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A projectile is shot directly away from Earth\'s surface. Neglect the rotation o

ID: 1723087 • Letter: A

Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth. What multiple of Earth's radius RE gives the radial distance a projectile reaches if (a) its initial speed is 0.500 of the escape speed from Earth and (b) its initial kinetic energy is 0.500 of the kinetic energy required to escape Earth? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth, if the projectile has a mass M ? Thank you so much for all your help! I will give a lifesaver rating to my lifesaver!

Explanation / Answer

(a). Initial speed v = 0.5 * escape speed                             = 0.5 * 11.2 km / s                             = 5600 m / s energy on the surface of the earth E = -GMm / R + (1/ 2) m v ^ 2 where G = gravitational constant = 6.67 * 10 ^ -11 N m ^2 / kg^ 2            M= mass of earth = 5.98 * 10 ^ 24 kg            m= mass of object energy at maximu altitude E ' = -GMm / ( R + h ) we know from law of conservation of energy , E = E'               -GMm / R+ ( 1/ 2) m v ^ 2 = -GMm / ( R + h )                                  ( 1/ 2) v ^ 2 = -GM [ ( 1/ ( R + h ) ) -( 1/R ) ]                                                      = -GM [ -h / ( R * ( R + h ) ) ]                                                      = GMh / [ R * ( R + h ) ]                                         R+ h = 2GMh / {R * v ^ 2 }                                               h= [ 2GMh / { R v ^ 2 } ] - R from this we find answer (b). escape kinetic energy K = ( 1/ 2) mv ^ 2 but initial speed v ' = v / 2 So, kinetic energy for this speed K ' = ( 1/ 2) m v' ^ 2                                                      = ( 1/ 4 ) ( 1/ 2) m v ^ 2                                                      = K / 4 i.e., initial kinetic energy is not equal to 0.5 of thekinetic energy required to escape earth                                                      
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