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Explain escape speed. |
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Answer» Solution :`implies` SUPPOSE a body is at distance `r (r gtR_E)`from the centre of earth. This body is stationary. E = K.E. + Potential energy `E = - (GM_(E)m)/r` `E=-(GM_(E) m)/((R_(E) +h))""....(1)` `implies`If by supply this amount of energy, its total energy becomes zero. It will go to infinite distance from the earth. This energy is known as escape energy. `implies` To speed to be given to the body to give the kinetic energy equal to the escape energy`:. 1/2 mv_(e)^(2) = (GM_(E)m)/((R_(E)+h) )""...(2)` `implies`Here, `v_(e)`is called escape speed. `implies` "The minimum amount of energy with which a body should be projected from the surface of the earth, so that it escape from the earth.s gravitational field is called escape speed (velocity). `implies` From equation (2), `v_(e)^2 =(2GM_E)/((R_(E)+h))` `:. v_(e) = sqrt((2GM_E)/((R_(E)+h)))""...(3)` `implies`If body is lying on the surface of earth h = 0, hence escape speed, `:. v_(e) = sqrt((2GM_E)/(R_(E)))""...(4)` ` :. v_(e)=sqrt((2GM_(E))/(R_E^2).R_E)` `:. v_(e)= sqrt(2g R_E)""...(5)` where g acceleration due to gravity of earth , `implies` In equation (4), TAKING mass of earth `M_(E) = 4/3 pi R_(E)^3 rho` ( `rho` = density of earth) `v_(e) = sqrt((2G)/R_(E)(4/3piR_(E)^2))` `:.v_(e) = sqrt((8pi)/3(GR_(E)^(2))rho)"".....(6)` `implies` Equation 3, 4, 5, are the necessary equations for escape velocity. `implies`It is cleared from all equation that escape velocity does not depend on the mass of the body. Heavier or LIGHTER, both bodies as same escape of velocity. `implies`In equation (4), `G = 6.67 x 10^(-11) (Nm^2)/(kg^2)` Putting the value of `M_E = 6 x 10^(24) kg, R_E = 6400` km escape velocity will be `v_e = 11.2 km//s.` |
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