You only have to do part C but please show steps clearly The equation mgy for gr
ID: 1563994 • Letter: Y
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You only have to do part C but please show steps clearly
The equation mgy for gravitational potential energy is valid only for objects near the surface of a planet. Consider two very large objects of mass m_1 and m_2, such as stars or planets, whose centers are separated by the large distance r. These two large objects exert gravitational forces on each other. The gravitational potential energy is U = - Gm_1m_2/r where G = 6.67 times 10^-11 Nm^2/kg^2 is the gravitational constant. (a) Sketch a graph of U versus r. The mathematical difficulty at r = 0 is not a physically significant difficulty because the masses will collide before they get that close together. (b) What separation r has been chosen as the point of zero potential energy? Does this make sense? Explain. (c) Two stars are at rest 1.0 times l0^14 m apart. This Ls about 10 times the diameter of the solar system. The first star is the size of our sun, with a mass of 2.0 times 10^30 kg and a radius of 7.0 times 10^8 m. The second star has mass 8.0 times 10^30 kg and radius of 11.0 times l0^8 in. Gravitational forces pull the two stars together. What is the speed of each star at the moment of impact?Explanation / Answer
Gravitational potential energy is converted into kinetic energy as they become closer. And just before the impact, P.E is completely coverted to K.E
=> P.E before pulling =K.E just before impact
=> G*M*m/R = 1/2*M*V^2 + 1/2*m*v^2
and frm conservation of momentum.
initial momentum =final momentum => 0 = m*v+M*V
=> V =-m*v/M ( - sign indicates m& M travel in opposite direction)
=> 6.67*10^-11*2*10^30*8*10^30/(1*10^14) = 1/2*2*10^30*v^2 +1/2*8*10^30*(2/8*v)^2
=> 1.0672*10^37 = 1*10^30*v^2 + 4*10^30*v^2/16
=> 1.0672*10^37 = v^2*( 1*10^30+ 10^30 /4 )
=> v^2 = 1.0672*10^37 / (1.25*10^30) =8537600
=> v= 2921.917 m/s is the velocity of smaller planet i.e 2*10^30kg
V =2* 10^30*2921.917 /(8*10^30) =730.479m/s
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