}{2}m\:v^2_f$

Gravitational potential;

$U_f=-\frac{GMm}{(R+h)}$

Therefore, total energy of the system at height h,

$E_f=\frac{1}{2}m\:v^2_e$–$\frac{GMm}{(R+h)}$

Now,

According to the law of conservation of energy;

Initial total energy = Final total energy

$\implies \frac{1}{2}m\:v^2_e-\frac{GMm}{R}=\frac{1}{2}m\:v^2_e-\frac{GMm}{(R+h)}$

$\implies \frac{1}{2}mv^2_f=(\frac{1}{2}mv^2_e-\frac{GMm}{R})\:+\:\frac{GMm}{R+h}$

The body will reach the max. height when,

$v_f$=0

Therefore, the body will escape through the earth's gravitational pull if,

$\frac{1}{2}mv^2_e-\frac{GMm}{R}$ ≥ 0

Or

vₑ ≥ $\sqrt{\frac{2GM}{R}}$

$\boxed{v_e=\sqrt{\frac{2GM}{R}}}$

________________________________

Factors on which escape velocity depends on:-

• Mass of the earth ‘ M ’.

• Radius of the earth ‘R’.

• Position from where the particle is projected.