What is the maximum height attained by a body projected with a velocity equal to one- third of the escape velocity from the surface of the earth? (Radius of the earth=R)
The escape velocity of a body from the earth's surface, v_("esc")= ……….. .
The escape velocity of a body from earth's surface is v_e . The escape velocity of the same body from a height equal to 7R from earth's surface will be
A particle of mass m is fired vertically upward with speed v_(0) (v_(0) lt " escape speed ") . Prove that (i) Maximum height attained by the particle is H = (v_(0)^(2)R)/(2gR - v_(0)^(2)) , where g is the acceleration due to gravity at the Earth's surface and R is the Earth's radius . (ii) When the particel is at height h , the increase in gravitational potential energy is (mgh)/(1+h/R)
A particle is projected vertically with speed V from the surface of the earth . Maximum height attained by the particle , in term of the radius of earth R,V and g is ( V lt escape velocity , g is the acceleration due to gravity on the surface of the earth )
A body mass 'm' is dropped from heinght R/2 , from earth's surface, where 'R' is the radius of earth. Its speed when it will hit the earth's surface is (v_(e)="escape velocity from earth's surface") .
A projectile is fired from the surface of earth of radius R with a velocity etav_(e) where v_(e) is the escape velocity and eta lt 1 . Neglecting air resistance, the orbital velocity of projectile is -
A particle of mass m is projected upwards with velocity v=(v_(e))/(2) , where v_(e) is the escape velocity then at the maximum height the potential energy of the particle is : (R is radius of earth and M is mass of earth)
