A planet of mass 4.00 times 10^25 kg is in a circular orbit of radius 6.00 times

Question

A planet of mass 4.00 times 10^25 kg is in a circular orbit of radius 6.00 times 10^11 m around a star. The star exerts a force on the planet of constant magnitude 5.96 times 10^22 N. The speed of the planet is 2.99 times 10^4 m/s. In half a "year" the planet goes half way around the star, what is the distance that the planet travels along the semicircle? distance = 1.885e12 m During this half "year", how much work is done on the planet by the gravitational force acting on the planet? work = -1.12346e35 What is the change in kinetic energy of the planet? Delta K = 11.341015e34 What is the magnitude of the change of momentum of the planet?

Explanation / Answer

c) Change in KE = 0 as speed of planet remain the same

d) The degrees covered by planet around Star = 360*1.855*10^12/(2*pi*6*10^11)

=177.23 degrees = around 180 degrees

Change in momentum =2*m*v

=2*4*10^25*2.99*10^4

=23.92*10^29 kg-m/s

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