The figure gives the potential energy function U(r) of a projectile, plotted out

Question

The figure gives the potential energy function U(r) of a projectile, plotted outward from the surface of a planet of radius Rs. If the projectile is launched radially outward from the surface with a mechanical energy of

The figure gives the potential energy function U(r) of a projectile, plotted outward from the surface of a planet of radius Rs. If the projectile is launched radially outward from the surface with a mechanical energy of -2.0 x 109 J, what are (a) its kinetic energy at radius r = 1.25Rs and (b) its turning point (in km; see Section 8-6) if Rs = 7800 km?

Explanation / Answer

The Mechanical energy ( ME ) = - 2 *109 J If r = 1.25R s , then U = - 4 .0 *109J Let at this time the K.E be Kr . (A) Kr = ME - U Kr = ( - 2 * 109 ) + (4*109 ) Kr = 2*109 J ( B ) The point at which the kinetic energy iszero is called turning point Forkinetic energy to be zero the potential energy should be equal tothe kinetic energy Then ( U ) = - 2 * 109 J Since , Kr = ME - U = - 2 * 109 +2*109 = 0 This ocurs at r = 2.5 Rs where Rs = 7800 km

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