Escape Velocity

How to Calculate Escape Velocity ?

If a particle of mass m, kept in an attractive gravitational field is given sufficient kinetic energy, it may escape the gravitational pull of the field.

A particle of mass m kept on the surface of the earth requires a minimum velocity ve so that it escapes to infinity. Since gravitation is conservative,

$ \displaystyle \frac{1}{2}m v_e^2 – \frac{G M m}{R} = 0 $

where M is the mass of the earth and R is its radius.

The gravitational potential energy of the system of masses shown in the adjacent figure is zero when they are at infinite separation.

The potential energy = − GMm/r at a separation r.

As the particle escapes to ∞, its potential energy increases by GMm/r , while its kinetic energy decreases from (1/2) mv2 to zero.

As the field is conservative, increase in P.E. = decrease in K.E.

$ \displaystyle \frac{1}{2}m v^2 – \frac{G M m}{r} = 0 $

$ \displaystyle v = \sqrt{\frac{2 G M}{r}} $

The particle m will escape forever from M. It will never return.

Therefore the minimum value of $ \displaystyle v = \sqrt{\frac{2 G M}{r}} $

is the escape velocity of the particle from a distance r from the mass M.

Illustration : A body of mass m is in equilibrium under the gravitational force of two identical heavy point masses each of mass M kept separated through a distance l . With what minimum speed should it be projected in order to escape to infinity ?

Solution: The body m is in equilibrium. Therefore, the net gravitational force acting on it is zero.

$\vec{F_A} + \vec{F_B} = 0 $

$F_A = F_B $

$ \frac{G M m}{r_A^2} = \frac{G M m}{r_B^2} $

rA = rB

Hence the body is situated at the mid point P.

rA = rB = l/2

Therefore, the gravitational potential at P = VP = VA + VB

$V_P = – \frac{2 G M}{l/2}$

Therefore the gravitational energy of the body at P

$U_P = m . V_P = – \frac{4 G M m}{l}$

If v is the escape speed, kinetic energy of the body at P

$K_P = \frac{1}{2} m v^2 $

Total energy of the body at P

$E = U_P + K_P $

$E = – \frac{4 G M m}{l} + \frac{1}{2} m v^2 $

Applying Conservation of Energy ,

$- \frac{4 G M m}{l} + \frac{1}{2} m v^2 = 0 $

$v = 2 \sqrt{\frac{2 G M}{l}}$

Illustration : Find the maximum height attained by a body projected with a speed v = ve /2 , where ve = escape velocity of any object at earth’s surface.

Solution : $\displaystyle \Delta K.E + \Delta P.E = 0 $

$\displaystyle – \frac{1}{2} m v^2 + G M m (\frac{1}{R} – \frac{1}{R+h} ) = 0 $

$\displaystyle \frac{1}{2} m v^2 = G M m (\frac{1}{R} – \frac{1}{R+h} ) $

Putting , $ v = v_e /2 = \frac{1}{2}\sqrt{\frac{2 G M}{R}} $

$\displaystyle \frac{1}{2} m (\frac{G M}{2 R}) = G M m (\frac{h}{R (R+h)}) $

$\displaystyle \frac{1}{4 R} = \frac{h}{R(R+h)} $

$\displaystyle 4 h = R + h $

$\displaystyle h = \frac{R}{3} $

Also Read :

∗ Newton’s Law of Gravitation & Gravitational Field Intensity
∗ Acceleration due to Gravity & Variation of g
∗ Work Done in Gravitational Field
∗ Gravitational Potential Energy
∗ Gravitational Potential
∗ Kepler’s Laws of Planetary Motion
∗ Motion of Satellites

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